Haskell: using function value

Question

I am writing my first Haskell program and I am having difficulties in making a function use a user-entered input value. I don't yet understand enough about the language, so I started off simply, by creating a function that reads in a user-input value and returns this as an integer. Then, in the main calling function I want to use this value as the input to another function. This is the code scenario:

module Main where

square :: Int -> Int
square n = n*n  

getInt :: IO Integer
getInt = do
    putStrLn "Enter a positive integer: "
    s <- getLine
    putStrLn("The number you entered is " ++ s)
    let num = read s :: Integer
    return num

main :: IO ()
main = do 
    num <- getInt
    print num -- works
    print $ square (getInt) -- breaks down here
    print $ square (num) -- also does not work
    print $ square (5) -- works

Bartek Banachewicz · Accepted Answer · 2015-10-15 11:46:03Z

6

Mismatch between `Int` and `Integer`

Let's look at the signatures here:

square :: Int -> Int
getInt :: IO Integer

It's clear that Int is not an Integer, so choose either, or use Num n => n:

square :: Num n => n -> n
getInt :: Num n => IO n

Mismatch between `Int` and `IO Int`

The second problem is a bit more complicated. Since getInt is an IO action, it needs to be extracted first to be used:

num <- getInt
print $ square num

There are other ways to compose a pure function with an IO action:

-- Control.Applicative (<$>)
square <$> getInt >>= print

-- Data.Functor (fmap)
fmap square getInt >>= print

-- directly composed with return
getInt >>= return . square >>= print

-- manual lambda extraction
getInt >>= (\x -> return $ square x) >>= print

edited Oct 15, 2015 at 11:46

answered Oct 15, 2015 at 9:39

Bartek Banachewicz

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3 Comments

chi Over a year ago

Actually, none of the last 4 examples type check.

Bartek Banachewicz Over a year ago

@chi I wrote that a bit too hastily. Fixed now.

jakubdaniel Over a year ago

Also the initial two signatures are not displayed as code (missing indentation). I would write an edit, but spaces do not count as characters apparently (so my edit won't go through).

C. Quilley · Accepted Answer · 2015-10-15 09:38:49Z

There are two things going on here.

Firstly, the reason you cannot do print $ square getInt is because getInt has the type IO Integer - you should extract the value via num <- getInt first, before running functions that can take and receive Integers.

The second is that you are actually using two different integer types in your program, and Haskell does not automatically coerce between them. Notice that getInt :: IO Integer whilst square :: Int -> Int. If you were to change the types to be consistent, your line print $ square num would work. The square function expects an Int, whilst you are giving it num, which is an Integer. The difference between these two types is that Int is bounded, as a machine integer, whilst Integer is unbounded.

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