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.

Solution

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);

$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