# Generic number type

### From HaskellWiki

## Contents |

## 1 Problem

Question:

Can I have a generic numeric data type in Haskell which coversAnswer: In principle you can define a type like

data GenericNumber = Integer Integer | Rational Rational | Double Double

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

A## 2 Idiomatic solutions

It is strongly advised to carefully check whether aSo 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

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 literals1 % 3 :: Rational 1 % floor pi :: Rational

### 2.3 isSquare

It may seem irksome thatisSquare :: (Integral a) => a -> Bool isSquare n = (round . sqrt $ fromIntegral n) ^ 2 == n

isSquare :: GenericNumber -> Bool isSquare n = (round . sqrt $ n) ^ 2 == n

### 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)

This will not work for several reasons:

- For a square number, may give a result slightly below an integer, whichsqrtwill round down to the next integer.floor
- will not preserve the (arbitrary high) precision offromIntegrals and thus will not give precise results.Integer
- may exceed the maximum exponent of the floating point representation and fail with an overflow error orfromIntegralresult.Infinity

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

- The discussion on haskell-cafe which provided the impetus for this page: http://www.haskell.org/pipermail/haskell-cafe/2007-June/027092.html