Question

In the “Programming Perl” book there is a snippet (cutted):

By default, when(EXPR) is treated as an implicit smartmatch of $_; that is, $_ ~~ EXPR. However, if the EXPR argument to when is one of the 10 exceptional forms listed below,
it is evaluated directly for a Boolean result, and no smartmatching occurs:

1. …

2. A regular expression match in the form of /REGEX/, $foo =~ /REGEX/, or $foo
   =~ EXPR.

What does it mean evaluated directly for a Boolean result?

Example:

#!/usr/bin/perl
use v5.14;
my @a = ('aaa', 'bbb', 'ccc');

given(@a) {
    when (/a/) { say '@a contains an a'; }
    default    { say '@a does not contain an a' }
}

when I run it the output varies from time to time:

@a does not contain an a

@a contains an a

@a does not contain an a

@a does not contain an a

I can’t understand what happens here, can anyone be so pleasant to help?

Appreciation in advance.

Answer 1

Read the documentation carefully:

Another useful shortcut is that, if you use a literal array or hash as
the argument to “given”, it is turned into a reference. So
“given(@foo)” is the same as “given(\@foo)”, for example.

Therefore, given (@a) is turned into given(\@a). There is no smart matching, because you use when (/a/), so you are trying to match

\@a =~ /a/

The reference is stringified. It sometimes contains “a”, as in ARRAY(0x9a4e7f8), but usually does not 🙂