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Generic number type

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1 Problem


Can I have a generic numeric data type in Haskell which covers
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
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 -
does not hold in general. Rationals are precise but
sqrt 2
are not rational. That is, when using
s you will encounter exactly the problems

that all scripting language users have encountered so far (or ignored :-).

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.

2 Idiomatic solutions

It is strongly advised to carefully check whether a
is indeed useful for your application.

So let's revisit some examples and their idiomatic solutions in plain Haskell 98.

2.1 average

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
and you can write
average :: Fractional a => [a] -> a
average xs = sum xs / genericLength xs

2.2 ratios

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
are interpreted as rationals itself. The second example fails, because
always returns an
number type, where
is not an instance. You should use
instead. This constructs a fraction out of two integers:
1 % 3 :: Rational
1 % floor pi :: Rational

2.3 isSquare

It may seem irksome that
is required in the function
isSquare :: (Integral a) => a -> Bool
isSquare n = (round . sqrt $ fromIntegral n) ^ 2 == n
With a
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
by calls to
, but that's no easier (and less type-safe) than just including the call to
in the first place. The point is that by using
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.

2.4 squareRoot

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
automatically in an implementation of

This will not work for several reasons:

  • For a square number,
    may give a result slightly below an integer, which
    will round down to the next integer.
  • fromIntegral
    will not preserve the (arbitrary high) precision of
    s and thus will not give precise results.
  • fromIntegral
    may exceed the maximum exponent of the floating point representation and fail with an overflow error or
That is,
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

3 See also