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==An evaluation of Philip Wadler's "Monads for functional programming"==
'''Note''' Since the size of the previous file was getting too big for a wiki, the tutorial has been divided into two parts: [[The Monadic Way Part I]] and [[The Monadic Way Part II]]. See below for some introductory remarks.


This tutorial is a "translation" of Philip Wedler's [http://homepages.inf.ed.ac.uk/wadler/papers/marktoberdorf/baastad.pdf "Monads for functional programming"].
'''Contents'''
(avail. from [http://homepages.inf.ed.ac.uk/wadler/topics/monads.html here])


I'm a Haskell newbie trying to grasp such a difficult concept as the
;[[Meet Bob The Monadic Lover]]
one of Monad and monadic computation.
:A (supposed-to-be) funny and short introduction to Monads, with code but without any reference to category theory: what monads look like and what they are useful for, from the perspective of a ... lover. It could be an introduction to "The Monadic Way" tutorial.


While [http://www.cs.utah.edu/~hal/htut/ "Yet Another Haskell Tutorial"]  
;[[The Monadic Way/Part I]]
gave me a good understanding of the type system when it
:In the first part of the tutorial we will start from a very simple evaluator that will be transformed into a monadic evaluator with an increasing number of features: output, exceptions, and state: a very simple counter for tracking the number of recursions of the evaluation precess.
comes to monads I find it almost unreadable.


But I had also Wedler's paper, and started reading it. Well, just
;[[The Monadic Way/Part II]]
wonderful! It explains how to ''create'' a monad!
:In the second part of the tutorial we will see how to take complexity out of our monadic evaluator with the use of monadic transformers, and specifically StateT. This part is just a skeleton, since, for the time being, it contains only the code.


So I decided to "translate it", in order to clarify to myself the
topic. And I'm now sharing this traslation ('''not completed yet'')
with the hope it will be useful to someone else.


Moreover, that's a wiki, so please improve it. And, specifically,
==Preliminary remarks==
correct my poor English. I'm Italian, after all.


'''Note: The source of this page can be used as a Literate Haskel
When I started writing this tutorial I though that the only way to
file and can be run with ghci or hugs: so cut paste change and run (in
explain monads to a newcomer was to show them from the inside, with
emacs for instance) while reading it...'''
the use of lambda abstraction. Not only because this is the way Philip
Wedler's
[http://homepages.inf.ed.ac.uk/wadler/papers/marktoberdorf/baastad.pdf paper]
adopts, but also because I believed, and still believe, that the
only way to understand what bind (>>=) does is to explain it as a
function that takes a monad and an anonymous function.


==A Simple Evaluator==
I had this feeling because I am a newcomer, and this is the way I came to understand monads.
 
Let's start with something simple: suppose we want to implement a new
programming language. We just finished with
[http://swiss.csail.mit.edu/classes/6.001/abelson-sussman-lectures/ Abelson and Sussman's Structure and Interpretation of Computer Programs]
and we want to test what we have learned.
 
Our programming language will be very simple: it will just compute the
sum of two terms.
 
So we have just one primitive operation (Add) that takes two constants
and calculates their sum.
 
Moreover we have just one kind of data type: Con a, which is an Int.
 
For instance, something like:
 
  (Add (Con 5) (Con 6))
 
should yeld:
 
  11
 
===The basic evaluator===
 
We will implement our language with the help of a data type
constructor such as:
 
<div id=BasicEval">
<haskell>
 
> module TheMonadicWay where
> data Term = Con Int
>          | Add Term Term
>            deriving (Show)
 
</haskell>
 
After that we build our interpreter:
 
<haskell>
 
> eval :: Term -> Int
> eval (Con a) = a
> eval (Add a b) = eval a + eval b
 
</haskell>
 
That's it. Just an example:
 
  *TheMonadicWay> eval (Add (Con 5) (Con 6))
  11
  *TheMonadicWay>
 
Very very simple. The evaluator checks if its argument is of type Con
Int: if it is it just returns the Int.
 
If the argument is not of type Con, but it is of type Term, it
evaluates the first Term and sums the result with the result of the
evaluation of the second Term.
 
== Some Output, Please!==
 
Now, that's fine, but we'd like to add some features, like providing
some output, to show how the computation was carried out.
 
Well, but Haskell is a pure functional language, with no side effects,
we were told.
 
Now we seem to be wanting to create a side effect of the computation,
its output, and be able to stare at it...
 
If we had some global variable to store the out that would be
simple...
 
But we can create the output and carry it along the computation,
concatenating it with the old one, and present it at the end of the
evaluation together with the evaluation of the expression given to our
evaluator/interpreter!
 
===The basic evaluator with output===
 
Simple and neat:
 
<haskell>
 
> type MOut a = (a, Output)
> type Output = String
>
> formatLine :: Term -> Int -> Output
> formatLine t a = "eval (" ++ show t ++ ") <= " ++ show a ++ " - "                                                     
>
> evalO :: Term -> MOut Int
> evalO (Con a) = (a, formatLine (Con a) a)
> evalO (Add t u) = ((a + b),(x ++ y ++ formatLine (Add t u) (a + b)))
>    where (a, x) = evalO t
>          (b, y) = evalO u
 
</haskell>
 
Now we have what we want. But we had to change our evaluator quite a
bit.
 
First we added a function, formatLine, that takes an argument of type
Term (the expression to be evaluated), one of type Int (the result of
the evaluation of Term) and gives back an output of type Output (that
is a synonymous of String). This is just a helper function to format
the string to output. Not very interesting at all.
 
The evaluator itself changed quite a lot! Now it has a different type
signature: it takes an argument of type Term and produces a new type,
we called it MOut, that is actually a compound pair of a variable (of
a variable) type a (an Int in our evaluator) and a type Output, a
string.
 
So our evaluator, now, will take a Term (the type of the expressions
in our new programming language) and will produce a pair, composed of
the result of the evaluation (an Int) and the Output, a string.
 
So far so good. But what's happening inside the evaluator?
 
The first part will just return a pair with the number evaluated and
the output formatted by formatLine.
 
The second part does something more complicated: it returns a pair
composed by
1. the result of the evaluation of the right Term summed to the result
of the evaluation of the second Term
2. the output: the concatenation of the output produced by the
evaluation of the right Term, the output produced by the evaluation of
the left Term (each this evaluation returns a pair with the number and
the output), and the formatted output of the evaluation.
 
Let's try it:
  *TheMonadicWay> evalO (Add (Con 5) (Con 6))
  (11,"eval (Con 5) <= 5 - eval (Con 6) <= 6 - eval (Add (Con 5) (Con 6)) <= 11 - ")
  *TheMonadicWay>
 
It works! Let's put the output this way:
  eval (Con 5) <= 5 -
  eval (Con 6) <= 6 -
  eval (Add (Con 5) (Con 6)) <= 11 -
 
Great! We are able to produce a side effect of our evaluation and
present it at the end of the computation, after all.
 
Let's have a closer look at this expression:
<haskell>
 
evalO (Add t u) = ((a + b),(x ++ y ++ formatLine (Add t u) (a + b)))
    where (a, x) = evalO t
          (b, y) = evalO u
 
</haskell>
 
Why all that? The problem is that we need:
* "a" and "b" to calculate their sum (a + b), that will be the first element of the compund pair rapresenting the type (MOut) our evaluator will return
* "x and "y" (the output of each evaluation) to be concatenated  with the ourput of formatLine by the expression (x ++ y ++ formatLine(...)): this will be the second element of the compound pair MOut, the string part.
 
So we need to separate the pairs produced by "evalO t" and "evalO u".
 
We do that within the where clause (remember: evalO now produces a value of type
MOut Int, i.e. a pair of an Int and a String).
 
Then we use the single element, "extraded" within the where clause, to
return a new MOut composed by
 
  ((a + b),(x ++ y ++ formatLine (Add t u) (a + b))).
  ------  -------------------------------------
    Int            Output = String
 
== Let's Go Monadic==
 
Is there a more general way of doing so?
 
Let's analyze the evaluator from another perspective. From the type
perspective.
 
We solved our problem by creating a new type, a pair of an Int (the
result of the evaluation) and a String (the output of the process of
evaluation).
 
The first part of the evaluator does nothing else but creating, from a
value of type Int, an object of type MOut Int (Int,Output). It does so
by creating a pair with that Int and some text produced by formatLine.
 
The second part evaluates the two Term(s) and "stores" the values thus
produced in some variables to be use later to compute the output.
 
Let's focus on the "stores" action. The correct term should be
"binds".
 
Take a function:
<haskell>
f x = x + x
</haskell>
"x" appears on both sides of the expression. We say that on the right
side "x" is bound to the value of x given on the left side.
 
So
<haskell>
f 3
</haskell>
binds x to 3 for the evaluation of the expression "x + x".
 
Our evaluator binds "a" and "x" / "b" and "y" with the evaluation of
"evalO t" and "evalO u" respectively.
 
Then "a","b","x" and "y" will be then used in the evaluation of
((a+b),(x++y++formatLine)), the will produce a value of type MOut Int:
 
<pre>
 
  ((a + b),(x ++ y ++ formatLine (Add t u) (a + b))).
  ------  -------------------------------------
    \ /          \                  /
    Int              Output = String
    ---------------------------------
                  \  /
                MOut Int           
</pre>
 
The binding happens in the where clause:
<haskell>
where (a, x) = evalO t
      (b, y) = evalO u
</haskell>
 
We know that there is an ad hoc operator for binding variables to a
value: lambda, or \.
 
Indeed f x = x + x is syntactic sugar for:
<haskell>
f = \x -> x + x
</haskell>
When we write f 3 we are actually binding "x" to 3 within what's next
"->", that will be used (substituted) for evaluating f 3.
 
So we can try to abstract this phenomenon.
 
===Monadic evaluator with output===
What we need is a function that takes our composed type MOut Int and a
function in order to produce a new MOut Int, concatenating the
output of the computation of the first with the output of the
computation of the second.
 
This is what bindM does:
 
<haskell>
 
> bindM :: MOut a -> (a -> MOut b) -> MOut b
> bindM m f = (b, x ++ y)
>            where (a, x) = m
>                  (b, y) = f a
 
</haskell>
 
It takes:
* "m": the compound type MOut Int carrying the result of an "eval Term",
* a function "f". This function will take the Int ("a") extracted by the evaluation of "m" ((a,x)=m). This function will produce anew pair: a new Int produced by a new evaluation; some new output.
 
bindM will return the new Int in pair with the concatenated outputs
resulting from the evaluation of "m" and "f a".
 
As you see, we took the binding part out from evalO and put it in this new function.
 
