Generic number type
Can I have a generic numeric data type in Haskell which covers
Double and so on, like it is done in scripting languages like Perl and MatLab?
Answer: In principle you can define a type like
data GenericNumber = Integer Integer | Rational Rational | Double Double
and define appropriate instances for
Num class et. al.
However you will find that it is difficult to implement these methods in a way that is appropriate for each use case.
There is simply no type that can emulate the others.
Floating point numbers are imprecise -
a/b*b==a does not hold in general.
Rationals are precise but
sqrt 2 are not rational.
That is, when using
GenericNumbers you will encounter exactly the problems
that all scripting language users have encountered so far (or ignored :-).
GenericNumber type would also negate the type safety that strongly typed numbers provide, putting the burden back on the programmer to make sure they are using numbers in a type-safe way. This can lead to subtle and hard-to-find bugs, for example, if some code ends up comparing two floating-point values for equality (usually a bad idea) without the programmer realizing it.
It is strongly advised to carefully check whether a
GenericNumber is indeed useful for your application.
So let's revisit some examples and their idiomatic solutions in plain Haskell 98.
You may find it cumbersome to manually convert integers to fractional number types like in
average :: Fractional a => [a] -> a average xs = sum xs / fromIntegral (length xs)
and you may prefer
average :: [GenericNumber] -> GenericNumber average xs = sum xs / genericNumberLength xs
with an appropriate implementation of
However, there is already
Data.List.genericLength and you can write
average :: Fractional a => [a] -> a average xs = sum xs / genericLength xs
You find it easy to write
1 / 3 :: Rational
but uncomfortable that
1 / floor pi :: Rational
does not work.
The first example works, because the numeric literals
3 are interpreted as rationals itself.
The second example fails, because
floor always returns an
Integral number type, where
Rational is not an instance.
You should use
% instead. This constructs a fraction out of two integers:
1 % 3 :: Rational 1 % floor pi :: Rational
It may seem irksome that
fromIntegral is required in the function
isSquare :: (Integral a) => a -> Bool isSquare n = (round . sqrt $ fromIntegral n) ^ 2 == n
GenericNumber type, one could instead write
isSquare :: GenericNumber -> Bool isSquare n = (round . sqrt $ n) ^ 2 == n
but there is a subtle problem here: if the input happens to be represented internally by a non-integral type, this function will probably not work properly. This could be fixed by wrapping all occurrences of
n by calls to
round, but that's no easier (and less type-safe) than just including the call to
fromIntegral in the first place. The point is that by using
GenericNumber here, all opportunities for the type checker to warn you of problems is lost; now you, the programmer, must ensure that the underlying numeric types are always used correctly, which is made even harder by the fact that you can't inspect them.
Closely related is the (floor of the) square root of integers. It is tempting to implement
squareRoot :: Integer -> Integer squareRoot = floor . sqrt . (fromIntegral :: Integer -> Double)
or to convert to
Double automatically in an implementation of
This will not work for several reasons:
- For a square number,
sqrtmay give a result slightly below an integer, which
floorwill round down to the next integer.
fromIntegralwill not preserve the (arbitrary high) precision of
Integers and thus will not give precise results.
fromIntegralmay exceed the maximum exponent of the floating point representation and fail with an overflow error or
fromIntegral is of no help here.
The most efficient way is to call the native implementation of the square root of GNU's multiprecision library.
(How to do that?)
The most portable way is to implement a square root algorithm from scratch.
(^!) :: Num a => a -> Int -> a (^!) x n = x^n squareRoot :: Integer -> Integer squareRoot 0 = 0 squareRoot 1 = 1 squareRoot n = let twopows = iterate (^!2) 2 (lowerRoot, lowerN) = last $ takeWhile ((n>=) . snd) $ zip (1:twopows) twopows newtonStep x = div (x + div n x) 2 iters = iterate newtonStep (squareRoot (div n lowerN) * lowerRoot) isRoot r = r^!2 <= n && n < (r+1)^!2 in head $ dropWhile (not . isRoot) iters
- Converting numbers
- The discussion on haskell-cafe which provided the impetus for this page: http://www.haskell.org/pipermail/haskell-cafe/2007-June/027092.html