@a = @b || @c;              # this is wrong
    @a = scalar(@b) || @c;      # really meant this

[download]
 Hang on, shouldn't you be using 'or' instead? '||' works in scalar context, while 'or' also works in list context. jweed didn't say that he used '||', he used 'or' instead in his program.

I have constructed the following test to show that you can use 'or' in list context.

 Update: I have realized my error in the code below. The line my @c = @a or @b was wrong, it was not behaving as what I have expected, it was really doing this instead: (my @c = @a) or @b;. Ok, another lesson for me on precedence and misconception about the 'or' operator.

use strict;
use Data::Dumper;

my @a = (1, 2, 3);
my @b = (4, 5, 6);
my @c = @a or @b;  # equivalent to (my @c = @a) or @b;

print Dumper(\@c);

[download]
And the output is as expected -
$VAR1 = [
          1,
          2,
          3
        ];

[download]
jweed's problem wasn't on how he used the 'or', but rather how he used the brackets. Consider the following example:

use strict;
use Data::Dumper;

my @a = (1, 2, 3);
my @b = (4, 5, 6);
my @c = @a or @b;
my @d = (@a or @b);

print Dumper(\@c);
print Dumper(\@d);

[download]
The line my @d = (@a or @b) is 'logically wrong', because the brackets force what's inside to be evaluated in scalar context. The output of the above program is -

$VAR1 = [
          1,
          2,
          3
        ];
$VAR1 = [
          3
        ];

[download]

In reply to Re: Re: Short circut operators in list context by Roger
in thread Short circut operators in list context by jweed

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