rules:

1. 指针是变量，该变量有类型 example: int * p;

2. 地址是一个值，该值是指针变量的值，该值有类型， example: (void*)0

3. 函数名是一个字符地址常量， 默认的类型为其对应的函数类型。 int a(int), int (*b)(int) b(指针）= a(与b同类型的函数的首地址）

4. 数组名是一个数组类型的字符地址常量，默认的类型为数组中下一维元素的类型。example:

   int a[10], a 字符地址常量，默认为int型指针值

   int a[10][10], a 字符地址常量，默认类型与a[0]相同 int (*b)[10]这中类型

5. 数组名取地址运算对应的地址的类型为数组型：

   int a [10], int *b; int (*c)[10]; b=a yes; c=a no; c=&a yes;

6. 函数形参中声明的数组等于一个数组下一维元素类型的指针，如int fun(int a[])与int fun(int *a)一样；其中的a就是一个指针，不在理解为一个函数名，这是函数传递数组中的一种强制方法，编译器就是这样处理的，没为什么。函数中的a++运算有意思，因为a是一个变量，int a[10] a++运算没有意义，因为a是一个字符地址常量

example:

#include
int fun(int a[])
{
        //infunction a = int *a;
 printf("data team name:\n"); 
 printf("0: %d\n", *(a++));
 printf("1: %d\n", *(a++));
 printf("2: %d\n", *(a++));
 
}
int fun1(int *a)
{
        //infunction a = int *a;
 printf("data team element pointer :\n");  
 printf("0: %d\n", *(a++));
 printf("1: %d\n", *(a++));
 printf("2: %d\n", *(a++));
 
}
int fun2(int a[])
{
        //infunction a = int *a;
 printf("data team name:\n"); 
 printf("0: %d\n", *(a++));
 printf("1: %d\n", *(a++));
 printf("2: %d\n", *(a++));
 
}
int fun3(int a[][3])
{
        //infunction a = int (*a)[3];
 printf("data team[3][3] name:\n"); 
 printf("0: %d\n", *(*(a)+0));
 printf("0: %d\n", *(*(a)+1));
 printf("0: %d\n", *(*(a)+2));
 
 printf("1: %d\n", *(*(++a)+0));
 printf("1: %d\n", *(*(a)+1));
 printf("1: %d\n", *(*(a)+2));
 
 printf("2: %d\n", *(*(++a)+0));
 printf("3: %d\n", *(*(a)+1));
 printf("3: %d\n", *(*(a)+2));
 
}
int fun4(int (*a)[3])
{
        //infunction a = int (*a)[3];
 printf("2data team[3][3] name:\n"); 
 printf("0: %d\n", *(*(a)+0));
 printf("0: %d\n", *(*(a)+1));
 printf("0: %d\n", *(*(a)+2));
 
 printf("1: %d\n", *(*(++a)+0));
 printf("1: %d\n", *(*(a)+1));
 printf("1: %d\n", *(*(a)+2));
 
 printf("2: %d\n", *(*(++a)+0));
 printf("3: %d\n", *(*(a)+1));
 printf("3: %d\n", *(*(a)+2));
 
}

int main()
{
 int a[]={1,2,3};
 int b[3][3]={{1,2,3},{4,5,6},{7,8,9}};
 double c;
 int* p2;
 char *p;
 int (*p1)[];//team type ponter
 p =(char*)a;//type problem
 p= (char*)&c;//type problem
 p1 = &a;// team tyep to team type pointer
 p2 = a;// a defualt type is element type
 //p2 = &a; cause type error, team pointer assign element type
 fun(a);
 fun1(a);
 //fun2(b);//b is int (*p)[3]type , a is int*p
 fun3(b);
 fun4(b);
 return(0);
}

result:

[zhliu@sdcsn05 ~]$ gcc -o test41 test41.c
[zhliu@sdcsn05 ~]$ test41
data team name:
0: 1
1: 2
2: 3
data team element pointer :
0: 1
1: 2
2: 3
data team[3][3] name:
0: 1
0: 2
0: 3
1: 4
1: 5
1: 6
2: 7
3: 8
3: 9
2data team[3][3] name:
0: 1
0: 2
0: 3
1: 4
1: 5
1: 6
2: 7
3: 8
3: 9