On Sat, Feb 23, 2008 at 1:05 AM, <[EMAIL PROTECTED]> wrote:
> Adding control effects (shift/reset) changes the expressivity
> results. Now all three calculi are distinct and none subsumes the
> other. For example, the expression
> reset( (\x -> 1) (abort 2))
> evaluates to 1 in call-by-name and evaluate to 2 in call-by-value.
> The expression
> reset ((\x -> x + x) (shift f f))
> has the type int->int in call-by-need (it is a function \x -> x + x)
> and it has the type int->int->int in call-by-name (and it is the
> curried addition function).
Aha. Okay, so shift/reset exposes evaluation order, amongst other
things. Hm... thank you very much!
--
Taral <[EMAIL PROTECTED]>
"Please let me know if there's any further trouble I can give you."
-- Unknown
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