PHP 全局和静态变量的引用问题?
哪位大神能从内存的角度解释下 php 官方手册上的这个例子,我是个初级 PHPer
搞不太明白,请前辈不吝赐教,晚辈不胜感谢!
最好能下面代码运行时的内存
的情况说明下。
类似的行为也适用于 static 语句。引用并不是静态地存储的:
function &get_instance_ref() { static $obj; echo 'Static object: '; var_dump($obj); if (!isset($obj)) { // 将一个引用赋值给静态变量 $obj = &new stdclass; } $obj->property++; return $obj; } function &get_instance_noref() { static $obj; echo 'Static object: '; var_dump($obj); if (!isset($obj)) { // 将一个对象赋值给静态变量 $obj = new stdclass; } $obj->property++; return $obj; } $obj1 = get_instance_ref(); $still_obj1 = get_instance_ref(); echo "\n"; $obj2 = get_instance_noref(); $still_obj2 = get_instance_noref();
以上例程会输出:
Static object: NULL Static object: NULL Static object: NULL Static object: object(stdClass)(1) { ["property"]=> int(1) }
上例演示了当把一个引用赋值给一个静态变量时,第二次调用 &get_instance_ref() 函数时其值并没有被记住。
本例出处:http://php.net/manual/en/language.variables.scope.php
回复内容:
哪位大神能从内存的角度解释下 php 官方手册上的这个例子,我是个初级 PHPer
搞不太明白,请前辈不吝赐教,晚辈不胜感谢!
最好能下面代码运行时的内存
的情况说明下。
类似的行为也适用于 static 语句。引用并不是静态地存储的:
function &get_instance_ref() { static $obj; echo 'Static object: '; var_dump($obj); if (!isset($obj)) { // 将一个引用赋值给静态变量 $obj = &new stdclass; } $obj->property++; return $obj; } function &get_instance_noref() { static $obj; echo 'Static object: '; var_dump($obj); if (!isset($obj)) { // 将一个对象赋值给静态变量 $obj = new stdclass; } $obj->property++; return $obj; } $obj1 = get_instance_ref(); $still_obj1 = get_instance_ref(); echo "\n"; $obj2 = get_instance_noref(); $still_obj2 = get_instance_noref();
以上例程会输出:
Static object: NULL Static object: NULL Static object: NULL Static object: object(stdClass)(1) { ["property"]=> int(1) }
上例演示了当把一个引用赋值给一个静态变量时,第二次调用 &get_instance_ref() 函数时其值并没有被记住。
本例出处:http://php.net/manual/en/language.variables.scope.php
用静态方式调用一个非静态方法会导致一个 E_STRICT 级别的错误。
就像其它所有的 PHP 静态变量一样,静态属性只能被初始化为文字或常量,不能使用表达式。所以可以把静态属性初始化为整数或数组,但不能初始化为另一个变量或函数返回值,也不能指向一个对象。
出处:http://php.net/manual/zh/language.oop5.static.php
所以global和static不用使用引用赋值,这是由php解释器决定的。

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