PHP 中的对象是按值分配还是按引用分配? [英] Are objects in PHP assigned by value or reference?

问题描述

在这段代码中:

<?php
class Foo
{
    var $value;

    function foo($value)
    {
        $this->setValue($value);
    }

    function setValue($value)
    {
        $this->value=$value;
    }
}

class Bar
{
    var $foos=array();

    function Bar()
    {
        for ($x=1; $x<=10; $x++)
        {
            $this->foos[$x]=new Foo("Foo # $x");
        }
    }

    function getFoo($index)
    {
        return $this->foos[$index];
    }

    function test()
    {
        $testFoo=$this->getFoo(5);
        $testFoo->setValue("My value has now changed");
    }
}
?>

当方法 Bar::test() 运行并且它改变了 foo 对象数组中 foo #5 的值时，数组中的实际 foo #5 会受到影响，或者$testFoo 变量是否只是一个局部变量，在函数结束时将不复存在?

When the method Bar::test() is run and it changes the value of foo # 5 in the array of foo objects, will the actual foo # 5 in the array be affected, or will the $testFoo variable be only a local variable which would cease to exist at the end of the function?

推荐答案

为什么不运行函数找出来?

Why not run the function and find out?

$b = new Bar;
echo $b->getFoo(5)->value;
$b->test();
echo $b->getFoo(5)->value;

对我来说，上面的代码(连同你的代码)产生了这个输出:

For me the above code (along with your code) produced this output:

Foo #5
My value has now changed

这不是由于通过引用传递"，而是由于通过引用赋值".在 PHP 5 中，通过引用赋值是对象的默认行为.如果您想按值分配，请使用 clone 关键字.

This isn't due to "passing by reference", however, it is due to "assignment by reference". In PHP 5 assignment by reference is the default behaviour with objects. If you want to assign by value instead, use the clone keyword.

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