Regex resetting the capture buffer

[Mario D'Alessio]

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

 

[anno4000]

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

 

[Mumia W.]

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.

 

[Mario D'Alessio]

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