[Haskell-cafe] Re: evaluate vs seq
viritrilbia at gmail.com
Thu Sep 14 22:24:23 EDT 2006
On 9/14/06, apfelmus at quantentunnel.de <apfelmus at quantentunnel.de> wrote:
> With this in mind the equations
> 1) return _|_ == Return _|_
> 2) _|_ `seq` (return _|_) == _|_
> can be interpreted:
> 1) when reducing a return-redex (i.e. evaluating it), we get weak-head
> normal form without evaluating the argument (which was _|_)
> 2) when reducing the `seq`-thing but not further evaluating the
> arguments (_|_ in our case), we get nowhere (i.e. only _|_)
> Thus, when evaluating the expressions (one step!, i.e. just to weak head
> normal form), number 1) returns immediately but in 2), the `seq` needs
> to evaluate its first argument. So the semantics of applying _|_ to a
> function determines its behavior with respect to how much of its
> arguments will be evaluated! I think this is the point our
> misunderstanding is about?
> For (evaluate :: a->IO a), I said something like
> 3) evaluate _|_ == ThrowException
> and meant, that when evaluating 3) one step to weak head normal form,
> the argument does not get evaluated. 2) is much different to that.
> Equation 3) is of course wrong and must be more like
> evaluate x == Evaluate x
> where (Evaluate x) is a constructor but with a particular behavior when
> executed in the IO-Monad.
Great, thank you! I think this finally answers my question.
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