C语言编程。 指针数组和指向数组的指针

Question

Why I'm getting compiling error in following code? and What is the difference between int (*p)[4] , int *p[4] , and int *(p)[4] ?

#include <stdio.h>
int main(){
   int (*p)[4];// what is the difference between int (*p)[4],int *p[4], and int *(p)[4]
   int x=0;
   int y=1;
   int z=2;
   p[0]=&x;
   p[1]=&y;
   p[2]=&z;
   for(int i=0;i<3;++i){
      printf("%i\n",*p[i]);
   }
   return 0;
}

Answer 1

There is no difference between

int *p[4];

and

int *(p)[4];

Both declare p to be an array of 4 pointers.

int x;
p[0] = &x;

is valid for both.

int (*p)[4];

declares p to be a pointer to an array of 4 int s.

You can get more details on the difference between

int *p[4];
int (*p)[4];

at C pointer to array/array of pointers disambiguation .

Answer 2

• int (*p)[4] : (*p) is an array of 4 int => p pointer to an array (array of 4 int )
• int *p[4] = int * p[4] : p is an array of 4 int *
• int *(p)[4] : same as the second

In your case, you should the second form.

Answer 3

This is array of 4 pointers: int *p[4] . So array will have 4 pointers, they point to int value. This is the same int *(p)[4] .

As for int (*p)[4]; this is pointer to an array of 4 integers.