重载operator []以替换向量元素

本文介绍了重载operator []以替换向量元素的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 我有一个 classA ，其中包含向量 vObject 。 class ClassA { public： ClassB ** operator []（int index）; private： vector< ClassB *> vObject }; 让我们假设vObject填充了一些classB *对象。 我想要做的是能够替换矢量的classB *元素： classA_obj [3] = classA_obj [5]; classA_obj [1] = classB_obj; 我试图返回一个ClassB Element的指针。 这是我当前的操作符实现： ClassB ** ClassA :: operator []（int index）{ return& vObject [index]; } 之后，我尝试了以下操作： * classA_obj [3] = * classA_obj [5] 使用向量完成所有工作的代码是： 向量< ClassB *> vObject; vObject.push_back（new ClassB（ARG1，ARG2））; vObject.push_back（new ClassB（ARG1，ARG2））; vObject [0] = vObject [1]; 我真的很困惑，我认为我的代码是正确的，但它实际上不工作。 上面的代码只是我实际代码的一个例子。 / p> 解决方案如果您返回参考，您将能够替换您要求的内容。 class ClassA { public： ClassB *& operator []（int index）{return vObject [index]; } private： std :: vector< ClassB *> vObject };但是，你描述你的用法的方式似乎表明你可以很容易地改变你的矢量持有 code> ClassB 对象而不是指针。 class ClassA { public ： ClassB& operator []（int index）{return vObject [index]; } private： std :: vector< ClassB> vObject }; 然后代替将 new ClassB 向量，你只需按 ClassB ： vObject.push_back ARG1，ARG2））; vObject.push_back（ClassB（ARG1，ARG2））; 这有你不需要显式访问你的容器删除指针的优势。否则，您需要更新 ClassA 才能服从第三条规则。 I have a classA which has a vector< classB* > vObject.class ClassA{public: ClassB** operator [] (int index);private: vector<ClassB*> vObject};Let's Assume that vObject is filled with some classB* objects.What I want to do is to be able to replace classB* elements of vector like that:classA_obj[3] = classA_obj[5];classA_obj[1] = classB_obj;I tried to return a pointer of ClassB Element.Here is my current operator implementation:ClassB** ClassA::operator [](int index){ return &vObject[index]; }After that i tried the following:*classA_obj[3] = *classA_obj[5]The code of doing all the work with just a vector would be:vector<ClassB*> vObject;vObject.push_back(new ClassB(ARG1,ARG2));vObject.push_back(new ClassB(ARG1,ARG2));vObject[0] = vObject[1];I'm really confused about this, I thought my code was right but it actually doesn't work. I would love if someone could tell me what I'm doing wrong.The above code is just a sample of my actual code. 解决方案 If you return a reference, you will be able to do the replacement you requested.class ClassA{public: ClassB*& operator [] (int index) { return vObject[index]; }private: std::vector<ClassB*> vObject};However, the way you have described your usage seems to indicate you can easily change your vector to hold ClassB objects instead of pointers.class ClassA{public: ClassB& operator [] (int index) { return vObject[index]; }private: std::vector<ClassB> vObject};Then instead of pushing new ClassB into the vector, you just push ClassB:vObject.push_back(ClassB(ARG1,ARG2));vObject.push_back(ClassB(ARG1,ARG2));This has the advantage you not needing to explicitly visit your container to delete the pointers. Otherwise, you will need to update ClassA to obey the rule of three. 这篇关于重载operator []以替换向量元素的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！