dt3358
dt3358
2015-01-11 04:36

尝试通过引用分配来理解运算符优先级

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)
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1条回答

  • dqnek0079 dqnek0079 6年前

    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 :)

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