是否有简单的 PHP 代码来区分“将对象作为引用传递”与“将对象引用作为值传递”？

Posted 2023-02-25

【中文标题】是否有简单的 PHP 代码来区分“将对象作为引用传递”与“将对象引用作为值传递”？

【英文标题】：Is there simple PHP code to distinguish "Passing the object as reference" vs "Passing the object reference as value"?

【发布时间】：2011-04-26 20:19:53

【问题描述】：

这与问题有关：How does the "&" operator work in a php function?

有没有简单的代码来显示两者的区别

将对象作为引用传递

对比

将对象的引用作为值传递？

【问题讨论】：

康拉德鲁道夫不是已经提供了an example吗？

很有趣...我认为在C社区中，通过引用传递对象与将其引用作为值传递相同...我在上面的代码示例中看到的将称为“传递它的参考作为参考”

你可以看到这个链接例如：***.com/questions/879/…

【参考方案1】：

您可以通过引用将变量传递给函数。该函数将能够修改原始变量。

可以在函数定义中通过引用来定义段落：

<?php
function changeValue(&$var)

    $var++;


$result=5;
changeValue($result);

echo $result; // $result is 6 here
?>

【参考方案2】：

<?php
class X 
    var $abc = 10; 

class Y 
    var $abc = 20; 
    function changeValue(&$obj)//1>here the object,$x is a reference to the object,$obj.hence it is "passing the object's reference as value"
        echo 'inside function :'.$obj->abc.'<br />';//2>it prints 10,bcz it accesses the $abc property of class X, since $x is a reference to $obj.
        $obj = new Y();//but here a new instance of class Y is created.hence $obj became the object of class Y.
        echo 'inside function :'.$obj->abc.'<br />';//3>hence here it accesses the $abc property of class Y.
    

$x = new X();
$y = new Y();

$y->changeValue($x);//here the object,$x is passed as value.hence it is "passing the object as value"
echo $x->abc; //4>As the value has been changed through it's reference ,hence it calls $abc property of class Y not class X.though $x is the object of class X
?>

o/p：

inside function :10
inside function :20
20

【参考方案3】：

这个怎么样：

<?php
class MyClass 
    public $value = 'original object and value';


function changeByValue($originalObject) 
    $newObject = new MyClass();
    $newObject->value = 'new object';

    $originalObject->value = 'changed value';

    // This line has no affect outside the function, and is
    // therefore redundant here (and so are the 2 lines at the
    // the top of this function), because the object
    // "reference" was passed "by value".
    $originalObject = $newObject;


function changeByReference(&$originalObject) 
    $newObject = new MyClass();
    $newObject->value = 'new object';

    $originalObject->value = 'changed value';

    // This line changes the object "reference" that was passed
    // in, because the "reference" was passed "by reference".
    // The passed in object is replaced by a new one, making the
    // previous line redundant here.
    $originalObject = $newObject;


$object = new MyClass();
echo $object->value;  // 'original object and value';

changeByValue($object)
echo $object->value;  // 'changed value';

$object = new MyClass();
echo $object->value;  // 'original object and value';

changeByReference($object)
echo $object->value;  // 'new object';