C Programming - Structures, Unions, Enums - Discussion

Total 4+4+1 = 9 bits (not bytes).
1 byte = 8 bits.
So the answer is 2 bytes.

9 bits = 2 bytes.

Actually, this is a bit field program in that, the bit field only accepts the int and char member not a float that's wise giving an error.

But in bit field we can't apply size of operator.

Then how it is?

Thanks @Piyush.

#include<stdio.h>

int main()
{
struct value
{
int bit1:2;
int bit3:9;
int bit4:6;
int d:1;
}bit;
printf("%d\n", sizeof(bit));
return 0;
}
output:4

How? Please explain.

4 also the correct answer because as mentioned it depends on the compiler.

Then the answer to given question should be 9 but it is given 2 why?

The amd size of the function returns the number of bytes and it is in bits.

Well said @Robert.

To clarify one aspect : run in gcc the following code (remember size of int = 4 bytes = 32 bits).

#include <stdio.h>

int main()
{
struct a
{
int bit1:1;
int bit3:4;
int bit4:4;
}bit;
printf("%d", sizeof(bit));

return 0;
}


* If you do simple math : 1 + 4 + 4 (number of bits afferent for each int variable ) = 9 bits.
* Since we are speaking about a machine that "thinks" in 4 bytes, it will adjust the value of bit structure to the size of 4 bytes.

Run below code to compare :

#include <stdio.h>

int main()
{
struct a
{
int bit1:1;
int bit3:4;
int bit4:4;
int bit5:24;
}bit;
printf("%d", sizeof(bit));

return 0;
}

The number of bits that the structure occupies is exactly 33.
The output of the program will be 8 ( bytes ) why? 32 + 1 = 4 bytes + another 4 bytes ( the machine thinks in its block of memory of 4 bytes ).