Xor returns true when two side are true - PHP

Question

As I learned Xor should return false when two variable are true

But I tried it and it doesn't work!

$n1 = true;
$n2 = true;
$res = $n1 xor $n2;
echo var_dump($res);

Answer 1

This because of PHP Operator Precedence. Following example is correct

$n1 = true;
$n2 = true;
$res = ($n1 xor $n2);
echo var_dump($res); // bool(false) 

Answer 2

Priority. Your third statement is evaluated as

($res = $n1) xor $n2;

I.e. first the assignment happens, and $res becomes what $n1 is (i.e. true); the assignment evaluates as the value being assigned, so as the next step true xor $n2 is evaluated, producing false; but this value is not stored or used in any way, and is discarded. Basically, your code is equal to

$n1 = true;
$n2 = true;
$res = $n1;
$res xor $n2;        // useless
echo var_dump($res);

To expound a bit more, note that there are two sets of logical operators, where the only difference is how tightly they bind. and, or and xor bind very loosely, while their counterparts && and || bind tighter. Unfortunately, there is no tight binding counterpart for xor. These operators have been brought over from Perl, where a common idiom was to use the loosely-binding operators for flow control:

$f = doSomethingThatMightBeFalsy() or die("I failed at my task");

which will assign $f, and if that value ends up being falsy, die. This is distinct from

$f = doSomethingThatMightBeFalsy() || die("I fail evrytym :(");

tl;dr: As Akam already wrote, you should explicitly use parentheses to make xor bind tighter than the assignment.