[Haskell-cafe] IO is not a monad

[Brian Hulley]

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.
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