On 9/14/06, [EMAIL PROTECTED] <[EMAIL PROTECTED]> 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. Mike _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
