On Nov 14, 2007 4:27 PM, Justin Bailey <[EMAIL PROTECTED]> wrote: > It's: > > f $! x = x `seq` f x > > That is, the argument to the right of $! is forced to evaluate, and > then that value is passed to the function on the left. The function > itself is not strictly evaluated (i.e., f x) I don't believe.
Unless you mean f -- which I still don't think would do much -- it wouldn't make sense to evaluate (f x) strictly. (x `seq` x) is equivalent to (x), for any x (including (f x)). (Right?) Shachaf _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe