C Programming - Structures, Unions, Enums - Discussion

answer is 8 Bytes. Because here structure padding is coming to picture.

1. Byte memory will allocate first.
2. Bytes allocate for int but after char int will stars from multiple of 2 memory location till now structure size is (1char+1 int padding+2 int)=4 bytes.

Now 4 bytes will allocate for long here no structure padding is required because here memory will starts from multiples of 4 already, So instantly 4 bytes memory for long will allocate for long.

So total size is:(1char+1 for int padding+2 int+4 for long) = 8 bytes.

Yes, I agree @Venkatesh.

#include<stdio.h>
struct ex
{
char ch;
int i;
long int a;
};

void main()
{
struct ex x;
printf("%d",sizeof(x));
}
answer is:16 for 64 bit,8 for 32 bit;

It allocates memory like char 2 bytes(1 for char + 1 for padding(holes)), and 2 for int and long int takes 4 bytes so, total it takes 8 bytes answer is NO;.

Can anyone tell me the correct answer with the explanation?

#include<stdio.h>
struct ex
{
char ch;
int i;
long int a;
};

void main()
{
struct ex x;
printf("%d",sizeof(x));
}
This output is 7 ,So answer is A.

Answer is 7 in turbo C.

I agree to the answer "yes".

The contradiction mentioned above didn't work with the most compiler. So the answer should be yes.

Adding of extra byte as wholes in structure called structure padding.

Can any one explain what is structure padding?

I hope that the given option is wrong because structure can store different that types and it should be 1+2+4 = 7.

There was a problem with answer of question number 11. It's output result should be 7. So the answer will option "A" means TRUE. It was tested on TURBO C++ compiler...