Brian Hulley wrote:
Yitzchak Gale wrote:
I wrote:
Prelude> let f = undefined :: Int -> IO Int
Prelude> f `seq` 42
*** Exception: Prelude.undefined
Prelude> ((>>= f) . return) `seq` 42
42
The monad laws say that (>>= f) . return must be
identical to f.
I thought it was:
return x >>= f = f x
so here the lhs is saturated, so will hit _|_ when the action is
executed just as the rhs will.
I think the problem you're encountering is just that the above law
doesn't imply:
(>>= f) . return = f
in Haskell
ok so far...
because the lhs is of the form \x -> _|_
No I got this wrong. Evaluating the lhs to WHNF doesn't hit the _|_.
(Incidentally the version using .! evaluates to exactly the same thing since
(>>= f) `seq` ((>>= f) . return) = ((>>= f) . return) since (\x -> x >>= f)
is already in WHNF afaiu)
Brian.
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