in reply to How to get split $var to work like split ' '?

You'll need to use an if statement:

my $split_pattern //= getOpt();

my @x;
unless( defined $split_pattern ) {
    @x = split ' ', $line;
}
else
    @x = split $split_pattern, $line;
}

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Not elegant, but the only way. The special behaviour of split ' ', is triggered by the use of exactly ' ' in the source code.

Substituting a variable set to a space, or even a conditional statement where one branch is ' ', will not do the same thing.

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Re^2: How to get split $var to work like split ' '?
by hdb (Monsignor) on Sep 10, 2013 at 20:19 UTC

    UPDATE: Pls ignore this post. Thanks to LanX above I now understand what the problem is.

    Can you explain what the supposedly special behavior of split ' ' is? I cannot find any documentation saying anything about it. For me this works: 

    use strict;
use warnings;
use Data::Dumper;

my $str = "I want a split pattern";
my $pattern = $ARGV[0] // ' ';

my @pieces = split /$pattern/, $str;
print Dumper \@pieces;

    [download]
    hdb$ perl split4.pl
$VAR1 = [
          'I',
          'want',
          'a',
          'split',
          'pattern'
        ];
hdb$ perl split4.pl a
$VAR1 = [
          'I w',
          'nt ',
          ' split p',
          'ttern'
        ];

    [download]
[reply]
[d/l]
[select]

      You may also want to use quotemeta. e.g.

      ski@anito:~$ perl -e 'use strict; use warnings; my $x = "a.*b.*c.*d"; my $pattern = ".*"; my @x = split /$pattern/, $x; print join " - ",@x'
      ski@anito:~$

      vs.

      ski@anito:~$ perl -le 'use strict; use warnings; my $x = "a.*b.*c.*d"; my $pattern = quotemeta(".*"); my @x = split /$pattern/, $x; print join " - ",@x'
      a - b - c - d
      ski@anito:~$

      perldoc -f quotemeta.
[reply]
      Search for "specifying a PATTERN of space" in perldoc -f split. The difference relates to leading empty strings.
[reply]

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