How to wrap char literal in IO monad in Haskell?

2
votes

I know you're supposed to wrap up the operations you want to perform on a result in a monad rather than unwrap things from the monad.

What I can't find are any idiot-friendly examples of how to do that.

For example, I want to do something like this:

myFunction = do
    c <- getChar
    if (c == 'q')
        then putStrLn "take action 1"
        else putStrLn "take action 2"

But you can't compare a char literal to an IO Char directly.

GHCi version is 8.4.4.

Error Message:

[1 of 2] Compiling Lib ( /Users/jamesstrieter/hask-tink/src/Lib.hs, interpreted )

/Users/jamesstrieter/hask-tink/src/Lib.hs:66:18: error: • Couldn't match expected type ‘IO char’ with actual type ‘Char’ • In the second argument of ‘(==)’, namely ‘'q'’ In the expression: x == 'q' In an equation for ‘what2do’: what2do x = x == 'q' • Relevant bindings include x :: IO char (bound at /Users/jamesstrieter/hask-tink/src/Lib.hs:66:9) what2do :: IO char -> Bool (bound at /Users/jamesstrieter/hask-tink/src/Lib.hs:66:1) | 66 | what2do x = x == 'q' | ^^^ Failed, no modules loaded.

3
haskellmonadsio-monaddo-notation
What is the error message? This code looks fine. – 4castle
you can't compare a char literal to an IO Char directly Apparently the code isn't doing that – n. 1.8e9-where's-my-share m.
Added error message to question – James Strieter
Your error message seems to be about completely different code than what you posted. Please have code exemplifying your actual problem in your question. – Cubic
That's a good point. The errant code was for a related but different issue, so I will create a separate question for that. My apologies I'm new at this. – James Strieter

3 Answers

8
votes

The code you posted looks perfectly correct and functional.

do-notation is a way of working with value in monad.

c <- getChar within the do block binds c to the char inside the IO Char you get with getChar. You can compare c == 'q' just fine here because c is a plain char, not an IO Char.

To answer you direct question, you can use the return function to put a pure value into any monad, including IO, so return 'q' "wraps" the character literal 'q' into a monad. This isn't what you want in this case, the code you already have is what you are looking for.

5
votes

But you can't compare a char literal to an IO Char directly.

Sure, but when you "bind" the result of the IO action it is no longer an IO Char but just a Char so that's why it works.

In more words:

Prelude> :t getChar
getChar :: IO Char
Prelude> c <- getChar
x
Prelude> :t c
c :: Char
2
votes

One of the most important things to understand about the IO monad is that the expression m >>= f does not execute the action m, nor does it every call the function f.

Instead, it just creates a new IO action that wraps both m and f and, when executed, finally executes m, extracts the return value, and calls f with the result.

That's it. Your entire Haskell program is nothing but a DSL for building a single IO action that gets assigned to main, which the Haskell runtime will execute for you.

So when you write

-- Rewritten slightly for brevity
myFunction = do
    c <- getChar
    putStrLn (if (c == 'q')
        then "take action 1"
        else "take action 2")

this is desugared to

myFunction = getChar >>= (\c -> putStrLn (if (c == 'q') then "take action 1" else "take action 2")

and what you are actually saying is "Build an IO action containing getChar and a function of type Char -> IO (), such that when this action is executed, it executes getChar and passes the resulting Char to the function to produce another IO action to be executed immediately."