On 11/10/06, Dougal Stanton <[EMAIL PROTECTED]> wrote:
As you noted that doesn't seem right --- how does compile capture its input? Well, the (.) operator is slightly different. It captures variables and passes them into the 'innermost' function, a bit like this:
That is a great explanation. I've got a much better understanding of the operator now - thanks very much! Justin _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
