在构造函数中调用虚方法：Java和C ++之间的区别 [英] Invoking virtual method in constructor: difference between Java and C++

问题描述

在Java中：

class Base {
    public Base() { System.out.println("Base::Base()"); virt(); }
    void virt()   { System.out.println("Base::virt()"); }
}

class Derived extends Base {
    public Derived() { System.out.println("Derived::Derived()"); virt(); }
    void virt()      { System.out.println("Derived::virt()"); }
}

public class Main {
    public static void main(String[] args) {
        new Derived();
    }
}

这将输出

Base::Base()
Derived::virt()
Derived::Derived()
Derived::virt()

但是，在C ++中，结果不同：

However, in C++ the result is different:

Base::Base()
Base::virt() // ← Not Derived::virt()
Derived::Derived()
Derived::virt()

（请参阅http://www.parashift.com/c++-faq-lite/calling-virtuals-from-ctors.html for C ++ code）

(See http://www.parashift.com/c++-faq-lite/calling-virtuals-from-ctors.html for C++ code)

Java和C ++之间有什么区别？是vtable初始化的时间吗？

What causes such a difference between Java and C++? Is it the time when vtable is initialized?

编辑：我明白Java和C ++机制。

I do understand Java and C++ mechanisms. What I want to know is the insights behind this design decision.

推荐答案

这两种方法都有明显的缺点：

Both approaches clearly have disadvatages:

• 在Java中，调用使用的方法不能正确使用 this ，因为它的成员


• 在C ++中，如果您不知道C ++如何构建类，则会调用非直观的方法（即不是派生类中的方法）。

• In Java, the call goes to a method which cannot use this properly because its members haven’t been initialised yet.
• In C++, an unintuitive method (i.e. not the one in the derived class) is called if you don’t know how C++ constructs classes.

为什么每种语言都做的是一个开放的问题，但两者都可能声称是更安全的选择：C ++的方式防止使用未初始化的成员; Java的方法允许在一个类的构造函数（这是一个完全有效的用例）内部的多态语义（在某种程度上）。

Why each language does what it does is an open question but both probably claim to be the "safer" option: C++’s way prevents the use of uninitialsed members; Java’s approach allows polymorphic semantics (to some extent) inside a class’ constructor (which is a perfectly valid use-case).

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