C Programming - Structures, Unions, Enums - Discussion

sizeof(p), sizeof(q)

Here, we are printing the size of a pointer variable, so it will be [2] in a 16-bit system, and [4] in a 32-bit system.

Because the pointer variable just stores the memory address.

The size of a pointer in C/C++ is not fixed. It depends upon different issues like Operating system, CPU architecture etc. Usually it depends upon the word size of underlying processor for example for a 32 bit computer the pointer size can be 4 bytes for a 64 bit computer the pointer size can be 8 bytes. So for a specific architecture pointer size will be fixed.

It is common to all data types like int *, float * etc.

It will be 4, 4 since int takes 2 bytes and pointer in 16-bit takes 2 bytes since structure 2+2=4. Therefore answer is 4, 4.

As structure has a size of 2 (i.e sizeof(int) as pointer doesn't have size).

As per me, the answer is 8, 8.

It gives the output as 8, 8 in Ubuntu. Which will be the correct answer.

Yes, Agree @Hardik.

It's 8,8 for 64 bit.

For 64-bit linux os answer comes to be 8, 8.

@ALL.

The modified program to understand it better.
int main()
{

int *d;
struct node
{
int data;
struct node *link;
};
struct node *p, *q;
p = (struct node *) malloc(sizeof(struct node));
q = (struct node *) malloc(sizeof(struct node));
printf("%d, %d\n", sizeof(p), sizeof(q)); // p and q both are pointers

printf("\nsize of struct node is %d ", sizeof(struct node)); // size =integer + pointer
printf("\nsize of any pointer is %d ", sizeof(d));

return 0;
}

The pointer contains only the integer format address value of any type of the variable, it does not depend on the type of data it is storing.

So, the size of the pointer is 2 bytes.