CODEWITHSUNDEEP
haskellmonadsassociativity
user1065942
user1065942

Reputation:

Monad's associativity rule in haskell

(m >>= f) >>= g = m >>= (\x -> f x >>= g)

what's different from f and \x->f x ??

I think they're the same type a -> m b. but it seems that the second >>= at right side of equation treats the type of \x->f x as m b. what's going wrong?

Upvotes: 10

Views: 567

Answers (1)

C. A. McCann
C. A. McCann

Reputation: 77384

The expressions f and \x -> f x do, for most purposes, mean the same thing. However, the scope of a lambda expression extends as far to the right as possible, i.e. m >>= (\x -> (f x >>= g)).

If the types are m :: m a, f :: a -> m b, and g :: b -> m c, then on the left we have (m >>= f) :: m b, and on the right we have (\x -> f x >>= g) :: a -> m c.

So, the difference between the two expressions is just which order the (>>=) operations are performed, much like the expressions 1 + (2 + 3) and (1 + 2) + 3 differ only in the order in which the additions are performed.

The monad laws require that, like addition, the answer should be the same for both.

Upvotes: 15

Related Questions