Divide Int to Int and return Int

Question

I need a function which gets two Ints (a and b) and returns A/B as Int. I am sure that A/B will always be an integer.

Here is my solution:

myDiv :: Int -> Int -> Int
myDiv a b = 
      let x = fromIntegral a
          y = fromIntegral b
      in truncate (x / y)

But want to find more simpler solution. Something like this:

myDiv :: Int -> Int -> Int
myDiv a b = a / b

How can I divide Int to Int and get Int ?

Solution 1

Why not just use quot?

quot a b

is the integer quotient of integers a and b truncated towards zero.

Solution 2

Here's what I did to make my own:

quot' a b
         | a<b = 0  -- base case
         | otherwise = 1 + quot' a-b b

Comments

[Admin]

Or a `quot` b for the infix lovers (wow, you can actually escape backticks insice backticks with backslash?).

[Antal Spector-Zabusky]

Also a `div` b; if I remember correctly, quot truncates (like demas wanted), and div rounds towards zero. So (-3) `quot` 4 == 0, and (-3) `div` 4 == -1.

[luqui]

+1 for div. More mathematically well-behaved, quot is a bitch when there are negative numbers around.

[MuffinTheMan]

This was a life-saver for me. I was wrestling with fromIntegral (ceiling (int1 / int2)) and other things--none of which gave me back an Int, but this one did.

[Lacuno]

div rounds towards negative inifinity, not zero.

[Ruud Helderman]

Nice as an exercise, but useless in production. On large numbers it is slow (linear time) and has a high memory footprint (not tail-recursive). For negative numbers, it is either wrong (a<0) or never ends (b<0).

[mLstudent33]

in ghci: <interactive>:12:1: error: • Non type-variable argument in the constraint: Fractional (a -> a) (Use FlexibleContexts to permit this) • When checking the inferred type it :: forall a. (Fractional (a -> a), Integral a) => a -> a

[Pointy]

@mLstudent33 I hope it's clear that you're responding to an 11 year old answer