Integer vs Int trouble in Haskell function

Question

I have the following function that will recursively multiply the digits of a number until it reaches a single digit number

multResWithCounter' :: (Int, [Int],[Integer]) -> (Int, [Int],[Integer])
multResWithCounter' (counter, listOfDigits,listOfProducts)
   | (length listOfDigits) <= 1 = (counter, listOfDigits, listOfProducts)
   | otherwise = multResWithCounter' ((counter + 1), newListOfDigits, digitProduct:listOfProducts)
   where newListOfDigits = map (\n -> read [n] :: Int) (show $ digitProduct)
         digitProduct = ((product listOfDigits) :: Integer)

Which of course doesn't compile in ghci, complaining that it couldn't match expected type Integer with actual type Int. If I back down and change the type declaration to Int and call it, this way for example,

multResWithCounter' (0,[2,7,7,7,7,7,7,8,8,8,8,8,8,9,9],[])

I get

(2,[0],[0,1191706624])

Now this "1191706624" is of course the result of

(product [2,7,7,7,7,7,7,8,8,8,8,8,8,9,9]) `mod` (maxBound :: Int)

and sure enough the result of the first part when run directly in ghci is the correct 4996238671872.

So, how can I have Haskell give me Integer product results (that I will turn to strings right away with show anyway)?

Thanks

Daniel Wagner · Accepted Answer · 2021-03-15 01:14:53Z

5

You may use

fromIntegral :: Int -> Integer

to promote an Int to an Integer. For similar future questions, you can use Hoogle to search for functions by type, which would have gotten you a closely related answer:

toInteger :: Int -> Integer

answered Mar 15, 2021 at 1:14

Daniel Wagner

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