`bind` Equivalent to join (fmap f m)?

Question

This excellent answer in this question demonstrates how bind can be written in terms of join and fmap:

(>>=) :: m v -> (v -> m w) -> m w

says "if you have a strategy to produce a v, and for each v a follow-on strategy to produce a w, then you have a strategy to produce a w". How can we capture that in terms of join?

mv >>= v2mw = join (fmap v2mw mv)

But, I don't understand how v2mw, which has a type of a -> m b type checks to the first argument of fmap.

fmap :: Functor f => (a -> b) -> f a -> f b

Let `b` be `m b` and `fmap v2mw :: f a -> f (m b)`. Then `join` forces `f` to equal `m` and collapses the layers. — J. Abrahamson, Aug 29 '14 at 23:39
Yep! Exactly. The two `b`s arise in different contexts and are not required to be the same. — J. Abrahamson, Aug 30 '14 at 01:22
`The two bs arise in different contexts ` you're talking about the `b`'s in `fmap`'s signature: `fmap :: Functor f => (a -> b) -> f a -> f b`? Surely the `b`'s must be the same type, no? — Kevin Meredith, Aug 30 '14 at 02:29
Sorry, no, I meant the `b` in `fmap`'s signature versus the `b` in the `m b` in `v2mw`'s/`(>>=)`'s signature — J. Abrahamson, Aug 30 '14 at 03:33

score 10 · Accepted Answer · answered Aug 29 '14 at 23:35

10

Let's say v2mw :: c -> m d, just so things aren't ambiguous, and

fmap :: Functor f => (a -> b) -> f a -> f b

Then fmap v2mw works out so that f ~ m, a ~ c and b ~ m d, so

fmap v2mw :: m c -> m (m d)

and join :: m (m e) -> m e, so join (fmap v2mw mv) has type m d as expected.

answered Aug 29 '14 at 23:35

Louis Wasserman

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"equals," or "is the same as." – Louis Wasserman Aug 29 '14 at 23:49

`bind` Equivalent to join (fmap f m)?

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