我试图了解PHP构造函数和析构函数的行为.一切都按照构造函数的预期进行,但是我无法隐式地触发析构函数.我已经完成了PHP.net和相关网站上的所有阅读,但我无法找到这个问题的答案.
如果我有一个简单的类,例如：
class test{
public function __construct(){
print "contructing<br>";
}
public function __destruct(){
print "destroying<br>";
}
}
我称之为：
$t = new test;
它打印构造函数消息.但是,我希望当脚本结束并呈现页面时,析构函数应该触发.当然不是.
如果我打电话未设置($t);当脚本结束时,析构函数当然会触发,但有没有办法让它隐式触发？
谢谢你的任何提示
这很容易测试.
<?PHP
class DestructTestDummy {
protected $name;
function __construct($name) {
echo "Constructing $name\n";
$this->name = $name;
}
function __destruct() {
echo "Destructing $this->name\n";
//exit;
}
}
echo "Start script\n";
register_shutdown_function(function() {
echo "Shutdown function\n";
//exit
});
$a = new DestructTestDummy("Mr. Unset");
$b = new DestructTestDummy("Terminator 1");
$c = new DestructTestDummy("Terminator 2");
echo "Before unset\n";
unset($a);
echo "After unset\n";
echo "Before func\n";
call_user_func(function() {
$c = new DestructTestDummy("Mrs. Scopee");
});
echo "After func\n";
$b->__destruct();
exit("Exiting\n");
在PHP 5.5.12中打印：
Start script Constructing Mr. Unset Constructing Terminator 1 Constructing Terminator 2 Before unset Destructing Mr. Unset After unset Before func Constructing Mrs. Scopee Destructing Mrs. Scopee After func Destructing Terminator 1 Exiting Shutdown function Destructing Terminator 2 Destructing Terminator 1
因此,我们可以看到,当我们显式取消设置对象,超出范围时以及脚本结束时,将调用析构函数.