A question about the comma operator:
(John and Chas deserve a rest from my questions, if they want it =). The binary comma operator in scalar context supposedly "evaluates its left argument, throws that value away, then evaluates its right argument and returns that value." So this: $_ = "dogs and cats"; $r = s/o/i/g, s/s/y/g; print "$r: $_\n"; ... prints "1: digy and caty". Why doesn't it print a "2" instead of a "1"? It did 2 replaces of s to y... If I change the second line to read: $r = s/s/y/g; ... it prints "2: dogy and caty". I thought given the definition above that the comma operator should do the first replace, then do the second replace which returns a "2", then assign that 2 to $r. Somehow $r is getting a 1. If instead I did: $r = s/o/i/g, s/s/y/g, s/d/cow/g; ... is that still a binary comma operator? It still does all the replaces, and still comes back with a 1. And why is: print $r, "\n"; ... not evaluated as a binary comma operator? Because it's in list context? Then why can't I do this with a list (without the parenthesis)? @a = 1, 2, 3; TIA. - Bryan -- To unsubscribe, e-mail: beginners-unsubscr...@perl.org For additional commands, e-mail: beginners-h...@perl.org http://learn.perl.org/
