C ++交换指针 [英] C++ Swapping Pointers

问题描述

我正在处理一个函数来交换指针，我不知道为什么这不工作。当我在交换函数中打印出r和s时，值被交换，这使我相信我操作一个我不明白的副本，因为我通过参考p和q。

  void swap（int * r，int * s）
 {
 int * pSwap = r; 
 r = s; 
 s = pSwap; 
 return; 
} 
 
 int main（）
 {
 int p = 7; 
 int q = 9; 
 swap（& p，& q）; 
 cout<< p =<< p < q =< q<< endl; 
 return 0; 
} 
  

列印：p = 7q = 9

<在你的 swap 函数里面，你只是改变指针的方向，即改变对象的指针指向（这里，具体来说，它是对象的地址 p 和 q ）。指针指向的对象根本不改变。

您可以直接使用 std :: swap 。或者编写你的交换函数，如下所示：

  void swap（int * r，int * s）
 {
 int temp = * r; 
 * r = * s; 
 * s = temp; 
 return; 
} 
  

I'm working on a function to swap pointers and I can't figure out why this isn't working. When I print out r and s in the swap function the values are swapped, which leads me to believe I'm manipulating a copy of which I don't understand because I pass by reference of p and q.

void swap(int *r, int *s)
{
    int *pSwap = r;
    r = s;
    s = pSwap;
    return;
}

int main()
{
    int p = 7;
    int q = 9;  
    swap(&p, &q);
    cout << "p = " << p << "q= " << q << endl;
    return 0;
}

Prints: p = 7q = 9

解决方案

Inside your swap function, you are just changing the direction of pointers, i.e., change the objects the pointer points to (here, specifically it is the address of the objects p and q). the objects pointed by the pointer are not changed at all.

You can use std::swap directly. Or code your swap function like the following:

void swap(int *r, int *s)
{
   int temp = *r;
   *r = *s;
   *s = temp;
   return;
} 

这篇关于C ++交换指针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IT屋！