Recently I played with the logical/defined or in a scenario where the result should be an lvalue. For the defined or (//) the documentation clearly states that the result is not an lvalue. While this is not mentioned for the logical or (||), it holds there too.

I don't know the reason for this behaviour, but it prevents the usage of these operators as a substitute for the non-existing (binary) "Elvis operator" (?:), where X ?: Z would act like X ? X : Z, without evaluating X twice.

To my surprise I realized that dereferencing a reference to the result of an or operator provides an lvalue, given the operands are lvalues. That is, the following expression does compile and works: ${\(X // Y)}++.
I could not find anything in the documentation that would forbid such a usage.

So my questions are:

• What is the reason for the logical/defined or operators not to provide an lvalue as result?
• Would changing this behaviour break any existing code?
• Is the usage of ${\(X // Y)} as an lvalue well-defined behaviour?

Here is a complete example:

#!/usr/bin/perl

use v5.24;
use warnings;
use experimental 'signatures';

say "This is perl $]";

sub f :lvalue ($k) {
    state $href = {a => 10};
    $href->{$k};
}

sub g :lvalue ($k) {
    state $href = {b => 20};
    $href->{$k};
}

eval q{
    (f($_) || g($_))++ for qw(a b);
    1;
} or warn $@;

${\(f($_) // g($_))}++ for qw(a b);

say "f(a) = ", f('a');
say "g(b) = ", g('b');

__DATA__
Can't modify logical or (||) in postincrement (++) at (eval 5) line 2,
+ near ")++"
This is perl 5.032001
f(a) = 11
g(b) = 21
[download]

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