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phpxor
Gargoyle
Gargoyle

Reputation: 10385

PHP xor returns wrong value

Using php 7.1.0 I'm running this little test:

<?php

$a = true;
$b = true;

$value = $a xor $b;
if ($value == true) {
    print "bad!\n";
} else {
    print "good\n";
}

and it's coming back and saying bad. Why? An xor of two true values should be FALSE, not true.

Upvotes: 6

Views: 760

Answers (1)

Barmar
Barmar

Reputation: 782785

The problem is operator precedence. The xor operator has lower precedence than =, so your statement is equivalent to:

($value = $a) xor $b;

You need to write:

$value = ($a xor $b);

or

$value = $a ^ $b;

The ^ operator is bit-wise XOR, not boolean. But true and false will be converted to 1 and 0, and the bit-wise results will be equivalent to the boolean results. But this won't work if the original values of the variables could be numbers -- all non-zero numbers are truthy, but when you perform bit-wise XOR with them you'll get a truthy result for any two numbers that are different.

See the PHP Operator Precedence Table

See the related Assignment in PHP with bool expression: strange behaviour

Upvotes: 11

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