当我以这种方式修改交换函数时会发生什么？

Question

what happens when we modify the swap function this way ? I know it doesn't work but what exactly is going on ? I'm not understanding what I actually did ?

#include <stdio.h>

void swap(int*, int*);

int main(){
   int x=5,y=10;
   swap(&x, &y);
   printf("x:%d,y:%d\n",x,y);
   return 0;
 }

void swap(int *x, int *y){ 
   int* temp;
   temp=x;
   x=y;
   y=temp;
}

Answer 1

In the function

void swap(int* x, int* y){ 
    int* temp;
    temp=x;
    x=y;
    y=temp;
}

you just swap the pointer values for the two function arguments.

If you want to swap their values you need to implement it like

void swap(int* x, int* y){ 
    int temp = *x;  // retrive the value that x poitns to
    *x = *y;        // write the value y points to, to the memory location x points to
    *y = temp;      // write the value of tmp, to the memory location x points to
}

that way the swap function swaps the value for the referenced memory locations.

Answer 2

void swap(int *x, int *y){ 
   int* temp;
   temp = x;
   x = y;
   y = temp;
}

swap()需要两个指向int此处的指针，但在内部仅交换本地指针的值x和y- 它们指向的对象的地址。- 表示在结束swap()之前，它返回之前，x指向y已经指向的对象和y指向x已经指向的对象。

在对象x和y指向的调用者中没有影响。

如果要交换 whichx和y指向的对象的值，则需要这样定义swap()：

void swap(int *x, int *y){ 
   int temp;
   temp = *x;     // temp gets the int value of the int object x is pointing to.
   *x = *y;       // x gets the int value of the int object y is pointing to.
   *y = temp;     // y gets the int value of temp (formerly x).
}

示例程序（在线示例）：

#include <stdio.h> 

void swap(int *x, int *y){ 
   int temp;
   temp = *x;     // temp gets the int value of the int object x is pointing to.
   *x = *y;       // x gets the int value y of the int object y is pointing to.
   *y = temp;     // y gets the int value of temp (formerly x).
}

int main (void)
{
    int a = 1, b = 2;
    printf("Before the swap() function:\n");
    printf("a = %d b = %d\n\n",a,b);

    swap(&a,&b);

    printf("After the swap() function:\n");
    printf("a = %d b = %d",a,b);

    return 0;
}

输出：

Before the swap() function:
a = 1 b = 2

After the swap() function:
a = 2 b = 1