Am Samstag, 8. November 2003, 13:13 schrieb Paul Hudak: > But note that x `seq` x is equivalent to x, even operationally. To see why > denotationally, note that if x evaluates to _|_, so does x `seq` x. And if > x evaluates to a value v, so does x `seq` x. To see why operationally, > consider the two lists: > > let x = 1+1 in [x `seq` x] > let x = 1+1 in [x] > > Using conventional lazy evaluation in both cases, the term "1+1" is not > evaluated until the head of the list is taken. In other words, x `seq` x in > no way hurries the evaluation of x.
Yes, you are right. > -Paul Wolfgang _______________________________________________ Haskell-Cafe mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell-cafe
