请教c函数参数入栈的问题

2018-08-20 08:44:43 494 阅读 AI 写同款 · GEO 优化AI 选题数据监测一键分发 ›

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直接上代码



char * add_normal(int num, ...) {   

    int *p;  

    char * ret;  

    p = &num + 1; //p指向参数列表下一个位置   

    ret = *(char **)p;  



    return ret;  

}  



char * add_normal2(short num, ...) {   

    short *p;  

    char * ret;  



    p = &num + 8; //p指向参数列表下一个位置    为什么要+8才能定位要下一个地址呢 ？

    ret = *(char **)p;



//  ret = ret + strlen(ret) + 1;

    return ret;  

}  



char * add_normal3(short num, ...) {   

    char * ret;  

    int  *p; 

    p = (int *)&num+4;

    ret = *(char **)p;



//  ret = ret + strlen(ret) + 1;

    return ret;  

}  



void show_test1(int num1, int num2)

{

    printf("num1=%d in address [%p]\n",num1, &num1);

    printf("num2=%d in address [%p]\n",num2, &num2);

    printf("address2-address1=%d\n",(int)&num2-(int)&num1);

}



void show_test2(short num1, int num2)

{

    printf("num1=%d in address [%p]\n",num1, &num1);

    printf("num2=%d in address [%p]\n",num2, &num2);

    printf("address2-address1=%d\n",(int)&num2-(int)&num1);

}



int main(int argc, char **argv) {   

    printf("%s\n", add_normal(5,"13dsfsdaf",2,3,4,5));  

    printf("%s\n", add_normal2(5,"13dsfsdaf","2fsdfs",3,4,5));

    printf("%s\n", add_normal3(5,"13dsfsdaf","2fsdfs",3,4,5));

    printf("%d\n", add_normal4(5,1,2,3,4,5));

    show_test1(1,2);

    show_test2(1,2);

    return 0;  

}



liunux 下



输出是

13dsfsdaf

13dsfsdaf

13dsfsdaf

num1=1 in address [0xbffe0fc0]

num2=2 in address [0xbffe0fc4]

address2-address1=4

num1=1 in address [0xbffe0fac]

num2=2 in address [0xbffe0fc4]

address2-address1=24