问几个比较弱的问题 ！！呵呵
主函数中有int *p[n],int (*p)[n];把这两个的实参传递出去。
写出外调函数fun的形参的类型

[解决办法]

C/C++ code

#include <stdio.h>void foo(int *p1[], int (*p2)[3]){    printf("%d\n", *p1[0]);    printf("%d\n", *p1[1]);    printf("%d\n", *p1[2]);    printf("===\n");    printf("%d\n", (*p2)[0]);    printf("%d\n", (*p2)[1]);    printf("%d\n", (*p2)[2]);}int main(void){    int a = 1, b = 2, c = 3;    int v[] = { a, b, c };    int *p1[] = { &a, &b, &c };    int (*p2)[3] = &v;    foo(p1, p2);    return 0;}123===123Press any key to continue