Question

所以我做了两个课程：

class Husky{
private: 
    int age;
    int weight;


public: 
    //Usual Constructor, set, get functions here
    //......
    void foo(); //<-----This is the problem function. In here, I change the weight variable.
}

这是第二堂课：

class Dog{
private:
    vector<Husky> h;

public:
    Husky getHusky(int index);  //Returns the Husky element at the index given.
    void setHusky(Husky x){h.push_back(x);}  //Adds a Husky element to the 'h' array.

}

在我的主要：

int main(){

    //Note: the initialized value of `weight` is 0.

    Husky p;
    p.foo();
    cout << p.getWeight() << endl;  //This couts 5. foo() DID change 'weight'.


    Dog g;
    Husky s;
    g.push_back(s);

    g.getHusky(0).foo();

    cout << g.getHusky(0).getWeight() << endl;  //This couts 0. foo() DID NOT change 'weight'. 
    // Why did this not print out 5?

}

请给我一些线索或其他东西，指出我正确的方向。

Answer 1

Dog::getHusky()正在返回所请求的Husky的副本，因为您要按值返回它。然后，您在副本上而不是在原始上调用foo()。这就是你看到你所看到的结果的原因。

解决方案是更改getHusky()以通过引用而不是按值返回请求的Huskey，例如：

Husky& getHusky(int index);

Husky& Dog::getHusky(int index) { return h[index]; }

请注意返回类型附加的&。

如何从另一个类中正确访问类函数？

1 个答案: