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Here's an example script:

my $a = '1 foo';
my $b = 'foo bar';

if ( $a =~ /^\d+\s*(\w+)/ ) # line 1
{
$b =~ s/bar/$1/; # line 2
}
print $b, "\n";

I am using the capture buffer in line 1 in the regex in line 2. But it
doesn't
work. The capture buffer is cleared before the $1 substitution can take
place.

I would have thought that Perl would perform the $1 substitution in the line
2
before it "recognizes" the substitution command and clears the capture
buffers.

Please explain the order in which Perl processes line 1.

Do I have to "eval" line 2 to get it to work the way that I want without
having
to assign $1 to a temp var?

Thanks.

Mario

Mario D'Alessio <> wrote in comp.lang.perl.misc:
> Here's an example script:
>
> my $a = '1 foo';
> my $b = 'foo bar';
>
> if ( $a =~ /^\d+\s*(\w+)/ ) # line 1
> {
> $b =~ s/bar/$1/; # line 2
> }
> print $b, "\n";
>
> I am using the capture buffer in line 1 in the regex in line 2. But it
> doesn't
> work. The capture buffer is cleared before the $1 substitution can take
> place.
>
> I would have thought that Perl would perform the $1 substitution in the line
> 2
> before it "recognizes" the substitution command and clears the capture
> buffers.
>
> Please explain the order in which Perl processes line 1.

Do you mean "line 2"?

Perl delays substitution in the replacement string until after the
match. It *must* do so for constructs like

s/x(.)y/y$1x/;

to work. The match must have happened before $1 has its intended value.

> Do I have to "eval" line 2 to get it to work the way that I want without
> having
> to assign $1 to a temp var?

How would "eval" help? You can use $_ for a temporary variable:

if ( $a =~ /^\d+\s*(\w+)/ ) # line 1
{
$b =~ s/bar/$_/ for $1; # line 2
}
print $b, "\n";

Anno

On 06/21/2007 10:34 AM, Mario D'Alessio wrote:
> Here's an example script:
>
> my $a = '1 foo';
> my $b = 'foo bar';
>
> if ( $a =~ /^\d+\s*(\w+)/ ) # line 1
> {
> $b =~ s/bar/$1/; # line 2
> }
> print $b, "\n";
>
> I am using the capture buffer in line 1 in the regex in line 2. But it
> doesn't
> work. The capture buffer is cleared before the $1 substitution can take
> place.
>
> I would have thought that Perl would perform the $1 substitution in the line
> 2
> before it "recognizes" the substitution command and clears the capture
> buffers.
>
> Please explain the order in which Perl processes line 1.
>

Line 1 is not the issue, and it works as you expect it. Line 2 is the
problem; when a successful match occurs, the match variables are
set/reset to new values.

> Do I have to "eval" line 2 to get it to work the way that I want without
> having
> to assign $1 to a temp var?
>
> Thanks.
>
> Mario
>
>
>

Using eval() is worse than just assigning to a temporary variable.

>> Please explain the order in which Perl processes line 1.
>
> Do you mean "line 2"?

Yes. I just typed a typo.


Thanks for all the reponses. I now fully understand.

Mario