C Programming - Structures, Unions, Enums - Discussion

#include<stdio.h>

int main()
{
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));
return 0;
}

As we know in GCC compiler integer takes 4 byte, but when we are finding out the sizeof structure internally structure padding occur and even though in structure only one member is integer type here in program another member is structure pointer it will not be accounted as when cause its structure type, it has a special function only holds the address not allocate any memory so that when we are finding out the sizeof structure it will give the first member size in this program.

Ans for GCC -p=4, p=4, better to try and compile by own you will understand.

For better understanding. Try this:

#include<stdio.h>
#include<string.h>
int main()
{
struct node
{
int data;
//struct node *link;
//Here size of embedded pointer doesn't matter in sizeof(strucutre)
};

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));

return 0;
}

Hello friends.

The size of pointer depends upon the system with 16/32 bit.

So for 16 bit it will take 2 bytes.

For 32 bit system it will take 4 bytes.

Thank you.

If the question is printf(%d,%d\n",sizeof(struct node));

The answer would be 4.

But he is asking sizeof (p) which is a pointer.

Sizeof (pointer variable) is always 2 only irrespective of whether int* or struct node*.

Thank you so much @Chandresh.

The program prints the sizeof(p) &sizeof(q),then wat is the use of this two.

p = (struct node *) malloc(sizeof(struct node));
q = (struct node *) malloc(sizeof(struct node));

Without this two the program will give the output, then what is the need to this two lines?

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.

@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;
}

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

Yes, Agree @Hardik.

It's 8,8 for 64 bit.