实例化派生类时，是否不会隐式调用抽象类构造函数? [英] Are abstract class constructors not implicitly called when a derived class is instantiated?

问题描述

以这个例子为例:

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，我希望输出是

Coming from a C# background, I expect the output to be

基础__construct
子__construct

Base __construct
Child __construct

然而，实际输出只是

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.

从你的子类的构造函数中，你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();

如果需要，传递它的参数.

Passing it parameters, if needed.

通常，您会在子类的构造函数的开头、任何特定代码之前执行此操作；这意味着，在您的情况下，您将有:

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/>';
    }
}

而且，作为参考，您可以查看 PHP 手册的这一页:Constructors和析构函数——它声明(quoting) :

注意:如果子类不隐式调用父构造函数定义一个构造函数.
为了运行父构造函数，调用parent::__construct() 在子构造函数是必需的.

Note: Parent constructors are not called implicitly if the child class defines a constructor.
In order to run a parent constructor, a call to parent::__construct() within the child constructor is required.

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