相关疑难解决方法(0)

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,结果是不同的:

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

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

什么导致Java和C++之间的这种差异？这是vtable初始化的时间吗？

编辑:我确实理解Java和C++机制.我想知道的是这个设计决定背后的见解.