Usually where Perl operators return false they do so by either returning an empty string or 0

Actually, they return both at the same time.

What is the actual value returned by a failed regular expression match?

Please see this table of regular expression return values. print Dumper( "a" =~ m/b/ ); is providing list context to its arguments, so it's the same as writing print Dumper( );

Why does ... !~ returns the more common empty string version?

Because "a" !~ m/a/ is the same as !('a' =~ /a/), and that returns Perl's special false value.

Update: Your initial issue can be solved by scalar or PerlX::Maybe:

warn Dumper(
    {
        a => scalar( "a" =~ m/b/ ),
        b => 'asdf'
    }
);

use PerlX::Maybe;
warn Dumper(
    {
        maybe a => "a" =~ m/b/,
        b => 'asdf'
    }
);

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Eventually I realised that the failed match for the regular expression was somehow tricking the first comma to be evaluated in a scalar context, rather than a list context (note that if the regular expression matches then it returns 1 and the comma is evaluated in a list context).

Sorry, no, your analysis is not correct - the hash assignment is entirely in list context, in the examples you showed there's no scalar context going on there at all; only in the my $a = ... assignments. Update 2: To nitpick myself a little, the arguments to the =~ operator are in scalar context.

In reply to Re: What does a failed regular expression match actually return? (updated) by haukex
in thread What does a failed regular expression match actually return? by Anonymous Monk

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