So let's write the new version of the evaluator, that we will call evalM_1:
 
<haskell>
 
> evalM_1 :: Term -> MOut Int
> evalM_1 (Con a) = (a, formatLine (Con a) a)
> evalM_1 (Add t u) = bindM (evalM_1 t) (\a ->
>                                      bindM (evalM_1 u) (\b ->
>                                                        ((a + b), formatLine (Add t u) (a + b))
>                                                    )
>                                    )
 
</haskell>
 
Ugly, isn't it?
 
Let's start from the outside:
 
<haskell>
bindM (evalM_1 u) (\b -> ((a + b), formatLine (Add t u) (a + b)))
</haskell>
 
bindM takes the result of the evaluation "evalM_1 u", a type Mout Int,
and a function. It will extract the Int from that type and use it to
bind "b".
 
So in bindM (evalM_1 u) (\b ->) "b" will be bound to the value
returned by evalM_1 u, and this bounded variable will be available in
what comes after "->" as a bounded variable (not free).
 
Then the outer part (bindM (evalM_1 t) (\a...) will bind "a" to the
value returned "evalM_1 t", the result of the evaluatuion of the first
Term. This value is needed to evaluate "((a+b), formatLine...) and
produce our final MOut Int.
 
S we can use lambda notation to write our evaluator in a more convinient way:
 
<haskell>
 
> evalM_2 :: Term -> MOut Int
> evalM_2 (Con a) = (a, formatLine (Con a) a)
> evalM_2 (Add t u) = evalM_2 t `bindM` \a ->
>                    evalM_2 u `bindM` \b ->
>                    ((a + b), (formatLine (Add t u) (a + b)))
 
</haskell>
 
Now, look at the first part:
 
<haskell>
evalM_2 (Con a) = (a, formatLine (Con a) a)
</haskell>
 
We could use a more general way of creating some output.
 
We can create a function that takes an Int and returns the type MOut
Int. we do that by pairing the received Int with an empty string "".
 
This will be a general way of creating an object with type MOut Int starting from an Int.
 
Or, more generaly, a function that takes something of a variable type
a, and return an object of type MOut a, a coumpunt object made up of
an element of type a, and one of type String.
 
There it is:
 
<haskell>
 
> mkM :: a -> MOut a
> mkM a = (a, "")
 
</haskell>
 
Then we need a method of inserting some text in our object of type
MOut. So we will take a string and return it paired with a void
element "()":
 
<haskell>
 
> outPut :: Output -> MOut ()
> outPut x = ((), x)
 
</haskell>
 
Very simple: we have a string "x" (Output) and create a pair with a ()
instead of an Int, and the output.
 
Now we can rewrite:
<haskell>
evalM_2 (Con a) = (a, formatLine (Con a) a)
</haskell>
using the bindM function:
<haskell>
evalM_2 (Con a) = outPut (formatLine (Con a) a) `bindM` \_ -> mkM a
</haskell>
 
First we create an object of type MOut with the Int part (). As you
see bindM will not use it ("\_"), but will concatenate the String part
with the result of mkM, which in turn is the empry string "".
 
In other words, first we insert the Output part (a string) in our MOut
Int type, and then we insert the Int.
 
Let's rewrite the evaluator:
 
<haskell>
 
> evalM_3 :: Term -> MOut Int
> evalM_3 (Con a) = outPut (formatLine (Con a) a) `bindM` \_ ->
>                  mkM a
> evalM_3 (Add t u) = evalM_3 t `bindM` \a ->
>                    evalM_3 u `bindM` \b ->
>                    outPut (formatLine (Add t u) (a + b)) `bindM` \_ ->
>                    mkM (a + b)
 
</haskell>
 
Well, this is fine, definetly better then before, anyway.
 
Still we use `bindM` \_ -> that binds something we do not use (_). We
could write something for this case, when we concatenate computations
without the need of binding variables. Let's call it `combineM`:
 
<haskell>
 
> combineM :: MOut a -> MOut b -> MOut b
> combineM m f = m `bindM` \_ -> f
 
</haskell>
 
So the new evaluator:
 
<haskell>
 
> evalM :: Term -> MOut Int
> evalM (Con a) = outPut (formatLine (Con a) a) `combineM`
>                mkM a
> evalM (Add t u) = evalM t `bindM` \a ->
>                  evalM u `bindM` \b ->
>                  outPut (formatLine (Add t u) (a + b)) `combineM`
>                  mkM (a + b)
 
</haskell>
 
Let's put everything together (and change some names changing M into
MO, so that this file will be still usable as a Literate Haskell
file):
 
<haskell>
 
> type MO a = (a, Out)
> type Out = String
 
> mkMO :: a -> MO a
> mkMO a = (a, "")
 
> bindMO :: MO a -> (a -> MO b) -> MO b
> bindMO m f = (b, x ++ y)
>              where (a, x) = m
>                    (b, y) = f a
 
> combineMO :: MO a -> MO b -> MO b
> combineMO m f = m `bindM` \_ -> f
 
> outMO :: Out -> MO ()
> outMO x = ((), x)
> evalMO :: Term -> MO Int
> evalMO (Con a) = outMO (formatLine (Con a) a) `combineMO`
>                  mkMO a
> evalMO (Add t u) = evalMO t `bindMO` \a ->
>                    evalMO u `bindMO` \b ->
>                    outMO (formatLine (Add t u) (a + b)) `combineMO`
>                    mkMO (a + b)
 
</haskell>
 
==What Does Bind Bind?==
 
<div id="Bind">
The evaluator looks like:
<haskell>
evalM t >>= \a -> evalM u >>= \b -> outPut "something" >>= \_ -> mkM (a +b)
</haskell>
where >>= is bindMO, obviously.
 
Let's do some substitution, writing the type of their output of each function:
* evalMO t => (a,Out) - where a is Int
* evalMO u => (b,Out) - where b is the same of a, an Int, but with a different value
* outMO Out = ((),Out)
* mkMO (a+b) => ((a+b),Out) - where (a+b) is the same of a and b, but with a different value from either a and b


I did not received very much feedback for this tutorial, and I must
admit that I would like to. But one person, on the haskell-cafe
mailing list, [http://www.haskell.org/pipermail/haskell-cafe/2006-September/017740.html told me]
that:
<pre>
<pre>
B | (a,Out) >>= \a -> (b,Out) >>= \b -> ((),Out) >>= \_ >>= ((a + b), Out)---\
imho your tutorial makes the error that is a very typical: when you
i |  V  V        V    V  V        V    V  V        ^      ^  ^    ^    |\
write your tutorial you already know what are monads and what the
n |  |__|________^    |  |        ^    |  |        |      |  |    |    |  MOut Int <=> ((a+b), Out)
program you will construct at the end. but your reader don't know all these!
d |_____|__(++)__|_Out_|__|__(++)__V_Out_|___|___(++)_|_(++)__|___|____|_____|/
for such fresh reader this looks as you made some strange steps, write
i |              |    |______(b)__|_____|_____(b)____|__(b)__|___|
some ugly code and he doesn't have chances to understand that this ugly
n |              |_________(a)___________|____________|__(a)__|
code is written just to show that this can be simplified by using monads.
g |                                      |_____()_____|
 
</pre>
</pre>


Clear, isn't it?
I believe that Bulat is right. In this tutorial you go through some
 
code that is '''really''' ugly and then, with some kind of magic, it
"bindMO" is just a function that takes care of gluing together, inside
turns out in the (redundant but) clean evaluator of the end of [[The Monadic Way/Part II]].
a data type, a sequence of computations!
 
== Some Sugar, Please!==
Now our evaluator has been completely transformed into a monadic
evaluator. That's what it is: a monad.
 
We have a function that constructs an object of type MO Int, formed by
a pair: the result of the evaluation and the accumulated
(concatenated) output.
 
The process of accumulation and the act of parting the MO Int into its
component is buried into bindMO, now, that can also preserve some
value for later uses.
 
So we have:
* MO a type constructor for a type carrying a pair composed by an Int and a String;
* bindMO, that gives a direction to the process of evaluation: it concatenates computations and captures some side effects we created (the direction is given by the changes in the Out part: there's a "before" when Out was something and there's a "later" when Out is something else).
* mkMO lets us create an object of type MO Int starting from an Int.
 
As you see this is all we need to create a monad. In other words
monads arise from the type system and the lambda calculus. Everything
else is just syntactic sugar.
 
So, let's have a look at that sugar: the famous do-notation!
 
===Basic monadic evaluator in do-notation===
 
We will now rewrite our [[The Monadic Way#BasicEval|basic evaluator]] using the do-notation.
 
Now we have to crate a new type: this is necessary in order to use
specific monadic notation and have at our disposal the more practical
do-notation ('''below we will see the consequences of doing so!'''):
 
<haskell>
 
> newtype Eval a = Eval a
>    deriving (Show)
 
</haskell>
 
So, our type will be an instance of the monad class. We will have to
define the methods of this class (>>= and return), but that will be
easy since we have already done that when defining bindMO and mkMO:
 
<haskell>
 
> instance Monad Eval where
>    return a = Eval a
>    Eval m >>= f = f m
 
</haskell>
 
You can see that return will create, from an argument of a variable
type a (in our case that will be an Int) an object of type Eval Int,
that carries inside just an Int, the result of the evaluation of a
Con.
 
Bind (>>=) will match for an object of type Eval, extracting what's
inside ("m") and will bind "m" in "f". We know that "f" must return an
object of type Eval with inside an Int resulted by the computations
made by "f" over "m" (that is to say, computations made by "f" where
"f" is a functions with variables, and one of those variables is bound
to the value resulting from the evaluation of "m").
 
In exchange for doing so we will now be able to take the old version
of our evaluator and substitute `bindMO` with >>= and `mkMO` with
return:
 
<haskell>
 
> evalM_4 :: Term -> Eval Int
> evalM_4 (Con a) = return a
> evalM_4 (Add t u) = evalM_4 t >>= \a ->
>                    evalM_4 u >>= \b ->
>                    return (a + b)
 
</haskell>
 
which is equivalent, in the do-notation, to:
 
<haskell>
 
> evalM_5 :: Term -> Eval Int
> evalM_5 (Con a) = return a
> evalM_5 (Add t u) = do a <- evalM_5 t
>                        b <- evalM_5 u
>                        return (a + b)
 
</haskell>
 
Simple: <hask>do</hask> binds the result of "eval_M5 t" to "a", binds
the result of "eval_M5 u" to "b" and then returns the sum of "a" and
"b". In a very imperative style.
 
===Monadic evaluator with output in do-notation===
 
We can now have an image of what our monad should be, if we want it to
produce output: it is out type (Eval) that is made up of a pair, and
Int and a String called Output.
 
During evaluation the first member of the pair (the Int) will "store"
the results of our computation (i.e.: the procedures to calculate the
final result). The second part, the String called Output, will get
filled up with the concatenated output of the computation.
 
The sequencing done by bindMO (now >>=) will take care of passing to
the next evaluation the needed (way to calculate the) Int and will do
some more side calculation to produce the output (concatenating
outputs resulting from computation of the new Int, for instance).
 
