[Haskell-cafe] IO is not a monad
Yitzchak Gale
gale at sefer.org
Tue Jan 23 17:33:09 EST 2007
Hi Brian,
Brian Hulley wrote:
> I thought it was:
> return x >>= f = f x
> ...I think the problem you're encountering is just
> that the above law doesn't imply:
> (>>= f) . return = f
Sorry, I was not clear.
For the purposes of this thread, I am using the
word "monad" in the category-theoretic sense.
The monad axioms are stated in terms of morphisms
and composition in the category. If we then define -
in the category - for each morphism f, a morphism
(>>= f) = join . fmap f
then we can derive the identities
(>>= return) = id
(>>= f) . return = f
(>>= g) . (>>= f) = (>>= (>>= g) . f)
directly from the monad axioms. Assume, for now,
that morphisms in the category will exactly correspond
to functions in Haskell. Then the above identities will be the
conditions for a monad in Haskell.
The conditions for a monad in Haskell are usually
stated in the non-points-free form, presumably for
clarity. But as you pointed out, that is _not_ equivalent
to the above when we are being careful about strictness.
My challenge is:
1. Find a way to model strictness/laziness properties
of Haskell functions in a category in a way that is
reasonably rich.
2. Map monads in that category to Haskell, and
see what we get.
3. Compare that to the traditional concept of
a monad in Haskell.
Is this possible? Any more ideas how to proceed?
I told you this was a troll. :)
Thanks,
Yitz
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