Why is the sizeof(v) showing 4 bytes??

Question

#include<stdio.h>
int main()
{
static struct my_struct{
unsigned a:1;
unsigned b:2;
unsigned c:3;
unsigned d:4;
unsigned :6;
}v={1,2,7,12};
printf("%d %d %d %d",v.a,v.b,v.c,v.d);
printf("\nSize-%d bytes",sizeof (v));
}

Answer 1

Alignment of my_struct is 4 bytes because of unsigned int and the accumulated bits only add up to 16 bits == 2 bytes so the size has to be 4 bytes.
As for the 3 bytes with unsigned chars, alignof(unsigned char) == 1, depending on how the compiler packs your structs and how the members are ordered it could be 2 or 3 bytes https://www.geeksforgeeks.org/is-sizeof-for-a-struct-equal-to-the-sum-of-sizeof-of-each-member/ is an example.

Answer 2

Type int, so 4 bytes.

Answer 3

Well then why it is giving 3 bytes when you specify the type of a,b,c,d as unsigned char??

Answer 4

Brother i am still confused about unsigned char..
Why is it giving 3 bytes and not 2 bytes..
Since we only have 16 bits in total.. so 2 bytes is enough for holding every element..
considering all elements including last as unsigned char..