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Perl xor returns unexpected result

What's going on here? Why not $ 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
+3


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2 answers


Always use

use strict;
use warnings qw( all );

It will detect your problem with priority.

Useless use of logical xor in void context

In particular,

$c = $a xor $b;

means

($c = $a) xor $b;


Operators and

, or

, not

and xor

have a very low priority. It does and

, or

and is not

very useful for flow control.

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

xor

is not short-circuited, so it is not suitable for flow control. xor

was created for a completely different reason. As or

, xor

is a logical operator. This means that the truth of its operand is integer, not a bitwise comparison. 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.
+11


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Destination priority is higher than xor. Just write:



$c = ($a xor $b);
+2


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