[英]PHP Get parent class variables from child class constructed in the parent constructor
With the following code snippet, why does $this->foo
return NULL
in the B child class constructor? 使用以下代码段,为什么
$this->foo
在B子类构造函数中返回NULL
? From what I've read I was under the impression that a child class inherits all its parent variables and methods 根据我的阅读,我给人的印象是子类继承了其所有父变量和方法
$a = new A();
class A {
protected $foo;
public function __construct() {
$this->foo = "Hello World";
$b = new B();
}
}
class B extends A {
public function __construct() {
var_dump($this->foo);
}
}
The selected answer of this question seems to suggest $this->foo
should be accessible. 该问题的选定答案似乎表明
$this->foo
应该可以访问。
Is it because B is being constructed from within its parent class? 是因为B是从其父类中构造的吗? If that is the case, how can I access the variable in the child class?
在这种情况下,如何访问子类中的变量?
You're doing new A()
and then separately you're doing new B()
. 您正在执行
new A()
,然后分别在执行new B()
。 You have two entirely independent object instances. 您有两个完全独立的对象实例。 It doesn't matter that one instantiation happens in the constructor of
A
, that's entirely irrelevant. 在
A
的构造函数中发生一个实例无关紧要,这是完全不相关的。 It also doesn't matter that B
extends A
. B
扩展A
也不重要。 You have two independent objects, and object instances do not share data implicitly, which is what you seem to be expecting. 您有两个独立的对象,并且对象实例不会隐式共享数据,这正是您所期望的。
Since B
's constructor overrides its parent's constructor, it's not executing any of A
's code either, so essentially nothing is happening when you instantiate a new B
. 由于
B
的构造函数会覆盖其父级的构造函数,因此它也不执行A
的任何代码,因此,当实例化一个新的B
时,实际上什么也没有发生。
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