C中的void指针交换函数

Question

i wrote a func swap_pointers meant to replace the memory adresses both pointers point to.

meaning if before calling swap, v_a points to the int a and v_b to b , afterwards v_a should point at b , and v_b at a .

Running main() , I do see the adresses switch, and callng printf() indeed prints 6 8 , but when the lines

a = *((int *) v_a);
b = *((int *) v_b);

are executed, both vars ( a and b ) recieve the value 6 .
I can't understand why this is happening.

void swap_pointers(void **a, void **b){
  void * tmp=*a;
  *a = *b;
  *b = tmp;
}

int main() {
  int a = 8;
  int b = 6;
  void *v_a = &a;
  void *v_b = &b;
  swap_pointers(&v_a, &v_b);
  printf("%d %d\n", *((int *) v_a), *((int *) v_b));
  a = *((int *) v_a);
  b = *((int *) v_b);

}

Answer 1

After you do the swap, v_a is pointing to the address of b , and v_b is pointing to the address of a .

Then, when you do:

a = *((int *) v_a);

you are actually assigning the value of b to a which makes it 6.

Then of course, when you do the assignment to b on the next line, you are assigning the new value of a , which is 6.

Answer 2

After calling the function swap_pointers

swap_pointers(&v_a, &v_b);

the pointer v_a points to the variable b amd the pointer v_b points to the variable a.

After this statement

a = *((int *) v_a);

the variable a has the value 6 . The pointer v_b points to the variable a that at this time contains the value 6 .. So after this statement

b = *((int *) v_b);

the variable b will get the value stored in the variable a obtained in the precedent assignment statement that is the same value 6 .

To swap the values in the variables a and b you need to use an intermediate variable or to write one more function that swaps values in objects of the type int .

For example

void swap_integers( int *a, int *b )
{
    int tmp = *a;
    *a = *b;
    *b = tmp;
}

The function can be called like

swap_integers( v_a, v_b );