Perl xor returns unexpected result

Question

What's going on here? Why isn't $c == 1 ?

$ cat y.pl
#!/usr/bin/perl
$a = 0;
$b = 1;
$c = $a xor $b;
print "$a ^ $b = $c\n";

$ ./y.pl 
0 ^ 1 = 0

Answer 1

Always use

use strict;
use warnings qw( all );

It would have detected your precedence issue.

Useless use of logical xor in void context

Specifically,

$c = $a xor $b;

means

($c = $a) xor $b;

The operators and , or , not and xor have very low precedence. This makes and , or and not very useful for flow control.

... or die ...;
... or next;

xor doesn't short-circuit, so it's not useful for flow control. xor was created for an entirely different reason. Like or , xor is a logical operator. That means considers the truth of its operand as whole instead of doing bit-by-bit comparisons. I don't believe this is the operation you want.

Fix:

$c = $a ^ $b;

---

                                    Low
                                    Precedence
Operation   Bitwise     Logical     Logical
----------  ----------  ----------  ----------
NOT         ~           !           not
AND         &           &&          and
OR          |           ||          or
OR*         N/A         //          N/A
XOR         ^           N/A         xor

OR* - Like OR, but only undefined is considered false.

Answer 2

Precedence of the assignment is higher than xor. Just write:

$c = ($a xor $b);