On Jan 5, Prahlad Vaidyanathan said: >> >my $leading_spaces = ($text =~ m/^(\s+)/) ; # This doesn't work >> >print $leading_spaces ; # Prints 1 >> >> You have executed the regex in scalar context. Why? Because the > >Does this mean the return value of the match is put into $leading_spaces >?
Yes. A regex returns different things in scalar or list context. In scalar context, it returns whether or not it match (1 or ''). In list context, it returns the $1, $2, $3 variables. Context is determined by what is on the left-hand side of the = sign. If there is NOT a list on the left-hand side, then the context is SCALAR. $foo = bar(); $a, $b, $c = bar(); # because '=' is tighter than ',' $a, $b, ($c = bar()); # same as above bar(); # NO = at all, this is void context Void context is a special type of scalar context. >Thanks. That worked. But, I still don't understand what "list" in this >context means. And, what does "one-element list" mean ? You might want to read my article "'List' is a Four-Letter Word", which you can read here: http://www.pobox.com/~japhy/articles/pm/2000-02.html -- Jeff "japhy" Pinyan [EMAIL PROTECTED] http://www.pobox.com/~japhy/ RPI Acacia brother #734 http://www.perlmonks.org/ http://www.cpan.org/ ** Look for "Regular Expressions in Perl" published by Manning, in 2002 ** <stu> what does y/// stand for? <tenderpuss> why, yansliterate of course. -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
