CODEWITHSUNDEEP
haskelloperatorsbindmonadsoperator-precedence
Ragoniis
Ragoniis

Reputation: 29

Understanding associativity of the Monad's bind operator in Haskell

I was trying to understanding how the Monad's bind operator works, but found an example which was odd, because the apparent associativity didn't make sense to me considering the fact that >>= is left-associative. Here is the example, testing at an interpreter's prompt:

> Just 3 >>= \x -> Just "!" >>= \y -> Just (show x ++ y)
Just "3!"

> Just 3 >>= (\x -> Just "!" >>= (\y -> Just (show x ++ y)))
Just "3!"

> (Just 3 >>= \x -> Just "!" )>>= \y -> Just (show x ++ y)
<interactive>:3:50: error: Variable not in scope: x :: ()

I don't understand it because the second example runs in opposition to the third, because it seems that it goes in the way which is contradictory to the known associativity. I know that I am missing something, but I don't know what.

Upvotes: 2

Views: 182

Answers (2)

Will Ness
Will Ness

Reputation: 71075

According to the maximal munch rule, lambdas are parsed as far to the right as possible, so the left associativity of the >>= operator doesn't have a chance to come into play. It is your second snippet that the first is parsed as, not the third (which of course is an invalid code).

Upvotes: 6

xilpex
xilpex

Reputation: 3237

This is because the parenthesis take x out of scope:

(Just 3 >>= \x -> Just "!" ) >>= \y -> Just (show x ++ y)

(Just 3 >>= \x -> Just "!" ) will become Just "!", and x will go out out of scope.

Upvotes: 1

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