Perl defined-or in a list context, why a scalar?

Question

use strict;
use warnings;
use Test::More;

subtest 'explicit array' => sub  {
    my @row = (1,2,3);
    # let's disassamble the array.
    # without default it works:
    my ($one, $two, $three) =  @row;
    is($one, 1, 'one');
    # this works too:
    ($one, $two, $three) =  @row ? @row : (10,20,30);
    is($one, 1, 'one');
    # and the default hits
    my @emptyness;
    ($one, $two, $three) = @emptyness ? @emptyness : (10,20,30);
    is($one, 10, 'one default');
    # however, this squashes the array to a scalar
    ($one, $two, $three) =  @row // (10,20,30);
    is($one, 3, 'scalar, length');
    is($two, undef, 'nothing else');
    # shouldn't 'defined-or' be equivalent to a ternary with a check against undef?
    # ($one, $two, $three) = defined @emptyness ? @emptyness : (10,20,30); # fails!
    # "Can't use 'defined(@array)' (Maybe you should just omit the defined()?)"
    # Probably @array // ... should fail in the same way, but instead it returns @array
    # in a scalar context.
    # so maybe this is a bug
};


done_testing();

Or can anybody give me a reasonable explanation for this behavior?

Answer 1

The behavior you are observing is the intended one. This is documented in perlop, in the section Logical Defined-Or:

EXPR1 // EXPR2 returns the value of EXPR1 if it's defined, otherwise, the value of EXPR2 is returned. (EXPR1 is evaluated in scalar context, EXPR2 in the context of // itself).

And, perldoc later provides the following example:

In particular, this means that you shouldn't use this for selecting between two aggregates for assignment:

@a = @b || @c;            # This doesn't do the right thing
@a = scalar(@b) || @c;    # because it really means this.
@a = @b ? @b : @c;        # This works fine, though.