So we can grasp the basic concept of a monad: it is like a label which
we attach to each step of the evaluation (the String attached to the
Int).
This label is persistent within the process of computation and at each
step bindMO can do some manipulation of it.
We are creating side-effects and propagating them within our monads.
 
Ok. Let's translate our output-producing evaluator in monadic
notation:
 
<div id="MonadicEvalIO">
<haskell>
 
> newtype Eval_IO a = Eval_IO (a, O)
>    deriving (Show)
> type O = String
 
> instance Monad Eval_IO where
>    return a = Eval_IO (a, "")
>    (>>=) m f = Eval_IO (b, x ++ y)
>                        where Eval_IO (a, x) = m
>                              Eval_IO (b, y) = f a
> print_IO :: O -> Eval_IO ()
> print_IO x = Eval_IO ((), x)
> eval_IO :: Term -> Eval_IO Int
> eval_IO (Con a) = do print_IO (formatLine (Con a) a)
>                      return a
> eval_IO (Add t u) = do a <- eval_IO t
>                        b <- eval_IO u
>                        print_IO (formatLine (Add t u) (a + b))
>                        return (a + b)
 
</haskell>
Let's see the evaluator with output in action:
  *TheMonadicWay> eval_IO (Add (Con 6) (Add (Con 16) (Add (Con 20) (Con 12))))
  Eval_IO (54,"eval (Con 6) <= 6 - eval (Con 16) <= 16 - eval (Con 20) <= 20 - eval (Con 12) <= 12 - \
      eval (Add (Con 20) (Con 12)) <= 32 - eval (Add (Con 16) (Add (Con 20) (Con 12))) <= 48 - \
      eval (Add (Con 6) (Add (Con 16) (Add (Con 20) (Con 12)))) <= 54 - ")
  *TheMonadicWay>
 
Let's format the output part:
  eval (Con 6) <= 6
  eval (Con 16) <= 16
  eval (Con 20) <= 20
  eval (Con 12) <= 12
  eval (Add (Con 20) (Con 12)) <= 32
  eval (Add (Con 16) (Add (Con 20) (Con 12))) <= 48
  eval (Add (Con 6) (Add (Con 16) (Add (Con 20) (Con 12)))) <= 54
 
==What Happened to Our Output??==
 
Well, actually something happened to the output. Let's compare the
output of evalMO (the monadic evaluator written without the
do-notation) and eval_IO:
 
  *TheMonadicWay> evalMO (Con 6)
  (6,"eval (Con 6) <= 6 - ")
  *TheMonadicWay> eval_IO (Con 6)
  Eval_IO (6,"eval (Con 6) <= 6 - ")
  *TheMonadicWay> 
 
They look almost the same, but they are not the same: the output of
eval_IO has the Eval_IO stuff. It must be related to the changes we
had to do to our evaluator in order to use the do-conation, obviously.
 
What's changed?
 
First the type definition. We have now:
 
<haskell>
newtype Eval_IO a = Eval_IO (a, O)
      deriving (Show)
</haskell>
 
instead of
 
<haskell>
type MO a = (a, Out)
</haskell>
 
Moreover our bindMO and mkMO functions changed too, to reflect the
change of the type definition:
 
<haskell>
instance Monad Eval_IO where
    return a = Eval_IO (a, "")
    (>>=) m f = Eval_IO (b, x ++ y)
                      where Eval_IO (a, x) = m
                            Eval_IO (b, y) = f a
</haskell>
 
Now <hask>return a</hask> is the product of the application of the
type constructor Eval_IO to the pair that are going to form our monad.
 
"return" takes an Int and insert it into our monad. It will also
insert an empty String "" that (>>=) or (>>) will then concatenate in
the sequence of computations they glue together.
 
The same for (>>=): it will now return something constructed by
Eval_IO: "b", the result of the application of "f" to "a" (better, the
binding of "a" in "f") and "x" (matched by <hask>Eval_IO (a, x)</hask> with
the evaluation of "m") and "y", (matched by "Eval_IO(b,y)" with the
evaluation of "f a".
 
That is to say: in the "where" clause, we are matching for the
elements paired in a type Eval_IO: this is indeed the type of "m"
(corresponding to "eval_IO t" in the body of the evaluator) and "f a"
(where "f" correspond to another application of "eval_IO" to the
result of the previous  application of "m").
 
And so, "Eval_IO (a,x) = m" means: match "a" and "x", paired in a type
Eval_IO, and that are produced by the evaluation of "m" (that is to
say: "eval_IO t"). The same for Eval_IO (b,y): match "b" and "y"
produced by the evaluation of "f a".
 
So the output of the evaluator is now not simply a pair made of and
Int and a String. It is a specific type (Eval_IO) that happens to
carry a pair of an Int and a String. But, if we want the Int and the
string, we have to extract them from the Eval_IO type, as we do in the
"where" clause: we ''unpack'' our type object (let's call it with its
name: our monad!) and take out the Int and the String to feed the next
function application and the output generation.
 
The same to insert something in our monad: if we want to create a pair
of an Int and a String, pair of type Eval_IO, we now have to ''pack''
the together by using our type constructor, feeding it with pair
composed by and Int and a String. This is what we do with the "return"
method of out monad and with "print_IO" function, where:
* return insert into the monad an Int;
* print_IO insert into the monad a String.
 
Notice that "combineM" disappeared. This is because it comes for free
by just defining our type Eval_IO as an instance of the Monad class.
 
Indeed, if we look at the definition of the Monad class in the Prelude we read:
<haskell>
class Monad m where
    return :: a -> m a
    (>>=)  :: m a -> (a -> m b) -> m b
    (>>)  :: m a -> m b -> m b
    fail  :: String -> m a
 
    -- Minimal complete definition: (>>=), return
    p >> q  = p >>= \ _ -> q
    fail s  = error s
</haskell>
You can see that the "combineM"" method (or (>>)) is automatically derived by
the "bindMO" (or >>=) method:
 
<haskell>
p >> q  = p >>= \ _ -> q
</haskell>
 
So, what the hell is the old <hask>type MO a = (a, Out)</hask> that
did not required all this additional work (apart the need to
specifically define (>>)?
 
Thanks the help of some nice guy of the
[http://www.haskell.org/mailman/listinfo/haskell-cafe haskell-cafe mailing list]
(look at the thread started by
[http://www.haskell.org/pipermail/haskell-cafe/2006-August/017634.html this silly question of mine])
we can answer.
 
Type MO is just a synonymous for (a,Out): the two can be substituted
one for the other. That's it.
 
We did not have to pack "a" and "Out" together with a type constructor
to have a new type MO.
 
As a consequence, we cannot use MO as an instance of Monad, and so, we
cannot use with it the syntactic sugar we needed: the do-notation.
 
That is to say: a type created with the "type" keyword cannot be an
instance of a class, and cannot inherits its methods (in our case
(>>=, >> and return). And without those methods the do-notation is not
usable.
 
Anyway we will better understand all the far reaching consequences of
this new approach later on.
 
==Errare Monadicum Est==
 
Now that we have a basic understanding of what a monad is, and does,
we will further explore it by making some changes to our evaluator.
 
In this section we will se how to handle exceptions in our monadic
evaluator.
 
Suppose that we want to stop the execution of our monad if some
conditions occurs. If our evaluator was to compute divisions, instead
of sums, then we would like to stop the evaluator when a division by
zero occurs, possibly producing some output, instead of the result of
the evaluation of the expression, that explains what happened.
 
Basic error handling.
 
We will do so starting from the beginning once again...


===The basic evaluator, non monadic, with exception===
I took that mail as a challenge and
[http://www.haskell.org/pipermail/haskell-cafe/2006-September/017778.html I responded]
by writing [[Meet Bob The Monadic Lover]].


We just take our basic evaluator, without any output, and write a
In "Meet Bob" the code is clean, variable names are very descriptive,
method to stop execution if a condition occurs:
and you see that I can create a monad without any use of lambda
abstraction.


<haskell>
Bind (in "Meet Bob" is askLover) now takes a monad and a partial
application, not an anonymous function.


> data M a = Raise Exception
Obviously you can see an anonymous function as a partial application.
>          | Return a
>            deriving (Show)
> type Exception = String


</haskell>
The problem, I think, is that, in "Meet Bob", you cannot understand
the strict relation between what I did before and what I do when I
start using the "do-notation". You see that the same functions are
being used ("tellMyself" and "newLove"), but "andrea <- tellMyself 1"
is not explained. It's just magic.


Now, our monad is of datatype "M a" which can either be constructed
The fact is that you cannot understand "andrea <- tellMyself 1"
with the "Raise" constructor, that takes a String (Exception is a
without the use of lambda abstraction.
synonymous of String), or by the "Return" constructor, that takes a
variable type ("a"), an Int in our case.


I should have written an intermediate step, something like this:
<haskell>
<haskell>
 
drunk = do newLove "Paula " >>
> evalE :: Term -> M Int
                      (tellLover 1 10) >>= \paula ->
> evalE (Con a) = Return a
                          (tellMyself 3) >>= \lorena -> tellMyself (paula + lorena)


</haskell>
</haskell>


If evalE matches a Con it will construct a type Return with, inside, the content of the Con.
With this approach I think you can understand '''why and how''' you
 
come to write something like this:
<haskell>
<haskell>
 
drunk = do newLove "Paula "
> evalE (Add a b) =  
          paula <- (tellLover 1 10)
>    case evalE a of
          lorena <- tellMyself 3
>      Raise e -> Raise e
          tellMyself (paula + lorena)
>      Return a ->
>          case evalE b of
>            Raise e -> Raise e
>            Return b ->
>                if (a+b) == 42
>                    then Raise "The Ultimate Answer Has Been Computed!! Now I'm tired!"
>                    else Return (a+b)
 
</haskell>
</haskell>


If evalE matches an Add it will check if evaluating the first part
That is to say, in this way you can see a do block as a series of
produces a "Raise" or a "Return": in the first case it will return a
nested anonymous functions whose arguments are bound to some value by
"Raise" whose content is the same received.  
the application of >>=. Anonymous functions that bind to some value
 
the variables appearing after the "->" ("paula" and "lorena").
If instead the evaluation produces a value of type "Return", the
evaluator will evaluate the second part of the Add type.
 
If this returns a "Raise", a "Raise" will be returned all the way up the
recursion, otherwise the evaluator will check whether a condition for
raising a "Raise" exists. If not, it will return a "Return" with the sum inside.
 
Test it with:


  evalE (Add (Con 10) (Add (Add (Con 20) (Con 10)) (Con 2)))
To summarize, I think that even if you can start using monads without
understanding that what happens inside a "do" block is strictly
related with lambda calculus, I don't think you can claim you
understand monads just because you are using them.


