#include <iostream>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() <<std::endl;
}

此时的输出为：

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