欢迎进入Linux社区论坛,与200万技术人员互动交流 >>进入 在PHP5中,对象间的赋值被默认视为地址引用,例如: ?php $objB = $objA; ? 针对上面这句代码,我们看一下内存分配结构: 从上面示意图中我们可以看出,定义$objA对象时,系统会为其分配一块内存(①
欢迎进入Linux社区论坛,与200万技术人员互动交流 >>进入
在PHP5中,对象间的赋值被默认视为地址引用,例如:
$objB = $objA;
?>
针对上面这句代码,我们看一下内存分配结构:
从上面示意图中我们可以看出,定义$objA对象时,系统会为其分配一块内存(①箭头),当$objB被赋值$objA时,系统没有再分配内存,而是将$objB的引用直接指向了$objA内存,实际上这两个对象指向的是同一块内存,这时,当我们修改$objB的值时,$objA的值也跟着变化了,这就使得如果我们需要两个同样的、相对独立的对象就需要分别定义,操作麻烦,而且程序的可读行定也不好。为了解决这个问题,PHP为我们提供了一种可以克隆对象的方法clone,在PHP手册里对clone方法介绍的比较简单,下面UncleToo用实例为大家介绍一下clone方法的使用。
示例一:
class subclass{
private $name;
private $age;
public function __construct(){
$this->name = '张三';
$this->age = '30';
}
public function __clone(){
$this->name = '李四';
$this->age = '29';
}
}
class myclass{
public $p1;
public $p2;
public function __construct(){
}
public function __clone(){
$this->p1 = clone $this->p1;
}
}
$obj = new myclass();
$obj->p1 = new subclass();
$obj->p2 = new subclass();
$obj2 = clone $obj;
echo '
'; <p> echo '这是obj<br>';</p> <p> var_dump($obj);</p> <p> echo '-----------------------';</p> <p> echo '<br>这是obj2<br>';</p> <p> var_dump($obj2);</p> <p> echo '</p>';
?>
输出:
这是objobject(myclass)#1 (2) {
["p1"]=>
object(subclass)#2 (2) {
["name":"subclass":private]=>
string(4) "张三"
["age":"subclass":private]=>
string(2) "30"
}
["p2"]=>
object(subclass)#3 (2) {
["name":"subclass":private]=>
string(4) "张三"
["age":"subclass":private]=>
string(2) "30"
}
}
-----------------------这是obj2object(myclass)#4 (2) {
["p1"]=>
object(subclass)#5 (2) {
["name":"subclass":private]=>
string(4) "李四"
["age":"subclass":private]=>
string(2) "29"
}
["p2"]=>
object(subclass)#3 (2) {
["name":"subclass":private]=>
string(4) "张三"
["age":"subclass":private]=>
string(2) "30"
}
}
可以看到,$obj2 clone $obj,$obj->p1 clone $obj->p1,执行了 __clone()方法。而在__clone方法中,对p1的name与 age属性进行修改,因此p1的name与age 发生变化。而p2因为没有执行clone()方法,所以新复制出来的$obj2->p2的属性与 $obj->p2一样。
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