Following a haskell tutorial, the author provides the following implementation of the withFile method:
withFile' :: FilePath -> IOMode -> (Handle -> IO a) -> IO a
withFile' path mode f = do
handle <- openFile path mode
result <- f handle
hClose handle
return result
But why do we need to wrap the result in a return? Doesn't the supplied function f already return an IO as can be seen by it's type Handle -> IO a?
You're right: f already returns an IO, so if the function were written like this:
withFile' path mode f = do
handle <- openFile path mode
f handle
there would be no need for a return. The problem is hClose handle comes in between, so we have to store the result first:
result <- f handle
and doing <- gets rid of the IO. So return puts it back.
This is one of the tricky little things that confused me when I first tried Haskell. You're misunderstanding the meaning of the <- construct in do-notation. result <- f handle doesn't mean "assign the value of f handle to result"; it means "bind result to a value 'extracted' from the monadic value of f handle" (where the 'extraction' happens in some way that's defined by the particular Monad instance that you're using, in this case the IO monad).
I.e., for some Monad typeclass m, the <- statement takes an expression of type m a in the right hand side and a variable of type a on the left hand side, and binds the variable to a value. Thus in your particular example, with result <- f handle, we have the types f result :: IO a, result :: a and return result :: IO a.
PS do-notation has also a special form of let (without the in keyword in this case!) that works as a pure counterpart to <-. So you could rewrite your example as:
withFile' :: FilePath -> IOMode -> (Handle -> IO a) -> IO a
withFile' path mode f = do
handle <- openFile path mode
let result = f handle
hClose handle
result
In this case, because the let is a straightforward assignment, the type of result is IO a.
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