Yitzchak Gale wrote:
I wrote:
Prelude> let f .! g = ((.) $! f) $! g
Prelude> let f = undefined :: Int -> IO Int
Prelude> f `seq` 42
*** Exception: Prelude.undefined
Prelude> ((>>= f) . return) `seq` 42
42
Prelude> ((>>= f) .! return) `seq` 42
42
Duncan Coutts wrote:
Perhaps I'm missing something but I don't see what's wrong.
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 because the lhs is of the form \x -> _|_ whereas the rhs, which
should really be of the form \x -> _|_, is actually _|_ already(!) so the
_|_ has effectively been allowed to jump to the left of the arrow hence the
"inequality" detected by seq.
Perhaps a solution would be to force _|_ to respect the shape of the type,
thus non-terminating values of type a -> b would be given the value \_ ->
_|_ instead of _|_ ?
Regards, Brian.
--
http://www.metamilk.com
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