Operator precedence tells that order should be: +, &, =. But this code execution shows that order is: &, =, +
$b = 1;
$a = & $b + print('print executed');
if ($a == 1)
echo ' but one was not added and error was not raised';
Output print executed but one was not added and error was not raised
Why precedence is changed for this case?
P.S.
$a = new stdClass();
$c = &$a instanceof $a;
var_dump($c); // class stdClass#1 (0) {}
$b = $a instanceof $a;
var_dump($b); // bool(true)
Arguably, this doesn't really answer your question but consider this code:
$b = 1;
$a = &$b + 123;
The opcodes reveal the following execution strategy:
compiled vars: !0 = $b, !1 = $a
line # * op fetch ext return operands
-----------------------------------------------------------------------------
3 0 > ASSIGN !0, 1
4 1 ASSIGN_REF $1 !1, !0
2 ADD ~2 $1, 123
3 FREE ~2
As you can see, the assignment by reference takes place and the addition gets stored in a temporary variable and then freed; basically, a no-op.
Perhaps the documentation could be clearer, but I can't imagine a scenario in which this particular code would ever make sense :)