And I'm quite sure that if a problem arises somewhere you can have a
very difficult time in trying to find out what the problem is. This is
even more true when you start doing monadic transformations.


===The basic evaluator, monadic, with exceptions===
==How to explain monads to newcomers?==


In order to produce a monadic version of the previous evaluator, the
Monads are not an impossible-to-understand-unless-you-have-a-phd
one that raises exceptions, we just need to abstract out from the
topic. I don't know if I can claim I'm a living proof of this
evaluator all that case analysis.
assumption, since I have a PhD. But please take into account that I
have a PhD in Comparative Private Law! Nothing related to computer
science. And I claim I understand monads.


<haskell>
Moreover, since I have come to understand them, I think I can try to
explain them to newcomers like me. This is why I started writing this
tutorial.


> data M1 a = Except Exception
Still, in order to understand them I studied, and I studied hard. I
>          | Ok {showM :: a }
wanted to understand, it was difficult, and I kept studying. Going
>            deriving (Show)
back to the foundation when needed.


</haskell>
So, I cannot explain monads to you unless you are willing to study. I can
show you the way, but you must take your time and follow that way.


The data type didn't change at all. Well, we changed the name of the Return type constructor (no Ok) so that
==What do I need to know to understand monads?==
this constructor can coexist with the previous one in the same Literate Haskell file.


<div id="MonadicEvalE">
===Functional programming===
<haskell>
> instance Monad M1 where
>    return a = Ok a
>    m >>= f = case m of
>                      Except e -> Except e
>                      Ok a -> f a


</haskell>
First you need at least a basic understanding of functional
programming.


Binding operations are now very easy. Basically we check:
This is going to be the easiest step, because there is an invaluable
* if the result of the evaluation of "m" produces an exception (first matct: Except e ->...), in which case we return its content by constructing our M1 Int with the "Raise" constructor".
resource available on line for free.
* if the result of the evaluation of "m" is matched with the "Ok" constructor, we get its content and use it to bind the argument of "f" to its value.


<hask>return a</hask> will just use the Ok type constructor for
This is what I did: I spent 20 hours of my life watching the Video
inserting "a" (in our case an Int) into M1 Int, the type of our monad.
Lectures by Hal Abelson and Gerald Jay Sussman on
[http://swiss.csail.mit.edu/classes/6.001/abelson-sussman-lectures/ Structure and Interpretation of Computer Programs].


<haskell>
This course, and the annexed text book (I did not read it entirely),
are an amazing introduction to computer science: you start by learning
some basic Scheme and then Abelson and Sussman will bring you, step by
step, to understand evaluation, interpretation and compilation of Lisp
(Scheme) code.


> raise :: Exception -> M1 a
The course is clear, interesting, funny and will provide you with a
> raise e = Except e
basic, but strong, understanding of functional programming.


</haskell>
Believe me: if you like programming but don't have a computer science
 
curriculum, you'll find out that spending 20 hours of your life to
This is just a helper function to construct our "M1 a" type with the
watch that course has been the most productive investment of your
Raise constructor. It takes a string and returns a type (M1 a) to be
learning life.
matched with the "Raise" constructor.


<haskell>
===My problem with Haskell===


> eval_ME :: Term -> M1 Int
I find Haskell an amazingly expressive programming language. It makes
> eval_ME (Con a) = do return a
it incredibly easy to perform very difficult tasks, just like monads,
> eval_ME (Add t u) = do a <- eval_ME t
for instance.
>                        b <- eval_ME u
>                        if (a+b) == 42
>                          then raise "The Ultimate Answer Has Been Computed!! Now I'm tired!"
>                          else return (a + b)


</haskell>
The problem is that, in doing so, it makes it difficult to understand
Haskell to newcomers.


The evaluator itself is very simple. We bind "a" with the result of
I must confess that tutorials and other learning material are quite
"eval_ME t", "b" with the result of "eval_ME u", and we check for a
dissatisfying on this regards too.
condition:
* if the condition is met we raise an exception, that is to say: we return a value constructed with the "Raise" constructor. This value will be matched by ">>=" in the next recursion. And >>= will just return it all the way up the recursion.
* if the condition is not me, we return a value constructed with the "Return" type constructor and go on with the recursion...


Run with:
Since Haskell seems to make easy what is not easy, these tutorial take
the "it's easy" approach.


  eval_ME (Add (Con 10) (Add (Add (Con 20) (Con 10)) (Con 2)))
I will give you an example. I think that the
[http://www.cs.utah.edu/~hal/htut/ "Yet Another Haskell Tutorial"] is
a really good attempt to explain Haskell, and if you want to
understand this tutorial you need to read it at least for getting a
good understanding of the Haskell type system.


It is noteworthy the fact that in our datatype definition we used a
The tutorial explains monads and explains the "do notation" in terms
label field with a label selector (we called it showM), even though it
of lambda abstraction (par. 9.1, p. 120). Still, to the lambda
was not used in our code. We will use this methodology later on.
operator ("\") the tutorial dedicates only 17 lines in par. 4.4.1.
 
Moreover "\" is explained just as another way of creating functions.
So, just to refresh your memory:
 
<haskell>
 
> data Person = Person {name :: String,
>                      age :: Int,
>                      hobby :: String
>                      } deriving (Show)
           
> andreaRossato = Person "Andrea" 37 "Haskell The Monadic Way"
> personName (Person a b c) = a
 
</haskell>
 
will produce:
  *TheMonadicWay> andreaRossato
  Person {name = "Andrea", age = 37, hobby = "Haskell The Monadic Way"}
  *TheMonadicWay> personName andreaRossato
  "Andrea"
  *TheMonadicWay> name andreaRossato
  "Andrea"
  *TheMonadicWay> age andreaRossato
  37
  *TheMonadicWay> hobby andreaRossato
  "Haskell The Monadic Way"
  *TheMonadicWay>
 
 
===Monadic evaluator with output and exceptions===
 
We will now try to combine the [[The Monadic Way#MonadicEvalIO|output-producing monadic evaluator]]
with [[The Monadic Way#MonadicEvalE| exception producing one]].
 
 
<haskell>
 
> data M2 a = Ex Exception
>          | Done {unpack :: (a,O) }
>            deriving (Show)
 
</haskell>
 
Now we need a datatype with two constructor: one to produce a value
type "Ex String" and one for value type "Done a".
 
'''Note''' that we changed the name of the exception type constructor
from "Raise" to "Ex" just to make the two coexist in the same Literate
Haskell file.
 
The constructor "Done a" is defined with a label sector: <hask>Done {unpack :: (a,O)}</hask>
is equivalent to <hask>Done (a,O)</hask>.
 
The only difference is that, this way, we are also defining a method
to retrieve the pair (a,O) (in our case "O" is a synonymous for
String, whereas "a" is a variable type) from an object of type "Done
a".
 
<haskell>
 
> instance Monad M2 where
>    return a = Done (a, "")
>    m >>= f = case m of
>                      Ex e -> Ex e
>                      Done (a, x) -> case (f a) of
>                                      Ex e1 -> Ex e1
>                                      Done (b, y) -> Done (b, x ++ y)
 
</haskell>
 
Now our binding operations gets more complicated by the fact that we
have to concatenate the output, as we did before, '''and''' check for
exceptions.
 
It is not possible to do has we did in the [[The Monadic Way#MonadicEvalE| exception producing evaluator]],
where we could check just for "m" (remember the "m" in the first run
stands for "eval t").
 
Since at the end we must return the output produced by the evaluation
of "m" ''concatenated'' with the output produced by the evaluation of
"f a" (where "a" is returned by "m", paired with "x" by "Done"), now
we must check if we '''do have''' an output from "f a" produced by
"Done".
 
Indeed, now, "f a" can also produce a value constructed by "Ex", and
this value does not contain the pair as the value produced by "Done".
 
So:
* we evaluate "m":
** if we match a value produced by type constructor "Ex" we return a value produced by type constructor "Ex" whose content is the one we extracted in the matching;
** if we match a value produced by "Done" we match the pair it carries "(a,x)" and we analyze what "f a" returns:
*** if "f a" returns a value produced by "Ex" we extract the exception and we return it, constructing a value with "Ex"
*** if "f a" returns a value produced by "Done" we return "b" and the concatenated "x" and "y".
 
And now the evaluator:
 
<haskell>
                     
> raise_IOE :: Exception -> M2 a
> raise_IOE e = Ex e
 
</haskell>
 
This is the function to insert in our monad (M2) an exception: we take
a String and produce a value applying the type constructor for
exception "Ex" to its value.
 
<haskell>
 
> print_IOE :: O -> M2 ()
> print_IOE x = Done ((), x)
 
</haskell>
 
The function to produce output is the very same of the one of the [[The Monadic Way#MonadicEvalIO|output-producing monadic evaluator]].
 
<haskell>
> eval_IOE :: Term -> M2 Int
> eval_IOE (Con a) = do print_IOE (formatLine (Con a) a)
>                      return a
> eval_IOE (Add t u) = do a <- eval_IOE t
>                        b <- eval_IOE u
>                        let out = formatLine (Add t u) (a + b)
>                        print_IOE out
>                        if (a+b) == 42
>                            then raise_IOE $ out ++ "The Ultimate Answer Has Been Computed!! Now I'm tired!"
>                            else return (a + b)
 
</haskell>
 
The evaluator procedure did not change very much from the one of the [[The Monadic Way#MonadicEvalIO|output-producing monadic evaluator]].
 
We just added the case analysis to see if the condition for raising an exception is met.
 
Running with
 
  eval_IOE (Add (Con 10) (Add (Add (Con 20) (Con 10)) (Con 2)))
 
will produce
 
  Ex "eval (Add (Con 10) (Add (Add (Con 20) (Con 10)) (Con 2))) <= 42 -
      The Ultimate Answer Has Been Computed!! Now I'm tired!"
 
Look at the <hask>let</hask> clause within the do-notation. We do not
need to use the "let ... in" construction: since all bound variables
remain bound within a <hask>do</hask> procedure (see [[The Monadic Wat#Bind|here]],
we do not need the "in" to specify "where" the variable "out" will be bound in!
 
==We Need  A State==
 
'''(Text to be done yet: just a summary)'''
 
We start adding complexity to our monadic evaluator. But in order to
add a counter we will start over again (to review out knowledeg).
 
===The basic evaluator, non monadic, with a counter===
The basic evaluator plus a counter: evalNM takes now the expression to
be evaluated plus an initial state (0) to start counting from.
 
<haskell>
 
> -- non monadic
> evalNMS :: Term -> MS Int
> evalNMS (Con a) x = (a, x + 1)
> evalNMS (Add t u) x = let (a, y) = evalNMS t x in
>                      let (b, z) = evalNMS u y in
>                      (a + b, z +1)
 
</haskell>
 
===The evaluator, monadic, with a counter, without do-notation===
The moadic version without do notation.
 
