协变返回类型和分派

Question

我被这个简单的程序弄糊涂了:

#include <iostream>
using namespace std;
struct Base {
  virtual ~Base() {}
  virtual Base* concrete() { cout << "concrete returning base - "; return this;}
  void stat() { cout << "I'm a BASEn";}
};
struct Derived: Base {
  Derived* concrete() override { cout << "concrete returning derived - "; return this;}
  void stat() { cout << "I'm a DERIVEDn";}
};

int main() {
  Base * bd = new Derived;
  bd->stat();  // prints: "I'm a BASE"
  bd->concrete()->stat(); //prints: "concrete returning derived - I'm a BASE"
  delete bd;
  return 0;
}

为什么最后一行不打印concrete returning derived - I'm a DERIVED ?这是怎么回事?

Answer 1

你误解了协变返回类型的效用…当调用上下文知道它正在处理一个Derived类型的对象时，它就会启动:

Derived d;
d->concrete()->stat();

在您的情况下，concrete在Base*上被调用，因此尽管它是多态调度并返回Derived*，但该指针在调用上下文中被视为Base*，并且它是Base*静态类型，确定静态调度到stat()。