CODEWITHSUNDEEP
regexperl
Sandra Schlichting
Sandra Schlichting

Reputation: 26026

Combination of while, array, and regex

The loop fails. What is wrong with the array?

I would like the regex to return B when it parses the first string, and M when it parses the second string.

How is such an regex constructed?

#!/usr/bin/perl

use warnings;
use strict;

my $a = "0.0 B/s";
my $b = "12.0 MiB/s";

while (defined (my $s = shift ("$a", "$b"))) {

    my $unit = $1 if ($a =~ m/.*([KMGT])i?B\/s$/);
    print "$unit\n";
}

Upvotes: 2

Views: 92

Answers (3)

Zaid
Zaid

Reputation: 37146

shift is meant to be used with arrays, not lists. If you want to use a while loop, you need to pre-declare an array containing $a and $b (which, by the way, are a bad choice for variable names).

Having said that, a for loop construct is the more natural choice here:

for my $s ( $var1, $var2 ) { ... }

And given that you're trying to extract the measurement unit, why not do things a slightly different way:

say for map { my ( $s ) = /$regex/; $s } $var1, $var2;

Upvotes: 2

Anirudh Ramanathan
Anirudh Ramanathan

Reputation: 46768

  1. Your while has issues.
  2. You are using variable $a inside loop, when you want to use $s.

I'd use it this way:

#!/usr/bin/perl
use warnings;
use strict;
my $a = "0.0 B/s";
my $b = "12.0 MiB/s";
foreach my $s($a, $b) {
    print $1 if ($s =~ m/.*([KMGT])i?B\/s$/);
}

Upvotes: 1

perreal
perreal

Reputation: 98078

You need another substitution:

for ($a, $b) {
  if (m!((?:[KMGT]i)?B)/s$!) {
    my $unit = $1; 
    $unit =~ s/(.).*/$1/;
    print "$unit\n" if $unit;
  }   
}

Upvotes: 1

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