【C++零散小记录】类中重载返回类型为常量引用和非常量引用的函数

#include 

class hyp
{
public:
    //这里后面必须要加上const修饰符，否则这两个test函数不能重载
    int const& test() const
    {
        std::cout << "const" << std::endl;
        return num;
    }

    //int &test()
    //{
    //   std::cout << "non-const" << std::endl;
    //    return num;
    //}
    
    int getNum()
    {
        return num;
    }

private:
    int num = 2;

};


int main()
{
    hyp h;
    int out = 3;
    std::cout << "out value = " << out <<std::endl;
    std::cout << "%%%%%%%%%%"  <<std::endl;
    
    //返回类型为non-const和const都可以用
    out = h.test();
    std::cout << "out value = " << out <<std::endl;
    std::cout << "num value = " << h.getNum() <<std::endl;
    std::cout << "%%%%%%%%%%"  <<std::endl;
    
    //返回类型为non-const和const都可以用
    const int conOut = h.test();
    std::cout << "conOut value = " << conOut <<std::endl;
    std::cout << "num value = " << h.getNum() <<std::endl;
    std::cout << "%%%%%%%%%%"  <<std::endl;
    
    //如果类中只有返回类型为const&的test函数，那么这个h.test()=4就会报错
    //也就是说这里只能调用h.test()的non-const reference的返回类型，就可成功赋值
    //h.test() = 4;
    //std::cout << "num value = " << h.getNum() <
}

此时的输出为：

如果把int &test() 和 h.test() = 4;的注释符都去掉：