__destruct PHP的可见性

Question

该__destruct()功能的"可见性"应该是公开的还是其他的？我正在尝试为我的小组编写标准文档,这个问题出现了.

Answer 1

除了Mark Biek的回答:

必须将__destruct()函数声明为public.否则,在脚本关闭时不会执行该函数:

Warning: Call to protected MyChild1::__destruct() from context '' during shutdown ignored in Unknown on line 0
Warning: Call to private MyChild2::__destruct() from context '' during shutdown ignored in Unknown on line 0

这可能不是有害的,而是不洁净的.

但最重要的是:如果析构函数被声明为private或protected,则运行时将在垃圾收集器尝试释放对象时抛出致命错误:

<?php
class MyParent
{
    private function __destruct()
    {
        echo 'Parent::__destruct';
    }
}

class MyChild extends MyParent
{
    private function __destruct()
    {
        echo 'Child::__destruct';
        parent::__destruct();
    }
}

$myChild = new MyChild();
$myChild = null;
$myChild = new MyChild();

?>

输出

Fatal error: Call to private MyChild::__destruct() from context '' in D:\www\scratchbook\destruct.php on line 20

(感谢Mark Biek的出色表现!)

Answer 2

我认为在子类需要显式调用父类的__destruct方法的情况下需要公开.

像这样的东西会引发错误:

<?php
class MyParent
{
    private function __destruct()
    {
        echo 'Parent::__destruct';
    }
}

class MyChild extends MyParent
{
    function __destruct()
    {
        echo 'Child::__destruct';
        parent::__destruct();
    }
}

$myChild = new MyChild();
?>