实例化派生类时,是否隐式调用了抽象类构造函数? [英] Are abstract class constructors not implicitly called when a derived class is instantiated?
问题描述
以这个示例为例:
abstract class Base {
function __construct() {
echo 'Base __construct<br/>';
}
}
class Child extends Base {
function __construct() {
echo 'Child __construct<br/>';
}
}
$c = new Child();
来自C#背景,我希望输出为
Coming from a C# background, I expect the output to be
基础__construct
子__construct
Base __construct
Child __construct
但是,实际输出是只是
孩子__construct
Child __construct
推荐答案
否,如果子类定义了构造函数,则不会调用父类的构造函数。
No, the constructor of the parent class is not called if the child class defines a constructor.
从子类的构造函数中,您必须调用父类的构造函数:
From the constructor of your child class, you have to call the constructor of the parent's class :
parent::__construct();
根据需要传递参数。
通常,您将在子类的构造函数的开始,在任何特定的代码之前执行此操作;这意味着,在您的情况下,您将:
Generally, you'll do so at the beginning of the constructor of the child class, before any specific code ; which means, in your case, you'd have :
class Child extends Base {
function __construct() {
parent::__construct();
echo 'Child __construct<br/>';
}
}
,您可以查看PHP手册的以下页面:构造函数和析构函数-声明(quoting):
注意:父级构造函数如果子类
定义了构造函数,则不会隐式调用。
为了使
运行父构造函数,在
子构造函数中调用
parent :: __ construct()
是
Note: Parent constructors are not called implicitly if the child class defines a constructor.
In order to run a parent constructor, a call toparent::__construct()
within the child constructor is required.
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