IO Char和[Char]之间的匹配类型错误

Question

I am trying to get input from user and print out depending on input (code modified from here ):

import Data.Char (toUpper)

isGreen = do
    putStrLn "Is green your favorite color?"
    inpStr <- getLine
    if (toUpper (head inpStr) == 'Y') then "YES" else "NO"
    -- following also does not work: 
    -- if (toUpper (head inpStr) == 'Y') then return "YES" else return "NO" 

main = do
    print isGreen

However, I am getting many errors:

testing.hs:7:44: error:
    • Couldn't match type ‘[]’ with ‘IO’
      Expected type: IO Char
        Actual type: [Char]
    • In the expression: "YES"
      In a stmt of a 'do' block:
        if (toUpper (head inpStr) == 'Y') then "YES" else "NO"
      In the expression:
        do { putStrLn "Is green your favorite color?";
             inpStr <- getLine;
             if (toUpper (head inpStr) == 'Y') then "YES" else "NO" }

testing.hs:7:55: error:
    • Couldn't match type ‘[]’ with ‘IO’
      Expected type: IO Char
        Actual type: [Char]
    • In the expression: "NO"
      In a stmt of a 'do' block:
        if (toUpper (head inpStr) == 'Y') then "YES" else "NO"
      In the expression:
        do { putStrLn "Is green your favorite color?";
             inpStr <- getLine;
             if (toUpper (head inpStr) == 'Y') then "YES" else "NO" }

testing.hs:12:5: error:
    • No instance for (Show (IO Char)) arising from a use of ‘print’
    • In a stmt of a 'do' block: print isGreen
      In the expression: do { print isGreen }
      In an equation for ‘main’: main = do { print isGreen }

Where is the problem and how can it be solved?

Answer 1

Your problem is, essentially, confusing IO code with non-IO, or "pure" code.

In isGreen , you have a do block involving putStrLn and getLine . This means the function must return a value of some type IO a . You can't get a "plain string" from it, which seems to be your intention. The final line of the do block must be a monadic value, in this case an IO value - so you can't simply have a string here.

However, you can use the return function to make an IO String out of a String , as in the commented out code that "also does not work":

if (toUpper (head inpStr) == 'Y') then return "YES" else return "NO" 

This works fine, and is necessary, as far as isGreen is concerned. The error comes because of what happens in main . You cannot print an IO value - an IO String is not actually a string, it's basically a "promise" of a future string, that will only materialise at runtime based on (in this case) the user's input.

But since main is an IO action anyway, this isn't a problem. Just do this instead:

main = do
    isItGreen <- isGreen
    print isItGreen

or, completely equivalently, but imo more succinctly,

main = isGreen >>= print

You might prefer putStrLn to print , which when printing a string will include the enclosing quotes.

Note that, if you had included a type signature for each top level value (here isGreen and main ), you'd have got more useful information from the compiler as to where the errors were.

Answer 2

isGreen uses IO , so the type needs to be IO String , not String . It is an IO action that can produce a string, not a string itself.

isGreen :: IO String
isGreen = do
    putStrLn "Is green your favorite color?"
    inpStr <- getLine
    return $ if (toUpper (head inpStr) == 'Y') then "YES" else "NO"

Likewise, print expects a String , not an IO String , so you need to use the IO monad instance. (Actually, you should use putStrLn , since you don't need the implicit show that print will use.)

main = isGreen >>= putStrLn

Here, >>= is creating a new IO action which will execute the IO action on the left ( isGreen ), and pass the value produced to the function on the right ( putStrLn ).

Answer 3

Since isGreen here is an IO a , you can not use if … then … else … as last line. You need to return an IO a .

You can however use pure here (or return , although there is nothing inherently wrong with that, it ):

import Data.Char(toUpper)

isGreen :: IO String
isGreen = do
    putStrLn "Is green your favorite color?"
    inpStr <- getLine
     (if "Y" == take 1 (map toUpper inpStr) then "YES" else "NO")

Here "Y" == take 1 (map toUpper inpStr) is more safe, since if the user has written an empty line, then head will error on the empty list.

Since isGreen has type IO String , then you can not use print isGreen , you here can use bind (>>=) :

main :: IO ()
main = isGreen  print