如何在兩種方法中通過引用傳遞？

Question

class foo { 
    public $value1 = 1; 

    public function setValue1(& $v) { 
     $this->value1 = $v; 
    } 
} 

class bar { 
    public $value2 = 2; 
    public $obj; 

    public function & getValue2() { 
     return $this->value2; 
    } 

    public function setValue2($v) { 
     $this->value2 = $v; 
} 
} 

$foo = new foo(); 
$bar = new bar(); 

$bar->obj = $foo; 

$foo->setValue1($bar->getValue2()); //<-- not works 

//$foo->value1 =& $bar->getValue2(); //<-- works 

$bar->setValue2(4); 


var_dump($foo, $bar) Prints: 
    object(foo)[124] 
     public 'value1' => int 2 
    object(bar)[123] 
     public 'value2' => int 4 
     public 'obj' => 
     object(foo)[124] 
      public 'value1' => int 2 

爲什麼當我使用$ foo-> setValue1（$ bar-> getValue2（））時，foo-> value1等於2，bar-> value2等於4？

我想通過引用傳遞和通過引用返回使用方法而不是屬性。

Answer 1

正確答案爲$方法foo->值1 =內部參考& $ V;

Answer 2

這工作：

class foo { 
    public $value1 = 1; 

    public function &setValue1($v) { // <-- ! 
     $this->value1 = &$v; // <-- ! 
    } 
} 

class bar { 
    public $value2 = 2; 
    public $obj; 

    public function &getValue2() { 
     return $this->value2; 
    } 

    public function &setValue2($v) { // <-- ! 
     $this->value2 = $v; 
} 
} 

$foo = new foo(); 
$bar = new bar(); 

$bar->obj = $foo; 

$tmp =& $bar->getValue2(); // <-- ! 
$foo->setValue1(&$tmp); // <-- ! 

$bar->setValue2(4); 

echo "<pre>" . print_r($foo, 1) . "</pre>"; 
echo "<pre>" . print_r($bar, 1) . "</pre>"; 

→ 輸出：

foo Object 
(
    [value1] => 4 
) 

bar Object 
(
    [value2] => 4 
    [obj] => foo Object 
     (
      [value1] => 4 
     ) 

)