I have a working program
main = do
inpStr <- getLine
putStrLn ( "Hello " ++ inpStr )
where
putStrLn :: String -> IO ()
and
getLine :: IO String
From this can I conclude that the type of <- is
IO a -> a
?
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1 Answers
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Unfortunately, it is not a regular function, but a language construct. You can use it only in do blocks to "extract" value from some context, such as IO.
do blocks and <- is just a syntax sugar for such things called Monads (and IO is one of them).
There are some other examples of such contexts, which you can use with do and <-, such as lists, optional, or nullable, values (such as Maybe Int), stateful computations and so on.
Some links:
- What is a monad?
- http://learnyouahaskell.com/a-fistful-of-monads#do-notation (you need to understand functors and monads here)
Yuri Kovalenko
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2To add to @Yuri's answer, `x <- foo` gets translated to `foo >>= \x -> ...`. As simple as that. – arrowd Nov 14 '19 at 07:21
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@arrowd, I thought that that's too into monads, so I told it is just a syntax sugar, and the second link explains the desugaring to `>>=` – Yuri Kovalenko Nov 14 '19 at 07:25
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@arrowd "... and `(>>=)` does in fact have a type, which is ..." etc. – Will Ness Nov 14 '19 at 08:26