c++ - 如何防止派生类中的方法被覆盖？

Question

如何强制基类方法不被派生类覆盖？

Answer 1

如果您能够使用 final来自 C++11 的说明符，您可以防止派生类覆盖该方法。 (Microsoft 编译器似乎支持具有相似语义的相似 sealed。)

这是一个例子:

#include <iostream>

struct base {
    // To derived class' developers: Thou shalt not override this method
    virtual void work() final {
        pre_work();
        do_work();
        post_work();
    }
    virtual void pre_work() {};
    virtual void do_work() = 0;
    virtual void post_work() {};
};

struct derived : public base {
    // this should trigger an error:
    void work() {
        std::cout << "doing derived work\n";
    }
    void do_work() {
        std::cout << "doing something really very important\n";
    }
};

int main() {
    derived d;
    d.work();
    base& b = d;
    b.work();
}

这是我尝试编译时得到的结果:

$ g++ test.cc -std=c++11
test.cc:17:14: error: virtual function ‘virtual void derived::work()’
test.cc:5:22: error: overriding final function ‘virtual void base::work()’