智能指针对象如何访问其他类的成员函数 [英] How object of smart pointer object accessing member funtion of other class

问题描述

我已经完成了智能指针的实现。在以下程序中：

I have gone through the smart pointer implementation. In the below program:

#include <iostream>

using namespace std;

class Car{

public:
    void Run(){
        cout<<"Car Running..."<<"\n";
    }
};

class CarSP{

    Car * sp;

public:
    //Initialize Car pointer when Car 
    //object is createdy dynamically
    CarSP(Car * cptr):sp(cptr)
    {
    }

    // Smart pointer destructor that will de-allocate
    //The memory allocated by Car class.
    ~CarSP(){       
        printf("Deleting dynamically allocated car object\n");
        delete sp;
    }

    //Overload -> operator that will be used to 
    //call functions of class car
    Car* operator-> ()
    {    
        return sp;
    }
};


//Test
int main(){

    //Create car object and initialize to smart pointer
    CarSP ptr(new Car());
    ptr.Run();

    //Memory allocated for car will be deleted
    //Once it goes out of scope.
    return 0;
}

此程序可以正常使用：

CarSP ptr(new Car());
ptr->Run();

但是 ptr 不是其对象的指针类 CarSP 中的一个现在，我的疑问是-> 如何用于与此一起访问Car成员函数。如果我使用的是 ptr.Run（）;
但它给出了错误，

But ptr is not a pointer its object of the class CarSP Now my doubt is how -> is used for accessing Car member function with this. If i am using ptr.Run(); But its giving error,

请帮助。

推荐答案

ptr->Run();

因此C ++定义了 a-> b 当且仅当 a 是指针时才为（* a）.b 。

so C++ defines a->b to be (*a).b if and only if a is a pointer.

在这种情况下， ptr 不是指针。当 a 不是指针时，它将定义为：

In this case, ptr is not a pointer. When a is not a pointer, it defines it to be:

(ptr.operator->())->b

在这种情况下， ptr 是具有 operator-> 的类类型。那个 operator-> 返回一个指向 Car 的指针。

in this case, ptr is a class type that has an operator->. That operator-> returns a pointer to Car.

所以我们有

(some pointer to Car)->b

，我们递归地将C ++规则应用于-> 。并且可以将-> Run（）作为指向汽车的指针。

and we recursively apply the C++ rule for ->. And as a pointer to car can be ->Run()'d, it works.

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