我们可以重新分配C ++中的引用吗？ [英] Can we reassign the reference in C++?

问题描述

为了测试我的理解，我写了一篇文章以下小程序。看来好像我已经成功地重新分配了一个引用。有人可以向我解释我的程序中实际发生了什么吗？

  #include< iostream> 
 #include< stdio.h> 
 #include< conio.h> 
 
 using namespace std; 
 
 int main（）
 {
 int i = 5，j = 9; 
 
 int& ri = i; 
 cout<< ri是：< ri <\ n; 
 
 i = 10; 
 cout<< ri是：< ri<< \\\
; 
 
 ri = j; //>>>>这不是重新分配参考？ << 
 cout<< ri是：< ri <\ n; 
 
 getch（）; 
 return 0; 
} 
  

代码编译正确，输出结果如下：

  ri is：5 
 ri is：10 
 ri is：9 
  

解决方案

ri = j; //>>>>这不是重新分配参考？ <<

否， ri 仍然是对 i 的引用 - 您可以通过打印& ri 和



您所做的是修改 i 参考 ri 。

此外，如果您创建 c $ c> const int& cri = i; 它不会让你分配。

I have read everywhere that a reference has to be initialized then and there and can't be re-initialized again.

To test my understanding, I have written the following small program. It seems as if I have actually succeeded in reassigning a reference. Can someone explain to me what is actually going on in my program?

#include <iostream>
#include <stdio.h>
#include <conio.h>

using namespace std;

int main()
{
    int i = 5, j = 9;

    int &ri = i;
    cout << " ri is : " << ri  <<"\n";

    i = 10;
    cout << " ri is : " << ri  << "\n";

    ri = j; // >>> Is this not reassigning the reference? <<<
    cout << " ri is : " << ri  <<"\n";

    getch();
    return 0;
}

The code compiles fine and the output is as I expect:

ri is : 5
ri is : 10
ri is : 9

解决方案

ri = j; // >>> Is this not reassigning the reference? <<<

No, ri is still a reference to i - you can prove this by printing &ri and &i and seeing they're the same address.

What you did is modify i through the reference ri. Print i after, and you'll see this.

Also, for comparison, if you create a const int &cri = i; it won't let you assign to that.

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