[Melbourne-pm] An old one but an important one

Jacinta Richardson jarich at perltraining.com.au
Sun Jan 20 23:31:29 PST 2008

Tim Connors wrote:

> But if you can't determine where $& is going to come from, why can you 
> determine where $1 is going to come from (given that $& is essentially 
> $0), such that you don't need to slow down all other regexps too?

Perl can't determine where $1 and friends is coming from at any given point in
your program (without running it).  Thus any regular expression which uses
capturing parentheses sets $1 and friends (and does so even if Perl can't find
you using $1 etc anywhere) (and does so even if you capture into variables or
not).  Of course this happens when you're merely using the parentheses for
grouping because Perl doesn't know any better.

For example consider this code:

	$_ = "happy happy joy joy";
	if(/(h.*y) joy/) {
		if(/(joy)/) {
		print "Matched: $1\n";   # Which $1 is this?

what if we wrote:

	$_ = "happy happy joy joy";
        if(/(h.*y) joy/) {
                if(/(pig)/)  {
                print "Matched: $1\n";   # Which $1 is this?

Perl knows that it has to record $1 for each expression that it successfully
evaluated, because we're using capturing parentheses.  It will try to clear $1
on an unsuccessful evaluation, but you shouldn't count on it (on my Perl 5.8.8
the second snippet of code prints "Matched: happy happy joy").

Perl doesn't care which expression actually set $1.  If we'd used $& and/or it's
friends it'd be the same story.  Perl would set it for each expression evaluated
because it cannot tell which expression you needed it from.

Perl can easily tell whether your regular expression is matching and capturing,
or just matching.  Thus using non-capturing parentheses instead of capturing
will speed up those expressions, while not using them will not slow down any others.

All the best,


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