<haskell>
 
> -- monadic
 
> type MS a = State -> (a, State)
> type State = Int
 
> mkMS :: a -> MS a
> mkMS a = \x -> (a, x)
 
> bindMS :: MS a -> (a -> MS b) -> MS b
> bindMS m f = \x ->
>              let (a, y) = m x in
>              let (b, z) = f a y in
>              (b, z)
 
> combineMS :: MS a -> MS b -> MS b
> combineMS m f = m `bindMS` \_ -> f
> incState :: MS ()
> incState = \s -> ((), s + 1)
 
> evalMS :: Term -> MS Int
> evalMS (Con a) = incState `combineMS` mkMS a
> evalMS (Add t u) = evalMS t `bindMS` \a ->
>                    evalMS u `bindMS` \b ->
>                    incState `combineMS` mkMS (a + b)
 
> --evalMS (Add (Con 6) (Add (Con 16) (Add (Con 20) (Con 12)))) 0
 
</haskell>
 
===The evaluator, monadic, with counter and output, without do-notation===
 
Now we'll add Output to the stateful evaluator:
 
<haskell>
 
> -- state and output
 
> type MSO a = State -> (a, State, Output)
 
> mkMSO :: a -> MSO a
> mkMSO a = \s -> (a, s, "")
 
> bindMSO :: MSO a -> (a -> MSO b) -> MSO b
> bindMSO m f = \x ->
>              let (a, y, s1) = m x in
>              let (b, z, s2) = f a y in
>              (b, z, s1 ++ s2)
 
> combineMSO :: MSO a -> MSO b -> MSO b
> combineMSO m f = m `bindMSO` \_ -> f
 
> incMSOstate :: MSO ()
> incMSOstate = \s -> ((), s + 1, "")
 
> outMSO :: Output -> MSO ()
> outMSO = \x s -> ((),s, x)
 
> evalMSO :: Term -> MSO Int
> evalMSO (Con a) = incMSOstate `combineMSO`
>                  outMSO (formatLine (Con a) a) `combineMSO`
>                  mkMSO a
> evalMSO (Add t u) =  evalMSO t `bindMSO` \a ->
>                      evalMSO u `bindMSO` \b ->
>                      incMSOstate `combineMSO`
>                      outMSO (formatLine (Add t u) (a + b)) `combineMSO`
>                      mkMSO (a + b)
 
> --evalMSO (Add (Con 6) (Add (Con 16) (Add (Con 20) (Con 12)))) 0
 
</haskell>
 
===The monadic evaluator with output and counter in do-notation=== 
 
State, Output in do-notation. Look at how much the complexity of our
(>>=) founction is increasing:
 
<haskell>
 
> -- thanks to  Brian Hulley
 
> newtype MSIO a = MSIO (State -> (a, State, Output))
> instance Monad MSIO where
>    return a = MSIO (\s -> (a, s, ""))
>    (MSIO m) >>= f = MSIO $ \x ->
>                      let (a, y, s1) = m x in
>                      let MSIO runNextStep = f a in
>                      let (b, z, s2) = runNextStep y in
>                      (b, z, s1 ++ s2)
 
 
> incMSOIstate :: MSIO ()
> incMSOIstate = MSIO (\s -> ((), s + 1, ""))
 
> print_MSOI :: Output -> MSIO ()
> print_MSOI x = MSIO (\s -> ((),s, x))
 
> eval_MSOI :: Term -> MSIO Int
> eval_MSOI (Con a) = do incMSOIstate
>                        print_MSOI (formatLine (Con a) a)
>                        return a
 
> eval_MSOI (Add t u) = do a <- eval_MSOI t
>                          b <- eval_MSOI u
>                          incMSOIstate
>                          print_MSOI (formatLine (Add t u) (a + b))
>                          return (a + b)
 
> run_MSOI :: MSIO a -> State -> (a, State, Output)
> run_MSOI (MSIO f) s = f s
 
> --run_MSOI (eval_MSOI (Add (Con 6) (Add (Con 16) (Add (Con 20) (Con 12))))) 0
 
</haskell>
 
===Another version of the monadic evaluator with output and counter, in do-notation===
 
This is e second version that exploit label fields in datatype to
decrease the complexity of the binding operations.
 
<haskell>
 
> -- Thanks Udo Stenzel
 
> newtype Eval_SIO a = Eval_SIO { unPackMSIOandRun :: State -> (a, State, Output) }
> instance Monad Eval_SIO where
>    return a = Eval_SIO (\s -> (a, s, ""))
>    (>>=) m f = Eval_SIO (\x ->
>                      let (a, y, s1) = unPackMSIOandRun m x in
>                      let (b, z, s2) = unPackMSIOandRun (f a) y in
>                      (b, z, s1 ++ s2))
 
> incSIOstate :: Eval_SIO ()
> incSIOstate = Eval_SIO (\s -> ((), s + 1, ""))
 
> print_SIO :: Output -> Eval_SIO ()
> print_SIO x = Eval_SIO (\s -> ((),s, x))
 
> eval_SIO :: Term -> Eval_SIO Int
> eval_SIO (Con a) = do incSIOstate
>                      print_SIO (formatLine (Con a) a)
>                      return a
> eval_SIO (Add t u) = do a <- eval_SIO t
>                        b <- eval_SIO u
>                        incSIOstate
>                        print_SIO (formatLine (Add t u) (a + b))
>                        return (a + b)
 
> --unPackMSIOandRun (eval_SIO (Add (Con 6) (Add (Con 16) (Add (Con 20) (Con 12))))) 0
 
</haskell>
 
 
==If There's A State We Need Some Discipline: Dealing With Complexity==
 
In order to increase the complexity of our monad now we will try to
mix State (counter), Exceptions and Output.
 
This is an email [http://www.haskell.org/pipermail/haskell-cafe/2006-August/017672.html I send to the haskell-cafe mailing list]:
 
<pre>
Now I'm trying to create a statefull evaluator, with output and
exception, but I'm facing a problem I seem not to be able to
conceptually solve.
 
Take the code below.
Now, in order to get it run (and try to debug) the Eval_SOI type has a
Raise constructor that produces the same type of SOIE. Suppose instead it
should be constructing something like Raise "something".  
Moreover, I wrote a second version of >>=, commented out.
This is just to help me illustrate to problem I'm facing.
 
Now, >>= is suppose to return Raise if "m" is matched against Raise
(second version commented out).
If "m" matches SOIE it must return a SOIE only if "f a" does not
returns a Raise (output must be concatenated).
 
I seem not to be able to find a way out. Moreover, I cannot understand
if a way out can be possibly found. Something suggests me it could be
related to that Raise "something".
But my feeling is that functional programming could be something out
of the reach of my mind... by the way, I teach Law, so perhaps you'll
forgive me...;-)
 
If you can help me to understand this problem all I can promise is
that I'll mention your help in the tutorial I'm trying to write on
"the monadic way"... that seems to lead me nowhere.
 
Thanks for your kind attention.
 
Andrea
</pre>
 
This was the code:
 
<haskell>
data Eval_SOI a = Raise { unPackMSOIandRun :: State -> (a, State, Output) }
                | SOIE { unPackMSOIandRun :: State -> (a, State, Output) }
 
instance Monad Eval_SOI where
    return a = SOIE (\s -> (a, s, ""))
    m >>= f =  SOIE (\x ->
                      let (a, y, s1) = unPackMSOIandRun m x in
                      case f a of
                        SOIE nextRun -> let (b, z, s2) = nextRun y in 
                                        (b, z, s1 ++ s2)
                        Raise e1 -> e1 y  --only this happens
 
                      )
--    (>>=) m f =  case m of
--                    Raise e -> error "ciao" -- why this is not going to happen?
--                    SOIE a -> SOIE (\x ->
--                                    let (a, y, s1) = unPackMSOIandRun m x in
--                                    let (b, z, s2) = unPackMSOIandRun (f a) y in
--                                    (b, z, s1 ++ s2))  
 
 
incSOIstate :: Eval_SOI ()
incSOIstate = SOIE (\s -> ((), s + 1, ""))
 
print_SOI :: Output -> Eval_SOI ()
print_SOI x = SOIE (\s -> ((),s, x))
 
raise x e = Raise (\s -> (x,s,e))
 
eval_SOI :: Term -> Eval_SOI Int
eval_SOI (Con a) = do incSOIstate
                      print_SOI (formatLine (Con a) a)
                      return a
eval_SOI (Add t u) = do a <- eval_SOI t
                        b <- eval_SOI u
                        incSOIstate
                        print_SOI (formatLine (Add t u) (a + b))
                        if (a + b)  ==  42
                          then raise (a+b) " = The Ultimate Answer!!"
                          else return (a + b)
 
runEval exp =  case eval_SOI exp of
                Raise a -> a 0
                SOIE p -> let (result, state, output) = p 0 in
                            (result,state,output)
 
 
 
--runEval (Add (Con 10) (Add (Con 28) (Add (Con 40) (Con 2))))
</haskell>
 
This code will produce
 
  eval (Con 10) <= 10 -
  eval (Con 28) <= 28 -
  eval (Con 40) <= 40 -
  eval (Con 2) <= 2 -  = The Ultimate Answer!!
  eval (Add (Con 28) (Add (Con 40) (Con 2))) <= 70 -
  eval (Add (Con 10) (Add (Con 28) (Add (Con 40) (Con 2)))) <= 80 -
 
The exception appears in the output, but executioon is not stopped.
 
===Monadic evaluator with output, counter and exception, in do-notation===
 
Brian Hulley [http://www.haskell.org/pipermail/haskell-cafe/2006-August/017680.html came up with this solution]:
 
<haskell>
 
> -- thanks to  Brian Hulley
> data Result a
>    = Good a State Output
>    | Bad State Output Exception
>    deriving Show
 
> newtype Eval_SIOE a = SIOE {runSIOE :: State -> Result a}
 
> instance Monad Eval_SIOE where
>    return a = SIOE (\s -> Good a s "")
>    m >>= f = SIOE $ \x ->
>              case runSIOE m x of
>                Good a y o1 ->
>                    case runSIOE (f a) y of
>                      Good b z o2 -> Good b z (o1 ++ o2)
>                      Bad z o2 e -> Bad z (o1 ++ o2) e
>                Bad z o2 e -> Bad z o2 e
 
> raise_SIOE e = SIOE (\s -> Bad s "" e)
 
> incSIOEstate :: Eval_SIOE ()
> incSIOEstate = SIOE (\s -> Good () (s + 1) "")
 
> print_SIOE :: Output -> Eval_SIOE ()
> print_SIOE x = SIOE (\s -> Good () s  x)
 
 
> eval_SIOE :: Term -> Eval_SIOE Int
> eval_SIOE (Con a) = do incSIOEstate
>                        print_SIOE (formatLine (Con a) a)
>                        return a
> eval_SIOE (Add t u) = do a <- eval_SIOE t
>                          b <- eval_SIOE u
>                          incSIOEstate
>                          let out = formatLine (Add t u) (a + b)
>                          print_SIOE out
>                          if (a+b) == 42
>                            then raise_SIOE $ out ++ "The Ultimate Answer Has Been Computed!! Now I'm tired!"
>                            else return (a + b)
 
> runEval exp =  case runSIOE (eval_SIOE exp) 0 of
>                  Bad s o e -> "Error at iteration n. " ++ show s ++
>                              " - Output stack = " ++ o ++
>                              " - Exception = " ++ e
>                  Good a s o -> "Result = " ++ show a ++
>                                " - Iterations = " ++ show s ++ " - Output = " ++ o
 
</haskell>
 
Run with runEval (Add (Con 18) (Add (Con 12) (Add (Con 10) (Con 2))))
 
==Taking Complexity Out of a Monad: Monadic Transformers==
 
We have seen how the complexity of (>>=) was growing by adding
operations to be done.
N
We will do the opposite: we will implement a state transformer (I
copied StateT).
 
