访问派生类中的基类成员 [英] accessing a base class member in derived class

问题描述

我有一个简单的类如下

class A {
        protected:
        int x;
        };

class B:public A
        {
        public:
        int y;
      void sety(int d)
        {
        y=d;
        }
        int gety(){ return y;}
        };

int main()  
{
B obj;
obj.sety(10);
cout<<obj.gety();
getch();
}   

因为变量A :: x具有受保护的访问权限，如果我想访问该类中的变量，并设置一个值，而不创建一个对象A.Is是可能的如果是，那么如何？

since the variable A::x has protected access and if i want to access that variable in class B and set a value to it without creating an object of A.Is it possible?if yes then how?

编辑：我们可以访问的值A :: x使用B的对象像obj.x

can we access the value of A::x using the object of B?like obj.x

推荐答案

B 是 A ，因此创建 B 的实例正在创建一个实例的 A 。也就是说，我不知道你的实际问题是什么，所以这里有一些代码，希望澄清的事情：

B is an A, so creating an instance of B is creating an instance of A. That being said, I'm not sure what your actual question is, so here's some code that will hopefully clarify things:

class A
{
protected:
    int x;
};

class B : public A
{
public:
    int y;

    int gety() const { return y; }
    void sety(int d) { y = d; }

    int getx() const { return x; }
    void setx(int d) { x = d; }
};

int main()
{
    B obj;

    // compiles cleanly because B::sety/gety are public
    obj.sety(10);
    std::cout << obj.gety() << '\n';

    // compiles cleanly because B::setx/getx are public, even though
    // they touch A::x which is protected
    obj.setx(42);
    std::cout << obj.getx() << '\n';

    // compiles cleanly because B::y is public
    obj.y = 20;
    std::cout << obj.y << '\n';

    // compilation errors because A::x is protected
    obj.x = 84;
    std::cout << obj.x << '\n';
}

obj 可以，因为 obj 的实例 A :: x $ c>隐含地是 A 的实例。

obj can access A::x just as an instance of A could, because obj is implicitly an instance of A.

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