如何通过子函数改变参数指向的地址

怎么通过子函数改变参数指向的地址

C/C++ code

void p_add(unsigned char *pt){    pt = pt + 1;}int main(){    unsigned char s[3] = {1, 2, 3};    unsigned char *p;    p = s;    printf("p = %d\n", *p);    p_add(p);        printf("p = %d\n", *p);    return 0;}

想让在main中执行完 p_add(p); 后， p 指向s[1], = 2。
上面这段代码显然不行，

C/C++ code

unsigned char *p_add(unsigned char *pt){    pt = pt + 1;    return pt;}int main(){    unsigned char s[3] = {1, 2, 3};    unsigned char *p;    p = s;    printf("p = %d\n", *p);    p = p_add(p);        printf("p = %d\n", *p);    return 0;}

这个可行，但 p_add 的返回值我要作其它用途。

C/C++ code

void p_add(unsigned char **pt){    unsigned char *pt1;    pt1 = *pt;    pt1 = pt1 + 1;}int main(){    unsigned char s[3] = {1, 2, 3};    unsigned char *p;    p = s;    printf("p = %d\n", *p);    p_add(&p);        printf("p = %d\n", *p);    return 0;}

不行，同第一种效果相同。

怎么实现在不使用 全局变量 或 p_add返回值的情况下，通过p_add的参数将main中p向后指？


[解决办法]

C/C++ code

#include<stdio.h>void p_add(unsigned char **pt){    unsigned char **pt1;    pt1 = pt;    *pt1 = *pt1 + 1;}int main(){    unsigned char s[3] = {1, 2, 3};    unsigned char *p;    p = s;    printf("p = %d\n", *p);    p_add(&p);        printf("p = %d\n", *p);    return 0;}
[解决办法]
void p_add(unsigned char **pt)
{
   (*pt)++;
}
int main()
{
   unsigned char s[3] = {1, 2, 3};
   unsigned char *p;

   p = s;

   printf("p = %d\n", *p);
   p_add(&p);   
   printf("p = %d\n", *p);
   return 0;
}


 

 

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