We will embed our monad in the StateT monad and we will start moving
state and output from the inner monad (our one) to the outer monad
(StateT).
 
===The StateT Monad: a Monad Container===
Let me introduce StateT with some useful functions:
 
<haskell>
 
> newtype StateT s m a  = StateT {runStateT :: s -> m (a,s) }  --StateT (s -> m (a,s))
 
> instance Monad m => Monad (StateT s m) where
>  return a  = StateT (\s -> return (a,s))
>  StateT m1 >>= k = StateT (\s -> do ~(a,s1)  <- m1 s
>                                      let StateT m2  = k a
>                                      m2 s1)
 
> -- | Execute a stateful computation, as a result we get
> -- the result of the computation, and the final state.
> runState          :: s -> StateT s m a -> m (a,s)
> runState s (StateT m)    = m s
 
> -- | Execute a stateful computation, ignoring the final state.
> evalState          :: Functor m => s -> StateT s m a -> m a
> evalState s m      = fmap fst (runState s m)
 
> -- | Execute a stateful computation, just for the side effect.
> execState          :: Functor m => s -> StateT s m a -> m s
> execState s m      = fmap snd (runState s m)
 
 
> lift :: (Monad m) => m a -> StateT s m a
> lift m  = StateT (\s -> do x <- m
>                            return (x,s))
 
</haskell>
 
StateT is pleased to meet you!.
 
===StateT as a counter, and monadic evaluator with output and exceptions===
And now out monad, with state out from it:
 
<haskell>
 
> data MTa a = FailTa Exception
>            | DoneTa {unpackDoneTa :: (a,O) }
>              deriving (Show)
 
 
> instance Monad MTa where
>    return a = DoneTa (a, "")
>    m >>= f = case m of
>                FailTa e -> FailTa e
>                DoneTa (a, x) -> case (f a) of
>                                    FailTa e1  -> FailTa e1
>                                    DoneTa (b, y) -> DoneTa (b, x ++ y)                     
 
> instance Functor MTa where
>  fmap _ (FailTa e) = FailTa e
>  fmap f (DoneTa (r,o)) = DoneTa ((f r),o)
 
> raiseTa_SIOE :: O -> StateT Int MTa a
> raiseTa_SIOE e = lift (FailTa e)
 
> printTa_SIOE :: O -> StateT Int MTa ()
> printTa_SIOE x = lift (DoneTa ((), x))
 
> incTaState :: StateT Int MTa ()
> incTaState = StateT (\s -> return ((), s + 1))
 
> evalTa_SIOE :: Term -> StateT Int MTa Int
> evalTa_SIOE (Con a) = do incTaState
>                          printTa_SIOE (formatLine (Con a) a)
>                          return a
> evalTa_SIOE (Add t u) = do a <- evalTa_SIOE t
>                            b <- evalTa_SIOE u
>                            incTaState
>                            let out = formatLine (Add t u) (a + b)
>                            printTa_SIOE out
>                            if (a+b) == 42
>                                then raiseTa_SIOE $
>                                        out ++ "The Ultimate Answer Has Been Computed!! Now I'm tired!"
>                                else return (a + b)
 
> runEvalTa :: Term -> String
> runEvalTa exp = case runStateT (evalTa_SIOE exp) 0 of
>                FailTa e -> e
>                DoneTa (~(r,s),o)-> "Result = " ++ show r ++
>                                    "; Iteration = " ++ show s ++
>                                      "; Output = " ++ o
 
> runEvalTa1 :: Term -> String
> runEvalTa1 exp = case runState 0 (evalTa_SIOE exp) of
>                    FailTa e -> e
>                    DoneTa ((r,s),o) -> "Result = " ++ show r ++
>                                        "; Iteration = "  ++ show s ++
>                                        "; Output = " ++ o
                               
> runEvalTa2 :: Term -> String
> runEvalTa2 exp = case evalState 0 (evalTa_SIOE exp) of
>                    FailTa e -> e
>                    DoneTa (r,o) -> "Result = "  ++ show r ++ "; Output = " ++ o
 
> runEvalTa3 :: Term -> String
> runEvalTa3 exp = case execState 0 (evalTa_SIOE exp) of
>                    FailTa e -> e
>                    DoneTa (s,o) -> "Iterations = "  ++ show s ++ "; Output = " ++ o
 
</haskell>
 
===StateT to keep output and counter, and monadic evaluator with (only) exceptions===
 
Now we take output away from the inner monad and place it in the outer
one (StateT):
 
<haskell>
 
 
> data MTb a = FailTb Exception
>            | DoneTb {unpackDoneTb :: a }
>              deriving (Show)
 
> type StateIO = (O,Int)
 
> instance Monad MTb where
>    return a = DoneTb a
>    m >>= f = case m of
>                      FailTb e -> FailTb e
>                      DoneTb a -> f a
 
> instance Functor MTb where
>  fmap _ (FailTb e) = FailTb e
>  fmap f (DoneTb b) = DoneTb (f b)
 
 
> raiseTb_SIOE :: O -> StateT StateIO MTb a
> raiseTb_SIOE e = lift (FailTb e)
 
> printTb_SIOE :: O -> StateT StateIO MTb ()
> printTb_SIOE x = StateT (\(o,s) -> return ((), (o ++ x,s)))
 
> incTbStateIO :: StateT StateIO MTb ()
> incTbStateIO = StateT (\(o,s) -> return ((), (o,s + 1)))
 
> evalTb_SIOE :: Term -> StateT StateIO MTb Int
> evalTb_SIOE (Con a) = do incTbStateIO
>                          printTb_SIOE (formatLine (Con a) a)
>                          return a
> evalTb_SIOE (Add t u) = do a <- evalTb_SIOE t
>                            b <- evalTb_SIOE u
>                            incTbStateIO
>                            let out = formatLine (Add t u) (a + b)
>                            printTb_SIOE out
>                            if (a+b) == 42
>                              then raiseTb_SIOE $
>                                        out ++ "The Ultimate Answer Has Been Computed!! Now I'm tired!"
>                              else return (a + b)
 
</haskell>
 
We take away complexity  from >>= and put it in the function we need
to use to manipulate content in our StateT monad.
 
These are some wrapper to the evaluator to get the result and the
side-effects produced by evaluation:
 
<haskell>
 
> runEvalTb :: Term -> String
> runEvalTb exp = case runStateT (evalTb_SIOE exp) ("",0) of
>                  FailTb e -> e
>                  DoneTb (r,~(o,s)) -> "Result = " ++ show r ++
>                                        "; Iteration = "  ++ show s ++
>                                        "; Output = " ++ o
 
 
> runEvalTb1 :: Term -> String
> runEvalTb1 exp = case runState ("",0)  (evalTb_SIOE exp) of
>                    FailTb e -> e
>                    DoneTb (r,~(o,s)) -> "Result = " ++ show r ++
>                                        "; Iteration = "  ++ show s ++
>                                        "; Output = " ++ o
                               
> runEvalTb2 :: Term -> String
> runEvalTb2 exp = case evalState ("",0)  (evalTb_SIOE exp) of
>                    FailTb e -> e
>                    DoneTb r -> "Result = "  ++ show r
 
> runEvalTb3 :: Term -> String
> runEvalTb3 exp = case execState ("",0)  (evalTb_SIOE exp) of
>                    FailTb e -> e
>                    DoneTb (o,s) -> "Iterations = "  ++ show s ++
>                                    " - Output = " ++ o
 
</haskell>
 
===StateT to keep output, counter and debug. The monadic evaluator is only for failable computations===
 
And now we will  keep in the inner monad only the result of the evaluation.
 
<haskell>
 
> data MT a = FailT Exc
>          | DoneT {unpackDoneT :: a }
>            deriving (Show)
 
> type Exc = String
> type IOstack = [Output]
> newtype StateTIO = StateTIO {unPackStateTIO :: (IOstack,Exc,Int)}
>    deriving(Show)
 
> instance Monad MT where
>    return a = DoneT a
>    m >>= f = case m of
>                FailT e -> FailT e
>                DoneT a -> f a
 
> instance Functor MT where
>    fmap _ (FailT a) = FailT a
>    fmap f (DoneT a) = DoneT (f a)
 
</haskell>
 
Simple isn't it?
 
The complexity is now below:
 
<haskell>
 
> stopExecT_SIOE :: Output -> StateT StateTIO MT Int
> stopExecT_SIOE exc = StateT (\s -> do x <- FailT exc
>                                      return (x, s))
 
> catchT_SIOE exc = StateT (\(StateTIO (o,e,s)) ->
>                          return ((), StateTIO (o ,"Exception at Iteration " ++
>                                                show s ++ ": " ++ exc ++ " - " ++ e,s)))
 
> printT_SIOE :: Output -> StateT StateTIO MT ()
> printT_SIOE x = StateT (\(StateTIO (o,e,s)) -> return ((), StateTIO (x:o,e,s)))
 
> incTstateIO :: StateT StateTIO MT ()
> incTstateIO = StateT (\(StateTIO (o,e,s)) -> return ((),StateTIO (o,e,s + 1)))
 
> evalT_SIOE :: Term -> StateT StateTIO MT Int
> evalT_SIOE (Con a) = do incTstateIO
>                        printT_SIOE (formatLine (Con a) a)
>                        return a
> evalT_SIOE (Add t u) = do a <- evalT_SIOE t
>                          b <- evalT_SIOE u
>                          incTstateIO
>                          let out = formatLine (Add t u) (a + b)
>                          printT_SIOE out
>                          case (a+b) of
>                            42 -> do catchT_SIOE "The Ultimate Answer Has Been Computed!! Now I'm tired!"
>                                      return (a+b)
>                            11 -> stopExecT_SIOE "11.... I do not like this number!"
>                            otherwise ->  return (a + b)
 
</haskell>
 
But now we have exceptions to stop execution and debugging output.
 
