在 PHP 5 中如何通过引用传递对象? [英] How do you pass objects by reference in PHP 5?

问题描述

在 PHP 5 中，您是否需要使用 & 修饰符来通过引用传递?例如，

In PHP 5, are you required to use the & modifier to pass by reference? For example,

class People() { }
$p = new People();
function one($a) { $a = null; }
function two(&$a) { $a = null; )

在 PHP4 中，您需要 & 修饰符在更改后保持引用，但我对我读过的有关 PHP5 自动使用传递引用的主题感到困惑，除非明确克隆对象.

In PHP4 you needed the & modifier to maintain reference after a change had been made, but I'm confused on the topics I have read regarding PHP5's automatic use of pass-by-reference, except when explicity cloning the object.

在 PHP5 中， & 修饰符是否需要为所有类型的对象(变量、类、数组等)通过引用传递?

推荐答案

您是否需要使用 &传递引用的修饰符?

are you required to use the & modifier to pass-by-reference?

从技术/语义上来说，答案是是，即使是对象也是如此.这是因为有两种方法可以传递/分配对象:通过引用或通过标识符.当函数声明包含 & 时，如:

Technically/semantically, the answer is yes, even with objects. This is because there are two ways to pass/assign an object: by reference or by identifier. When a function declaration contains an &, as in:

function func(&$obj) {}

无论如何，参数将通过引用传递.如果你声明没有 &

The argument will be passed by reference, no matter what. If you declare without the &

function func($obj) {}

所有东西都将通过值传递，对象和资源除外，它们将通过标识符传递.什么是标识符?好吧，您可以将其视为对引用的引用.举个例子:

Everything will be passed by value, with the exception of objects and resources, which will then be passed via identifier. What's an identifier? Well, you can think of it as a reference to a reference. Take the following example:

class A
{
    public $v = 1;
}

function change($obj)
{
    $obj->v = 2;
}

function makezero($obj)
{
    $obj = 0;
}

$a = new A();

change($a);

var_dump($a); 

/* 
output:

object(A)#1 (1) {
  ["v"]=>
  int(2)
}

*/

makezero($a);

var_dump($a);

/* 
output (same as before):

object(A)#1 (1) {
  ["v"]=>
  int(2)
}

*/

那么为什么$a 传递给makezero 后不会突然变成整数呢?这是因为我们只覆盖了标识符.如果我们通过引用传递:

So why doesn't $a suddenly become an integer after passing it to makezero? It's because we only overwrote the identifier. If we had passed by reference:

function makezero(&$obj)
{
    $obj = 0;
}

makezero($a);

var_dump($a);

/* 
output:

int(0) 

*/

现在 $a 是一个整数.因此，通过 identifier 传递和通过 reference 传递是有区别的.

Now $a is an integer. So, there is a difference between passing via identifier and passing via reference.

这篇关于在 PHP 5 中如何通过引用传递对象?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！