On 9/13/06, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
So `seq` forces its first argument. When we define f x = x `seq` (Return x) we thereby get f _|_ == _|_ f [] == Return [] f (x:xs) == Return (x:xs) To compare, the semantics of (evaluate) is evaluate _|_ == ThrowException =/= _|_ evaluate [] == Return [] evaluate (x:xs) == Return (x:xs) That should answer your question.
I must not be phrasing my question very well; I feel like we're talking past each other. It seems to me that when writing actual programs (rather than reasoning about denotational semantics) the reason one would use `seq' or `evaluate' is to force something to be evaluated "now" rather than "later", i.e. to get around Haskell's default lazy execution. Your semantics say that (x `seq` return x) and (evaluate x) have the same result when x is anything other than _|_. All well and good, but (return x) *also* has those same results when x is not _|_. Why would one use the former two rather than (return x), if x is known not to be _|_? Because they evaluate x at different "times", right? Even though the eventual return value is the same, and thus the *semantics* are the same. So laying aside the formal semantics, what is the difference in terms of actual, real, Haskell programs? Mike _______________________________________________ Haskell-Cafe mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell-cafe