Some helper functions:
 
<haskell>
 
> runEvalT :: Term -> String
> runEvalT exp = case runStateT (evalT_SIOE exp) (StateTIO ([],"",0)) of
>                  FailT e -> e
>                  DoneT (r,StateTIO (o,e,s)) -> "Result = " ++ show r ++ "; Iteration = "  ++ show s ++
>                                              "; Output = " ++ show o ++ " - Exceptions = " ++ e
 
 
> runEvalT1 :: Term -> String
> runEvalT1 exp = case runState (StateTIO ([],"",0))  (evalT_SIOE exp) of
>                  FailT e -> e
>                  DoneT (r,StateTIO(o,e,s)) -> "Result = " ++ show r ++ "; Iteration = "  ++ show s
>                                                ++ "; Output = " ++ show o ++ " - Exceptions = " ++ e
                               
> runEvalT2 :: Term -> String
> runEvalT2 exp = case evalState (StateTIO ([],"",0))  (evalT_SIOE exp) of
>                  FailT e -> e
>                  DoneT r -> "Result = "  ++ show r
 
> runEvalT3 :: Term -> String
> runEvalT3 exp = case execState (StateTIO ([],"",0)) (evalT_SIOE exp) of
>                  FailT e -> e
>                  DoneT (StateTIO (o,e,s)) -> "Iterations = "  ++ show s ++
 
>                                              " - Output = " ++ show o ++ " - Exceptions = " ++ e
> showOut :: [String] -> IO ()
> showOut [] = return ()
> showOut (a:xs) = do print a
>                    showOut xs
 
> runMyEval :: Term -> IO ()
> runMyEval exp = let StateTIO (a,b,c) = unpackDoneT $ execState (StateTIO ([],"",0)) (evalT_SIOE exp) in
>                showOut $ reverse a
 
</haskell>
 
Some tests:
<pre>
  *TheMonadicWay> runEvalT (Add (Con 18) (Add (Con 12) (Add (Con 10) (Con 2))))
  "Result = 42; Iteration = 7; Output = [\"eval (Add (Con 18) (Add (Con 12) (Add (Con 10) (Con 2)))) <= 42 - \",
    \"eval (Add (Con 12) (Add (Con 10) (Con 2))) <= 24 - \",
    \"eval (Add (Con 10) (Con 2)) <= 12 - \",
    \"eval (Con 2) <= 2 - \",
    \"eval (Con 10) <= 10 - \",
    \"eval (Con 12) <= 12 - \",
    \"eval (Con 18) <= 18 - \"] -
    Exceptions = Exception at Iteration 7: The Ultimate Answer Has Been Computed!! Now I'm tired! - "
  *TheMonadicWay> runEvalT2 (Add (Con 18) (Add (Con 12) (Add (Con 10) (Con 2))))
  "Result = 42"
  *TheMonadicWay> runEvalT3 (Add (Con 18) (Add (Con 12) (Add (Con 10) (Con 2))))
  "Iterations = 7 - Output = [\"eval (Add (Con 18) (Add (Con 12) (Add (Con 10) (Con 2)))) <= 42 - \",
    \"eval (Add (Con 12) (Add (Con 10) (Con 2))) <= 24 - \",
    \"eval (Add (Con 10) (Con 2)) <= 12 - \",
    \"eval (Con 2) <= 2 - \",
    \"eval (Con 10) <= 10 - \",
    \"eval (Con 12) <= 12 - \",
    \"eval (Con 18) <= 18 - \"] -
    Exceptions = Exception at Iteration 7: The Ultimate Answer Has Been Computed!! Now I'm tired! - "
  *TheMonadicWay> runEvalT3 (Add (Con 1) (Add (Con 7) (Add (Con 1) (Con 2))))
  "Iterations = 7 - Output = [\"eval (Add (Con 1) (Add (Con 5) (Add (Con 1) (Con 2)))) <= 9 - \",
    \"eval (Add (Con 5) (Add (Con 1) (Con 2))) <= 8 - \",
    \"eval (Add (Con 1) (Con 2)) <= 3 - \",
    \"eval (Con 2) <= 2 - \",
    \"eval (Con 1) <= 1 - \",
    \"eval (Con 5) <= 5 - \",
    \"eval (Con 1) <= 1 - \"] - Exceptions = "
  *TheMonadicWay> runEvalT3 (Add (Con 1) (Add (Con 7) (Add (Con 1) (Con 2))))
  "11.... I do not like this number!"
  *TheMonadicWay> runMyEval (Add (Add (Add (Add (Con 10) (Con 2)) (Add (Con 12) (Con 3))) (Con 3)) (Con 10))
  "eval (Con 10) <= 10 - "
  "eval (Con 2) <= 2 - "
  "eval (Add (Con 10) (Con 2)) <= 12 - "
  "eval (Con 12) <= 12 - "
  "eval (Con 3) <= 3 - "
  "eval (Add (Con 12) (Con 3)) <= 15 - "
  "eval (Add (Add (Con 10) (Con 2)) (Add (Con 12) (Con 3))) <= 27 - "
  "eval (Con 3) <= 3 - "
  "eval (Add (Add (Add (Con 10) (Con 2)) (Add (Con 12) (Con 3))) (Con 3)) <= 30 - "
  "eval (Con 10) <= 10 - "
  "eval (Add (Add (Add (Add (Con 10) (Con 2)) (Add (Con 12) (Con 3))) (Con 3)) (Con 10)) <= 40 - "
  *TheMonadicWay>
</pre>
 
 
==The Final Cut==
 
===StateT for output, counter, debug, using the Standard Library===
 
<haskell>
module MyStateT where
import Control.Monad.State hiding (State)
 
data Term = Con Int
          | Add Term Term
            deriving (Show)
 
type IOStack = [Output]
type Output = String
type Debug = [String]
data EvalST = State {getIOS :: IOStack, getDebug :: Debug, getCount:: Int}
    deriving(Show)
 
 
type Exception = String
data MT a = Fail Exception
          | Done {unpackDone :: a }
            deriving (Show)
 
type Eval s a = StateT s MT a
 
instance Monad MT where
    return a = Done a
    m >>= f = case m of
                Fail e -> Fail e
                Done a -> f a
 
instance Functor MT where
    fmap _ (Fail a) = Fail a
    fmap f (Done a) = Done (f a)
 
emptyState = State [] [] 0
 
stopExecT exc = lift $ Fail exc
 
catchT e = do st <- get
              let s = getCount st
              let es = getDebug st
              let o = getIOS st
              let exc = "Debug msg at Iteration " ++ show s ++ ": " ++ e
              put $ State o (exc:es) s
 
printT :: Output -> Eval EvalST ()
printT o = do st <- get
              let s = getCount st
              let e = getDebug st
              let os = getIOS st
              let out = show s ++ " - " ++ o
              put $ State (out:os) e s
 
incTcounter :: Eval EvalST ()
incTcounter = do st <- get
                let s = getCount st
                let e = getDebug st
                let o = getIOS st
                put $ State o e (s+1)
 
evalT :: Term -> Eval EvalST Int
evalT (Con a) = do incTcounter
                  printT (formatLine (Con a) a)
                  return a
evalT (Add t u) = do a <- evalT t
                    b <- evalT u
                    incTcounter
                    let out = formatLine (Add t u) (a + b)
                    printT out
                    case (a+b) of
                      42 -> do catchT "The Ultimate Answer Has Been Computed!! Now I'm tired!"
                                return (a+b)
                      11 -> stopExecT "11.... I do not like this number!"
                      otherwise ->  return (a + b)
 
 
formatLine :: Term -> Int -> Output
formatLine t a = "eval (" ++ show t ++ ") <= " ++ show a                                                     
 
printAll :: [String] -> IO ()
printAll [] = return ()
printAll (a:xs) = do putStrLn a
                    printAll xs
 
eval :: Term -> IO ()
eval exp = case execStateT (evalT exp) emptyState of
            Fail e -> putStrLn e
            Done (State a b c )
                -> do printAll $ reverse a
                      putStrLn $ show $ unpackDone $
                            evalStateT (evalT exp) emptyState
                      case b of
                        [] -> putStrLn $ "Iterations: " ++ show c
                        _ -> do printAll $ reverse b
                                putStrLn $ "Iterations: " ++ show c
 
---- testing functions ----
runEvalT :: Term -> String
runEvalT exp = case runStateT (evalT exp) emptyState of
                Fail e -> e
                Done (r,State o e s) -> "Result = " ++ show r ++
                                        "; Iteration = " ++ show s ++
                                        "; Output = " ++ show o ++
                                        " - Exceptions = " ++ show e
                               
getEvalResult :: Term -> String
getEvalResult exp = case evalStateT (evalT exp) emptyState of
                      Fail e -> e
                      Done r -> "Result = "  ++ show r
 
 
getSideEffects :: Term -> String
getSideEffects exp = case execStateT (evalT exp) emptyState of
                      Fail e -> e
                      Done (State o e s) -> "Iterations = "  ++ show s ++                       
                                            " - Output = " ++ show o ++
                                            " - Exceptions = " ++ show e
 
{-
Some runs:
 
*MyStateT> eval (Add (Add (Add (Add (Con 40) (Con 2)) (Add (Con 12) (Con 30))) (Con 3)) (Con 10))
1 - eval (Con 40) <= 40
2 - eval (Con 2) <= 2
3 - eval (Add (Con 40) (Con 2)) <= 42
4 - eval (Con 12) <= 12
5 - eval (Con 30) <= 30
6 - eval (Add (Con 12) (Con 30)) <= 42
7 - eval (Add (Add (Con 40) (Con 2)) (Add (Con 12) (Con 30))) <= 84
8 - eval (Con 3) <= 3
9 - eval (Add (Add (Add (Con 40) (Con 2)) (Add (Con 12) (Con 30))) (Con 3)) <= 87
10 - eval (Con 10) <= 10
11 - eval (Add (Add (Add (Add (Con 40) (Con 2)) (Add (Con 12) (Con 30))) (Con 3)) (Con 10)) <= 97
97
Debug msg at Iteration 3: The Ultimate Answer Has Been Computed!! Now I'm tired!
Debug msg at Iteration 6: The Ultimate Answer Has Been Computed!! Now I'm tired!
Iterations: 11
*MyStateT>
 
*MyStateT> eval (Add (Add (Add (Add (Con 10) (Con 2)) (Add (Con 12) (Con 3))) (Add (Con 5) (Con 2))) (Con 2))
1 - eval (Con 10) <= 10
2 - eval (Con 2) <= 2
3 - eval (Add (Con 10) (Con 2)) <= 12
4 - eval (Con 12) <= 12
5 - eval (Con 3) <= 3
6 - eval (Add (Con 12) (Con 3)) <= 15
7 - eval (Add (Add (Con 10) (Con 2)) (Add (Con 12) (Con 3))) <= 27
8 - eval (Con 5) <= 5
9 - eval (Con 2) <= 2
10 - eval (Add (Con 5) (Con 2)) <= 7
11 - eval (Add (Add (Add (Con 10) (Con 2)) (Add (Con 12) (Con 3))) (Add (Con 5) (Con 2))) <= 34
12 - eval (Con 2) <= 2
13 - eval (Add (Add (Add (Add (Con 10) (Con 2)) (Add (Con 12) (Con 3))) (Add (Con 5) (Con 2))) (Con 2)) <= 36
36
Iterations: 13
*MyStateT>
 
*MyStateT> eval (Add (Con 5) (Con 6))
11.... I do not like this number!
*MyStateT>
-}
</haskell>


== Next?==
This probably makes sense for a functional programmer who is studying
We need a parser to get a string from input and turn into something of type Term!
Haskell for the first time. But, on the other side it will make the
newcomer, who's not a functional programmer, believe that explaining
monads in terms of "\" is just ugly, and definitely not Haskell.


