左值和右值的C ++运算符重载[] [英] C++ Operator Overloading [ ] for lvalue and rvalue

问题描述

我制作了一个包含整数数组的Array类.从main函数中，我试图像使用main中声明的数组那样使用[]获取Array中的数组元素.我在下面的代码中重载了运算符[]；第一个函数返回一个左值，第二个函数返回一个右值(未显示构造函数和其他成员函数.)

I made a class Array which holds an integer array. From the main function, I'm trying to get an element of the array in Array using [ ] as we do for arrays declared in main. I overloaded the operator [ ] as in the following code; the first function returns an lvalue and the second an rvalue (Constructors and other member functions are not shown.)

#include <iostream>
using namespace std;

class Array {
public:
   int& operator[] (const int index)
   {
      return a[index];
   }
   int operator[] (const int index) const
   {
      return a[index];
   }

private:
   int* a;
}

但是，当我尝试从main调用这两个函数时，即使未将变量用作左值，也只能访问第一个函数.如果可以仅通过使用lvalue函数来处理所有事情，那么我看不到为rvalue创建单独的函数的意义.

However, when I try to call those two functions from main, only the first function is accessed even when the variable is not used as an lvalue. I can't see the point of creating a separate function for an rvalue if everything can be taken care of just by using the lvalue function.

以下代码是我使用的主要功能(运算符<<已适当地重载.):

The following code is the main function I used (Operator << is appropriately overloaded.):

#include "array.h"
#include <iostream>
using namespace std;

int main() {
   Array array;
   array[3] = 5;                // lvalue function called
   cout << array[3] << endl;    // lvalue function called
   array[4] = array[3]          // lvalue function called for both
}

有什么方法可以调用右值函数?是否还需要为左值和右值定义函数?

Is there any way I can call the rvalue function? Is it also necessary to define functions for both lvalue and rvalue?

推荐答案

第二个函数是const member function，如果您有const实例，它将被调用:

The second function is a const member function and it will be called if you have a const instance:

const Array array;
cout << array[3] << endl;  // rvalue function called

调用这些"lvalue"和"rvalue"函数并不常见.如果需要，可以定义一个const来返回const引用.

It isn't conventional to call these "lvalue" and "rvalue" functions. And you could define the const one to return a const reference if you want.

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