* Victor S. Miller <victorsmil...@gmail.com> [2012-01-21 12:29:32-0500]
> The "do" notation translates
>
> do {x <- a;f} into
>
> a>>=(\x -> f)
>
> However when we're working in the IO monad the semantics we want
> requires that the lambda expression be strict in its argument.
I'm not aware of any semantics that would require that.
According to a monad law,
return x >>= f
should be equivalent to (f x). In particular,
return x >>= const (return ())
is equivalent to (const (return ()) x) or simply (return ()).
So, const is non-strict in its second argument even when used in (>>=).
--
Roman I. Cheplyaka :: http://ro-che.info/
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