Let's see if we'll time for it... Fist we must complete the text above!!
This is my problem with Haskell: it hides complexity with
constructions like "let...in", "where", "do".


==Suggested Readings==
Still I believe that if you want to use those constructions you must
know what they do. And they do ugly stuff like the one you'll be
looking at in this tutorial.


Cale Gibbard, [http://haskell.org/haskellwiki/Monads_as_Containers Monads as Containers]
I am not saying that this is ''the'' way to learn Haskell. I'm just
saying that this is the way I'm learning Haskell. And this is the way
this tutorial has been written.


Jeff Newbern, [http://www.nomaware.com/monads/html/index.html All About Monads]
===Starting from the inside or the outside?===


[http://haskell.org/haskellwiki/IO_inside IO Inside]
In This tutorial I show monads from their inside. In "Meet Bob" I do
the opposite. As I said I think (and I did it on purpose) that after
reading "Meet Bob" you can have a general idea of what a monad is, but
you need to go inside a monad to see how it works.


[http://sigfpe.blogspot.com/2006/08/you-could-have-invented-monads-and.html You Could Have Invented Monads! (And Maybe You Already Have.) by  sigfpe]
As a suggestion, I'd invite you to start with "Meet Bob", and then
procede with the tutorial.


I hope the these two approaches will give you an overall image of a
monad.


==Acknowledgments==
===Prerequisites===


Thanks to Neil Mitchell, Daniel Fisher, Bulat Ziganzhin, Brian Hulley
As I said, in order to understand this tutorial you need:
and Udo Stenzel for the invaluable help they gave, in the
* a basic understanding of functional programming (this is required to understand Haskell after all)
[http://www.haskell.org/mailman/listinfo/haskell-cafe haskell-cafe mailing list],
* a basic understanding of the type system:
in understanding this topic.
** how to create new types (with "data" and "newtype")
** how to use type constructors to construct and to match types and their internal components
** how to use a label field to retrieve a type internal component
* a basic understanding of lambda abstraction


I couldn't do it without their help.
I hope I'll be able to write a short introductory summary to these
topics (I will put it below this introductory remarks).


Obviously errors are totally mine. But this is a wiki so, please,
Have fun with Haskell!
correct them!


- [[User:AndreaRossato|AndreaRossato]]
- [[User:AndreaRossato|Andrea Rossato]]


[[Category:Tutorials]]
[[Category:Tutorials]]
[[Category:Idioms]]
[[Category:Idioms]]
[[Category:Monad]]

Latest revision as of 15:16, 6 February 2021

Note Since the size of the previous file was getting too big for a wiki, the tutorial has been divided into two parts: The Monadic Way Part I and The Monadic Way Part II. See below for some introductory remarks.

Contents

Meet Bob The Monadic Lover
A (supposed-to-be) funny and short introduction to Monads, with code but without any reference to category theory: what monads look like and what they are useful for, from the perspective of a ... lover. It could be an introduction to "The Monadic Way" tutorial.
The Monadic Way/Part I
In the first part of the tutorial we will start from a very simple evaluator that will be transformed into a monadic evaluator with an increasing number of features: output, exceptions, and state: a very simple counter for tracking the number of recursions of the evaluation precess.
The Monadic Way/Part II
In the second part of the tutorial we will see how to take complexity out of our monadic evaluator with the use of monadic transformers, and specifically StateT. This part is just a skeleton, since, for the time being, it contains only the code.


Preliminary remarks

When I started writing this tutorial I though that the only way to explain monads to a newcomer was to show them from the inside, with the use of lambda abstraction. Not only because this is the way Philip Wedler's paper adopts, but also because I believed, and still believe, that the only way to understand what bind (>>=) does is to explain it as a function that takes a monad and an anonymous function.

I had this feeling because I am a newcomer, and this is the way I came to understand monads.

I did not received very much feedback for this tutorial, and I must admit that I would like to. But one person, on the haskell-cafe mailing list, told me that: 

imho your tutorial makes the error that is a very typical: when you
write your tutorial you already know what are monads and what the
program you will construct at the end. but your reader don't know all these!
for such fresh reader this looks as you made some strange steps, write
some ugly code and he doesn't have chances to understand that this ugly
code is written just to show that this can be simplified by using monads.

I believe that Bulat is right. In this tutorial you go through some code that is really ugly and then, with some kind of magic, it turns out in the (redundant but) clean evaluator of the end of The Monadic Way/Part II.

I took that mail as a challenge and I responded by writing Meet Bob The Monadic Lover.

In "Meet Bob" the code is clean, variable names are very descriptive, and you see that I can create a monad without any use of lambda abstraction.

Bind (in "Meet Bob" is askLover) now takes a monad and a partial application, not an anonymous function.

Obviously you can see an anonymous function as a partial application.

The problem, I think, is that, in "Meet Bob", you cannot understand the strict relation between what I did before and what I do when I start using the "do-notation". You see that the same functions are being used ("tellMyself" and "newLove"), but "andrea <- tellMyself 1" is not explained. It's just magic.

The fact is that you cannot understand "andrea <- tellMyself 1" without the use of lambda abstraction.

I should have written an intermediate step, something like this: 

drunk = do newLove "Paula " >>
                       (tellLover 1 10) >>= \paula ->
                           (tellMyself 3) >>= \lorena -> tellMyself (paula + lorena)

With this approach I think you can understand why and how you come to write something like this: 

drunk = do newLove "Paula "
           paula <- (tellLover 1 10)
           lorena <- tellMyself 3
           tellMyself (paula + lorena)

That is to say, in this way you can see a do block as a series of nested anonymous functions whose arguments are bound to some value by the application of >>=. Anonymous functions that bind to some value the variables appearing after the "->" ("paula" and "lorena").

To summarize, I think that even if you can start using monads without understanding that what happens inside a "do" block is strictly related with lambda calculus, I don't think you can claim you understand monads just because you are using them.

And I'm quite sure that if a problem arises somewhere you can have a very difficult time in trying to find out what the problem is. This is even more true when you start doing monadic transformations.

How to explain monads to newcomers?

Monads are not an impossible-to-understand-unless-you-have-a-phd topic. I don't know if I can claim I'm a living proof of this assumption, since I have a PhD. But please take into account that I have a PhD in Comparative Private Law! Nothing related to computer science. And I claim I understand monads.

Moreover, since I have come to understand them, I think I can try to explain them to newcomers like me. This is why I started writing this tutorial.

Still, in order to understand them I studied, and I studied hard. I wanted to understand, it was difficult, and I kept studying. Going back to the foundation when needed.

So, I cannot explain monads to you unless you are willing to study. I can show you the way, but you must take your time and follow that way.

What do I need to know to understand monads?

Functional programming

First you need at least a basic understanding of functional programming.

This is going to be the easiest step, because there is an invaluable resource available on line for free.

This is what I did: I spent 20 hours of my life watching the Video Lectures by Hal Abelson and Gerald Jay Sussman on Structure and Interpretation of Computer Programs.

This course, and the annexed text book (I did not read it entirely), are an amazing introduction to computer science: you start by learning some basic Scheme and then Abelson and Sussman will bring you, step by step, to understand evaluation, interpretation and compilation of Lisp (Scheme) code.

The course is clear, interesting, funny and will provide you with a basic, but strong, understanding of functional programming.

Believe me: if you like programming but don't have a computer science curriculum, you'll find out that spending 20 hours of your life to watch that course has been the most productive investment of your learning life.

My problem with Haskell

I find Haskell an amazingly expressive programming language. It makes it incredibly easy to perform very difficult tasks, just like monads, for instance.

The problem is that, in doing so, it makes it difficult to understand Haskell to newcomers.

I must confess that tutorials and other learning material are quite dissatisfying on this regards too.

Since Haskell seems to make easy what is not easy, these tutorial take the "it's easy" approach.

I will give you an example. I think that the "Yet Another Haskell Tutorial" is a really good attempt to explain Haskell, and if you want to understand this tutorial you need to read it at least for getting a good understanding of the Haskell type system.

The tutorial explains monads and explains the "do notation" in terms of lambda abstraction (par. 9.1, p. 120). Still, to the lambda operator ("\") the tutorial dedicates only 17 lines in par. 4.4.1. Moreover "\" is explained just as another way of creating functions.

This probably makes sense for a functional programmer who is studying Haskell for the first time. But, on the other side it will make the newcomer, who's not a functional programmer, believe that explaining monads in terms of "\" is just ugly, and definitely not Haskell.

This is my problem with Haskell: it hides complexity with constructions like "let...in", "where", "do".

Still I believe that if you want to use those constructions you must know what they do. And they do ugly stuff like the one you'll be looking at in this tutorial.

I am not saying that this is the way to learn Haskell. I'm just saying that this is the way I'm learning Haskell. And this is the way this tutorial has been written.

Starting from the inside or the outside?

In This tutorial I show monads from their inside. In "Meet Bob" I do the opposite. As I said I think (and I did it on purpose) that after reading "Meet Bob" you can have a general idea of what a monad is, but you need to go inside a monad to see how it works.

As a suggestion, I'd invite you to start with "Meet Bob", and then procede with the tutorial.

I hope the these two approaches will give you an overall image of a monad.

Prerequisites

As I said, in order to understand this tutorial you need:

  • a basic understanding of functional programming (this is required to understand Haskell after all)
  • a basic understanding of the type system:
    • how to create new types (with "data" and "newtype")
    • how to use type constructors to construct and to match types and their internal components
    • how to use a label field to retrieve a type internal component
  • a basic understanding of lambda abstraction

I hope I'll be able to write a short introductory summary to these topics (I will put it below this introductory remarks).

Have fun with Haskell!

- Andrea Rossato

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