# Generic number type

## Problem

Question:
Can I have a generic numeric data type in Haskell which covers `Integer`

, `Rational`

, `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 `pi`

and `sqrt 2`

are not rational.
That is, when using `GenericNumber`

s you will encounter exactly the problems
that all scripting language users have encountered so far (or ignored :-).

A `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.

## Idiomatic solutions

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.

### 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 `genericNumberLength`

.
However, there is already `Data.List.genericLength`

and you can write

```
average :: Fractional a => [a] -> a
average xs = sum xs / genericLength xs
```

### 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 `1`

and `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
```

### isSquare

It may seem irksome that `fromIntegral`

is required in the function

```
isSquare :: (Integral a) => a -> Bool
isSquare n = (round . sqrt $ fromIntegral n) ^ 2 == n
```

With a `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.

### 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 `Double`

automatically in an implementation of `sqrt`

for `GenericNumber`

.
This will not work for several reasons:

- For a square number,
`sqrt`

may give a result slightly below an integer, which`floor`

will round down to the next integer. `fromIntegral`

will not preserve the (arbitrary high) precision of`Integer`

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`Infinity`

result.

That is, `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
```

## See also

- 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