尝试通过引用赋值来理解运算符优先级

Question

Operator precedence 表示顺序应为：+, &, =。但是这段代码执行显示顺序是：&, =, +

$b = 1;
$a = & $b + print('print executed');
if ($a == 1)
    echo ' but one was not added and error was not raised';

输出print executed but one was not added and error was not raised

为什么更改了这种情况的优先级？

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)

Answer 1

可以说，这并不能真正回答您的问题，但请考虑以下代码：

$b = 1;
$a = &$b + 123;

操作码揭示了以下执行策略：

compiled vars:  !0 = $b, !1 = $a
line     # *  op                       fetch          ext  return  operands
-----------------------------------------------------------------------------
   3     0  >   ASSIGN                                              !0, 1
   4     1      ASSIGN_REF                                         !1, !0
         2      ADD                                          ~2      , 123
         3      FREE                                                 ~2

如您所见，通过引用进行赋值，加法存储在临时变量中，然后释放；基本上，一个空操作。

也许文档可以更清楚，但我无法想象这个特定代码会有意义的场景:)

尝试通过引用赋值来理解运算符优先级

Trying to understand operator precedence with assign by reference

php

reference

variable-assignment