Haskell - More efficient way to complete algorithm?

Question

I have created an algorithm which takes a non-negative Int value, representing a total number of minutes, and returns a triple that gives (days, hours, minutes) that this corresponds to.

Here is my code:

calcdays :: Int -> Int
calcdays x = x `div` (24*60)

calchours :: Int -> Int
calchours x = (x - ((calcdays x)*24*60)) `div` 60

calcmins :: Int -> Int
calcmins x = (x - ((calcdays x)*24*60) - ((calchours x)*60))

dayshoursmins :: Int -> (Int,Int,Int)
dayshoursmins x = (calcdays x, calchours x, calcmins x)

Using only basic Haskell operations (guards, divs, mods etc.), is there a more simple way of programming this function?

EDIT:

I have used a suggestion below to make this code simpler, while not as simple as the qoutRem solution, I thought I might post it:

calcdays :: Int -> Int
calcdays x = x `div` (24*60)

calchours :: Int -> Int
calchours x = (x `mod` (24*60)) `div` 60

calcmins :: Int -> Int
calcmins x = (x `mod` (24*60)) `mod` 60

dayshoursmins :: Int -> (Int,Int,Int)
dayshoursmins x = (calcdays x, calchours x, calcmins x)

I don't speak Haskell but the algorithm normally goes like this: take the input, divide it by 60, the remainder is the minute part, the result is then divided by 24, the remainder is the hours part, the result is the days part. So you'd need to chain the calculations rather than execute them in isolation. — biziclop
– biziclop, Commented Nov 18, 2015 at 13:21
Yes, however, in Haskell, you cannot set variables, hence I am unsure how to achieve this. — James Powell
– James Powell, Commented Nov 18, 2015 at 13:23
You don't need to set variables, just feed the result into the next step straight away. I'm unsure about the syntax though. — biziclop
– biziclop, Commented Nov 18, 2015 at 13:25

Luka Rahne · Accepted Answer · 2015-11-18 13:30:31Z

13

I think somethink like this

dayshoursmins x = (d,hr,mr) where
 (h,mr) = quotRem x 60
 (d,hr) = quotRem h 24

answered Nov 18, 2015 at 13:30

Luka Rahne

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3 Comments

Daniel Wagner Over a year ago

Big +1 from me -- this is the right way. I was even going to make the usual quibble about quotRem vs divMod, but I think in this case quotRem actually does what I expect on negative numbers!

James Powell Over a year ago

As a begginer in Haskell, I was not aware of quotRem. This is great, much simpler, thanks.

Luka Rahne Over a year ago

I used hoogle to find div. Similar functions are usually listed together and prelude lists quotRem in same section.

Karoly Horvath · Accepted Answer · 2015-11-18 13:25:56Z

3

You can use

a `mod` b

to directly get the remainder.

And you can use let or where to calculate a sub-expression.

answered Nov 18, 2015 at 13:25

Karoly Horvath

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