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;
}
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 ints.
You can get more details on the difference between
int *p[4];
int (*p)[4];
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 secondIn your case, you should the second form.
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.
