Pointer variables in parantheses

Asked Aug 25 '17 at 12:52

Active Aug 25 '17 at 13:00

Viewed 51 times

Why does the below code work in the way it has?How many elements do *p,*p1,*p2 pick when used in sizeof()?

#include<stdio.h>
//consider integer to be 4 bytes size   
#define R 10
#define C 20

int main()
{
    int (*p)[R][C];
    int *p1[R][C];
    int p2[R][C];
    printf("%d\n",sizeof(*p)); //output is 800
    printf("%d\n",sizeof(*p1)); //output is 160
    printf("%d",sizeof(*p2)); //output is 80
    getchar();
    return 0;
}

edited Aug 25 '17 at 13:00

mch

9,424
2
28
42

asked Aug 25 '17 at 12:52

Shivani

3

Input [the declarations here](https://cdecl.org/) and check. – Iharob Al Asimi Aug 25 '17 at 12:56
4

first case is `R * C * sizeof(int)`, second is `C * sizeof(int *)` and third `C * sizeof(int)`. – mch Aug 25 '17 at 12:58
2

`*p` is an `int[10][20]`. `*p1` is equivalent to `p1[0]`, which is an `int*[20]`. `*p2` is `p2[0]`; an `int[20]`. – molbdnilo Aug 25 '17 at 13:00
1

1) `sizeof(int[R][C])` 2) `sizeof(int *[C])`, 3) `sizeof(int[C])`, Also Use `%zu` instead of `%d`. – BLUEPIXY Aug 25 '17 at 13:01

Pointer variables in parantheses

0 Answers0