问题描述
似乎在 PHP 中对象是通过引用传递的.甚至赋值运算符似乎也没有创建 Object 的副本.
It appears that in PHP objects are passed by reference. Even assignment operators do not appear to be creating a copy of the Object.
这是一个简单的、人为的证明:
Here's a simple, contrived proof:
<?php
class A {
public $b;
}
function set_b($obj) { $obj->b = "after"; }
$a = new A();
$a->b = "before";
$c = $a; //i would especially expect this to create a copy.
set_b($a);
print $a->b; //i would expect this to show 'before'
print $c->b; //i would ESPECIALLY expect this to show 'before'
?>
在这两种打印情况下，我都得到了之后"
In both print cases I am getting 'after'
那么，我如何将 $a 通过值而不是引用传递给 set_b()?
So, how do I pass $a to set_b() by value, not by reference?
推荐答案
在 PHP 5+ 中，对象是通过引用传递的.在 PHP 4 中，它们是按值传递的(这就是为什么它在运行时通过引用传递，这已被弃用).
In PHP 5+ objects are passed by reference. In PHP 4 they are passed by value (that's why it had runtime pass by reference, which became deprecated).
您可以使用 PHP5 中的克隆"运算符来复制对象:
You can use the 'clone' operator in PHP5 to copy objects:
$objectB = clone $objectA;
此外，它只是通过引用传递的对象，而不是您在问题中所说的所有内容......
Also, it's just objects that are passed by reference, not everything as you've said in your question...
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