为什么我的 PHP 对象是通过引用传递的 [英] Why are my PHP objects passed by reference

问题描述

我看到一些与 PHP 中的引用/赋值相关的令人困惑的行为...

私有函数 doSomething($things){foreach ($things as $thing) {echo $thing->property;//'foobar'$copyThing = $thing;未设置($copyThing-> 属性)；echo $thing->property;//不明确的

我希望在通过引用传递变量时会出现这种行为 (&$thing)，但我并不打算在这里这样做，而且无论如何它似乎都在发生.我错过了什么?

解决方案

只是解释我的评论:

<块引用>

foreach 循环中的对象总是通过引用传递

当您对对象数组使用 foreach 循环时，您在循环内使用的变量是指向该对象的指针，因此它用作引用，循环内对象的任何更改都是对象的更改外部.这是因为:

<块引用>

对象总是通过引用传递(@user3137702 引用)

详细的官方解释这里.

<小时>

当您复制和取消设置变量时:

<块引用>

$copyThing = $thing;未设置($copyThing-> 属性)；

您正在创建另一个指针并取消设置它，因此原始值消失了.事实上，由于 foreach 循环也使用了一个指针，$things 数组也会受到影响.

检查这个 ideone(注意 vardump ['a' 属性消失的地方]，因为输出是和你得到的一样)

<小时>

我不知道它在哪个版本中改变了，如果有的话，因为它似乎是默认的对象/指针行为

<小时>

作为一种解决方法(一些想法):

1. 复制你的初始数组
2. 使用克隆:$x = clone($obj);(只要默认的复制构造函数适用于您的对象)

I'm seeing some confusing behavior related to reference/assignment in PHP...

private function doSomething($things)
{
    foreach ($things as $thing) {
        echo $thing->property; // 'foobar'
        $copyThing = $thing;
        unset($copyThing->property);
        echo $thing->property; // undefined

I expect this behavior when passing variables by reference (&$thing) but I'm not trying to do that here and it seems to be happening anyway. What am I missing?

解决方案

Just explaining my comment:

objects in foreach loops are always passed by reference

When you use a foreach loop for an array of objects the variable that you are using inside the loop is a pointer to that object so it works as a reference, any change on the object inside the loop is a change on the object outside. This is because:

objects are always passed by reference (@user3137702 quote)

Detailed and official explanation here.

---

When you copy and unset your variable:

$copyThing = $thing;
unset($copyThing->property);

you are creating another pointer and unseting it, so the original value is a gone. As a matter of fact, since the foreach loop also uses a pointer the $things array is also affected.

check this ideone (notice the vardump [where the 'a' property is gone], as the output is the same as you got)

---

I do not know in which version it changed, if ever, as it seems like default object/pointer behavior

---

As a workaround (some ideas):

1. Copy your initial array
2. Use clone: $x = clone($obj); (As long as the default copy constructor works for your objects)

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