对象内部的对象出现问题，并维护一个对象的一个​​实例

Question

在其他对象中传递对象实例时遇到麻烦。在主应用程序的结尾，我希望iDogClassOne和iDogClassTwo引用同一对象，并且我希望该对象包含放置在variableA中的数字。我在这里做错了什么？ tempDogClass，iDogClassOne和iDOgClassTwo的地址都不同，所以我可以肯定，我当前正在创建3个我不想做的dog类实例。..任何帮助将不胜感激。

class dogClass{
public:
    void setVariableA(std::vector<double> newNum) {
        variableA.push_back(newNum);
    }
    void printVariableA() {
        if (variableA.size() != 0) {
            std::cout << variableA[0][0] << std::endl;
            std::cout << variableA[0][1] << std::endl;
        }
    }
    dogClass &dogClass::operator=(const dogClass &src) {
        variableA = src.variableA;
        return *this;
    }
private:
    std::vector<std::vector<double>> variableA;
};

class animalClass{
public:
    void addNumberToDog(std::vector<double> num) {
        dogClass tempDogClass;
        getDogOne(tempDogClass);
        tempDogClass.setVariableA(num);
        std::cout << "Address of temp Dog: " << &tempDogClass << std::endl;
        std::cout << "Variables in tempDog:\n";
        tempDogClass.printVariableA();
        std::cout << std::endl;
    }
    void getDogOne(dogClass &dogOne) {
        dogOne = dogs[0];
    }
    void setDogOne(dogClass &dogOne) {
        dogs.push_back(dogOne);
    }
private:
    std::vector<dogClass> dogs;
};


int main() {
    animalClass iAnimalClass;
    dogClass iDogClassOne;

    iAnimalClass.setDogOne(iDogClassOne);

    std::vector<double> newNum;
    newNum.push_back(12);
    newNum.push_back(15);

    iAnimalClass.addNumberToDog(newNum);

    dogClass iDogClassTwo;

    iAnimalClass.getDogOne(iDogClassTwo);

    //Why are these not equal.  Also why does dog Two not contain the values in dog One?
    std::cout << "Address of dog One: " << &iDogClassOne << std::endl;
    std::cout << "Variables in dog One:\n";
    iDogClassOne.printVariableA();
    std::cout << std::endl;
    std::cout << "Address of dog Two: " << &iDogClassTwo << std::endl;
    std::cout << "Variables in dog Two:\n";
    iDogClassTwo.printVariableA();
    std::cout << std::endl;
    return 0;

}

控制台应用程序的输出（为什么iDogClassOne和iDogClassTwo不包含数字？为什么tempDog，iDogClassOne和iDogClassTwo不是同一对象？）

Address of temp Dog: 00000000002AF8A0
Variables in tempDog:
12
15

Address of dog One: 00000000002AF938
Variables in dog One:

Address of dog Two: 00000000002AF920
Variables in dog Two:

Answer 1

因为您正在许多地方制作dogClass对象的副本。 请考虑值，引用和指针的区别。 例如，在main()函数中，您调用iAnimalClass.setDogOne(iDogClassOne);，该参数作为引用传递，可以。但是随后您将其复制到向量中。这有点棘手，因为您不能在向量中传递引用。在这种情况下，您将必须使用指向dogClass的指针向量。

现在您已经有两个dogClass的实例。主要功能中的一个，animalClass向量中的另一个。

下一个问题是在main()函数中，您可以实例化 iDogClassTwo。

现在您有三个dogClass的实例。

然后调用iAnimalClass.getDogOne(iDogClassTwo);，它将狗从向量复制到主函数中的狗。

这是更正的版本：

#include <vector>
#include <iostream>

class dogClass{
public:
    void setVariableA(std::vector<double> newNum) {
        variableA.push_back(newNum);
    }
    void printVariableA() {
        if (variableA.size() != 0) {
            std::cout << variableA[0][0] << std::endl;
            std::cout << variableA[0][1] << std::endl;
        }
    }
    dogClass & operator=(const dogClass &src) {
        variableA = src.variableA;
        return *this;
    }
private:
    std::vector<std::vector<double>> variableA;
};

class animalClass{
public:
    void addNumberToDog(std::vector<double> num) {
        dogClass& tempDogClass = getDogOne();
        tempDogClass.setVariableA(num);
        std::cout << "Address of temp Dog: " << &tempDogClass << std::endl;
        std::cout << "Variables in tempDog:\n";
        tempDogClass.printVariableA();
        std::cout << std::endl;
    }
    dogClass& getDogOne() {
        return *dogs[0];
    }
    void setDogOne(dogClass &dogOne) {
        dogs.push_back(&dogOne);
    }
private:
    std::vector<dogClass*> dogs;
};


int main() {
    animalClass iAnimalClass;
    dogClass iDogClassOne;

    iAnimalClass.setDogOne(iDogClassOne);

    std::vector<double> newNum;
    newNum.push_back(12);
    newNum.push_back(15);

    iAnimalClass.addNumberToDog(newNum);

    dogClass& iDogClassTwo = iAnimalClass.getDogOne();

    //Why are these not equal.  Also why does dog Two not contain the values in dog One?
    std::cout << "Address of dog One: " << &iDogClassOne << std::endl;
    std::cout << "Variables in dog One:\n";
    iDogClassOne.printVariableA();
    std::cout << std::endl;
    std::cout << "Address of dog Two: " << &iDogClassTwo << std::endl;
    std::cout << "Variables in dog Two:\n";
    iDogClassTwo.printVariableA();
    std::cout << std::endl;
    return 0;

}

Answer 2

  

为什么iDogClassOne和iDogClassTwo不包含数字？

int main()
{
    animalClass iAnimalClass;
    dogClass iDogClassOne;

    iAnimalClass.setDogOne(iDogClassOne);  // pushes a copy of iDogClassOne 
                                           // into a vector animalClass::dogs

    std::vector<double> newNum;
    newNum.push_back(12);
    newNum.push_back(15);

    iAnimalClass.addNumberToDog(newNum);  // Only operates on function local variable
                                          // variable tempDogClass without having
                                          // any effect on the outside world.

    dogClass iDogClassTwo;

    iAnimalClass.getDogOne(iDogClassTwo); // copies the copy of iDogClassOne earlier pushed 
                                          // into the vector animalClass::dogs to iDogClassTwo

    std::cout << "Address of dog One: " << &iDogClassOne << std::endl;
    std::cout << "Variables in dog One:\n";
    iDogClassOne.printVariableA();
    std::cout << "\nAddress of dog Two: " << &iDogClassTwo << std::endl;
    std::cout << "Variables in dog Two:\n";
    iDogClassTwo.printVariableA();
}

  

为什么tempDog，iDogClassOne和iDogClassTwo不是同一对象？

在下面的示例中，为什么a，b和c不是同一对象？

void func()
{
    int c;
}

int main()
{
    int a;
    int b;
    func();
}