I'm learning to use input and output in Haskell. I'm trying to generate a random number and output it to another file. The problem is that the random number seems to be returning an IO Int, something that I can't convert to a String using show.
Could someone give me a pointer here?
It's helpful if you show us the code you've written that isn't working.
Anyway, you are in a do block and have written something like this, yes?
main = do
...
writeFile "some-file.txt" (show generateRandomNumberSomehow)
...
You should instead do something like this:
main = do
...
randomNumber <- generateRandomNumberSomehow
writeFile "some-file.txt" (show randomNumber)
...
The <- operator binds the result of the IO Int value on the right to the Int-valued variable on the left. (Yes, you can also use this to bind the result of an IO String value to a String-valued variable, etc.)
This syntax is only valid inside a do block. It's important to note that the do block will itself result in an IO value --- you can't launder away the IO-ness.
dave4420's answer is what you want here. It uses the fact that IO is a Monad; that's why you can use the do notation.
However, I think it's worth mentioning that the concept of "applying a function to a value that's not 'open', but inside some wrapper" is actually more general than IO and more general than monads. It's what we have the Functor class for.
For any functor f (this could, for instance, be Maybe or [] or IO), when you have some value
wrapped :: f t (for instance wrapped :: Maybe Int), you can use fmap to apply a function
t -> t' to it (like show :: Int -> String) and get a
wrappedApplied :: f t' (like wrappedApplied :: Maybe String).
In your example, it would be
genRandomNumAsString :: IO String
genRandomNumAsString = fmap show genRandomNumPlain
fmap is really handy in IO.