----- Original Message ----
From: Ovid <[EMAIL PROTECTED]>
To: [email protected]
Sent: Sunday, January 21, 2007 10:26:20 PM
Subject: Re: putting ";" as a replacement in the substitution.
--- Michael Alipio <[EMAIL PROTECTED]> wrote:
<snip>
> No, not correct. The regular expression is what's being matched. Period.
> The capturing parentheses merely capture some of all of the regular
> expression into a 'dollar digit' variable ($1, $2, and so on).
> So for this:
> $var =~ s/foo(bar)/$1/;
> The 'foo(bar)' is what is being matched and the 'bar' is captured to the $1
> variable. For this:
> if ( $var =~ /foo(bar)/ ) { ... }
> The 'foo(bar)' is *still* what is being matched and the 'bar' is *still* what
> is being captured to the $1 variable.
> The first version is when you want to alter the string you're matching. The
> second version is good when you want to take action based upon a match and
> possibly extract data out of the string.
I see.. Now it's a lot more clearer.
It would be pointless to put () in my regexp when testing with if, unless I'm
grouping something or I want to do something with $1.
if /(^\w+)\s+/
But if I am assigning something, like:
my $captured =~ /^(\w+)\s+/
I should put it inside parenthesis.
I also noticed that $capture here will always contain the first catched match
($1).
The, (?:) as suggested by someone is also good when I want to avoid something
being stored in $n... I have read about it and a lot more(particularly the
extended regexp features) in perlre but not quite sure what they mean. The
topic on backtracking when using quantifier is also a good read.
Anyway, thanks for your help!
Have a nice day!
> Cheers,
> Ovid
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