子类中using引入基类函数时需要注意的情况
class base{public: void test(){ cout << "base::test()" << endl; } void test(int){ cout << "base::test(int)" << endl; }};
class derived : public base{public: void test(){ cout << "derived::test()" << endl; }};此时derived::test()会隐藏（hide）父类中的两个test重载函数（base::test()和base::test(int)），因此我们为子类中加上一个using声明：class derived : public base{public: void test(){ cout << "derived::test()" << endl; } using base::test;//此声明放在test前面和后面效果都一样};
现在会不会出现下面所述的情况呢？---------------------------------------------------------------------------------------------------------------既然using base::test将父类中的两个test函数都引入子类，则子类中就相当于有了一个void test()函数，所以我们在子类中重新定义的void test()函数将会和从父类中引入的void test()函数发生冲突，进而出现“重定义”错误。---------------------------------------------------------------------------------------------------------------答案是：不会！此时，子类中重新定义的void test()函数将“顶替”从父类中引入的void test()函数。（PS:从父类中引入的另外一个void test(int)函数则没有发生变化（仍然是父类中的函数实现）。）
类似的另外一种情况如下，此时加入了virtual:class base{public: virtual void test(){ cout << "base::test()" << endl; } virtual void test(double){ cout << "base::test(double)" << endl; } void test(int){ cout << "base::test(int)" << endl; }};
class derived : public base{public: void test(){ cout << "derived::test()" << endl; }};此时derived::test()虽然重写（override）了base::test()，但是同时也隐藏（hide）父类中的两个test重载函数（一个virtual函数base::test(double)和一个nonvirtual函数base::test(int)）。现在，我们为子类中加上一个using声明：class derived : public base{public: void test(){ cout << "derived::test()" << endl; } using base::test;//此声明放在test前面和后面效果都一样};与上面的类似，此时derived::test()“仍然重写”了父类的base::test()，并且与父类中的base::test(double)和base::test(int)[在子类的域]中形成重载集合。
最后，留一个思考题目，如下：class base{public: virtual void test(){ cout << "base::test()" << endl; } virtual void test(double){ cout << "base::test(double)" << endl; } void test(int){ cout << "base::test(int)" << endl; }};
class derived : public base{public: void test(){ cout << "derived::test()" << endl; } //using base::test;};
class A : public derived{public: void test(double){ cout << "A::test(double)" << endl; }};
int main(int argc, char **argv){ base *pb = new A; pb->test(2.4); return 0;}
问题：derived中的using base::test加上与否，对程序的结果有什么影响？答：没有影响。（关键点：名字解析是编译时期的事情，而virtual函数动态绑定是运行时期的事情。）(PS：但是将main函数改成“derived *pd = new A; pd->test(2.4);”，则有区别了：如果将using base::test去掉，则编译失败。)
