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Sunday, November 6, 2022

[FIXED] Why do I get the first capture group only?

 November 06, 2022     perl, regex, regex-group     No comments   

Issue

(https://stackoverflow.com/a/2304626/6607497 and https://stackoverflow.com/a/37004214/6607497 did not help me)

Analyzing a problem with /proc/stat in Linux I started to write a small utility, but I can't get the capture groups the way I wanted. Here is the code:

#!/usr/bin/perl
use strict;
use warnings;

if (open(my $fh, '<', my $file = '/proc/stat')) {
    while (<$fh>) {
        if (my ($cpu, @vals) = /^cpu(\d*)(?:\s+(\d+))+$/) {
            print "$cpu $#vals\n";
        }
    }
    close($fh);
} else {
    die "$file: $!\n";
}

For example with these input lines I get the output:

> cat /proc/stat
cpu  2709779 13999 551920 11622773 135610 0 194680 0 0 0
cpu0 677679 3082 124900 11507188 134042 0 164081 0 0 0
cpu1 775182 3866 147044 38910 135 0 15026 0 0 0
cpu2 704411 3024 143057 37674 1272 0 8403 0 0 0
cpu3 552506 4025 136918 38999 160 0 7169 0 0 0
intr 176332106  ...
 0
0 0
1 0
2 0
3 0

So the match actually works, but I don't get the capture groups into @vals (perls 5.18.2 and 5.26.1).


Solution

Replacing

    while (<$fh>) {
        if (my ($cpu, @vals) = /^cpu(\d*)(?:\s+(\d+))+$/) {

with

    while (<$fh>) {
        my @vals;
        if (my ($cpu) = /^cpu(\d*)(?:\s+(\d+)(?{ push(@vals, $^N) }))+$/) {

does what I wanted (requires perl 5.8 or newer).



Answered By - U. Windl
Answer Checked By - Terry (PHPFixing Volunteer)
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