Haskell: Divide Integer by Integer and get Double

Question

I want to write a Haskell function which takes two Integers divedes them and outputs a Double.

The signature should look like this:

divide :: Integer -> Integer -> Double

The function I would like to have is:

divide x y = x / y

The error message I get wit this function is:

Couldn't match expected type ‘Double’ with actual type ‘Integer’

How to get a correct double result for this function?

what type does (/) have?

user1198582
– user1198582

2020-03-19 17:39:02 +00:00
Commented Mar 19, 2020 at 17:39 — user1198582
– user1198582, Commented Mar 19, 2020 at 17:39
fromInteger is your friend.

Redu
– Redu

2020-03-19 17:39:03 +00:00
Commented Mar 19, 2020 at 17:39 — Redu
– Redu, Commented Mar 19, 2020 at 17:39

willeM_ Van Onsem · Accepted Answer · 2020-03-19 18:40:30Z

11

Haskell is a strongly typed language. That means that no implicit conversions re done. You can first convert the two numbers to a Double, for example with fromIntegral :: (Integral a, Num b) => a -> b, and then use (/) :: Fractional a => a -> a -> a:

divide :: Integer -> Integer -> Double
divide x y = fromIntegral x / fromIntegral y

Convertin a number to a Double can result in loss of precision however.

It might be better to return a Ratio, and thus use fractions, for example with the (%) :: Integral i => i -> i -> Ratio i, so then divide is just divide = (%).

You can, like @DanielWagner says use fromRational to covert the Rational to any Fractional type:

import Data.Ratio((%))

divide :: Fractional a => Integer -> Integer -> a
divide x y = fromRational (x % y)

So then you can still convert it to a Double:

Prelude Data.Ratio> divide 5 2 :: Double
2.5

edited Mar 19, 2020 at 18:40

answered Mar 19, 2020 at 17:42

willeM_ Van Onsem

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2 Comments

Daniel Wagner Over a year ago

You can get the good behavior of % (namely, dealing correctly with numbers too large to fit in a Double but whose ratio is small enough or vice versa) and the "good" type of divide by combining the two: divide' :: Integer -> Integer -> Double; divide' x y = fromRational (x % y).

eugenkaltenegger Over a year ago

Thank you for your answer! The signature is given from an example I need to do for university.

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