C Programming - Pointers - Discussion

The answer will respect with bitwise.

If 16 bit then 10, 2, 2.

If 32bit then 20, 4, 4. Here we assume that it is 32bits.

@All.

#include<stdio.h>

int main()
{
int arr[] = {12, 13, 14, 15, 16};
printf("%d, %d, %d\n", sizeof(arr), sizeof(*arr), sizeof(arr[0]));

/* (i) sizeof(arr) -> there are 5 elements in the arr[] , and it contains integers and size of integer is 4 bytes...
-> 5 * 4 = 20 bytes
(ii) (*arr) -> it will return 12
sizeof(*arr) -> As , 12 is an integer of 4 bytes...

(iii) sizeof(arr[0])) -> size of first element of arr[] (i.e. 4 bytes)...

*/

return 0;
}

Thank you @N Prathyusha.

By size of array, it means the no of elements it can have rather than the complete size.

Actually C language int datatype 2 bytes. So 2*5 elements then 10, and array size 2 and first element size of array is 2. So 10, 2, 2.

*arr is same as arr[0], coz *arr is *(arr+0), So it is the size of int. Not the size of pointer. Try printing *arr, it will print the first element.

And if you make it a char array, it will print:

5,1,1

Where 5 is 1*5.

And rest 1 are size of char not the char pointers.

There are 5 elements in an array. Each having size of 4 bytes.

Hence ans 20, ii)*ptr = declaration size of each array element=4,

iii)Array size of arr(0) = 4 bytes.
Hence answer would be 20, 4, 4.

Thank you.

The no of elements in the array are 5, each one size is 4 bytes.

So 4*5=20, and arr[0], *arr are same.. first element size is 4 bytes.

Thank you neeraj.

What if the array was declared as

int *arr[] = { 1,2,3,4,5 } ;

What would sizeof(arr) return ?