用C编程语言切换函数中的指针

Question

如何在函数中切换指针？

void ChangePointers(int *p_intP1, int *p_intP2); 

int main() {

int i = 100,  j = 500;
int *intP1, *intP2; /* pointers */
intP1 = &i;
intP2 = &j;
printf("%d\n", *intP1); /* prints 100 (i) */
printf("%d\n", *intP2); /* prints 500 (j) */
ChangePointers(intP1, intP2);


printf("%d\n", *intP1); /* still prints  100, would like it swapped by now */
printf("%d\n", *intP2); /* still prints  500 would like it swapped by now */
}/* end main */

void ChangePointers(int *p_intP1, int *p_intP2) {
int *l_intP3; /* local for swap */
l_intP3 = p_intP2;
p_intP2 = p_intP1;
p_intP1= l_intP3;
}

Answer 1

在C中,参数始终按值传递.虽然您要更改被调用函数内指针变量的值,但更改不会反映回调用函数.试着这样做:

void ChangePointers(int **p_intP1, int **p_intP2); /*Prototype*/

void ChangePointers(int **p_intP1, int **p_intP2) /*Definition*/
{
    int *l_intP3; /* local for swap */
    l_intP3 = *p_intP2;
    *p_intP2 = *p_intP1;
    *p_intP1= l_intP3;
}

来自main()的相应调用应该是:

ChangePointers(&intP1, &intP2);/*Passing in the address of the pointers instead of their values*/

Answer 2

你需要一个指向指针的指针.

ChangePointers(&intP1, &intP2);

void ChangePointers(int **p_intP1, int **p_intP2) {
        int *l_intP3;
        l_intP3 = *p_intP2;
        *p_intP2 = *p_intP1;
        *p_intP1 = l_intP3;
}

Answer 3

更改签名以获取指针指针.

void ChangePointers(int **p_intP1, int **p_intP2) {
int *l_intP3; /* local for swap */
l_intP3 = *p_intP2;
*p_intP2 = *p_intP1;
*p_intP1= l_intP3;
}

或者,如果您希望调用代码看起来相同,类似于C++引用,请使用宏.

void ChangePointersImpl(int **p_intP1, int **p_intP2) {
int *l_intP3; /* local for swap */
l_intP3 = *p_intP2;
*p_intP2 = *p_intP1;
*p_intP1= l_intP3;
}

#define ChangePointers(a,b) ChangePointersImpl(&a, &b)