Why is this struct size 3 instead of 2?

Question

I have defined this struct:

typedef struct {     char A:3;     char B:3;     char C:3;     char D:3;     char E:3; } col;  

The sizeof(col) give me the output of 3, but shouldn't it be 2? If I comment just one element, the sizeof is 2. I don't understand why: five element of 3 bits are equal to 15 bits, and that's less than 2 bytes.

Is there an "internal size" in defining a structure like this one? I just need a clarification, because from my notion of the language so far, I expected a size of 2 byte, not 3.

Answer 1

Because you are using char as the underlying type for your fields, the compiler tries to group bits by bytes, and since it cannot put more than eight bits in each byte, it can only store two fields per byte.

The total sum of bits your struct uses is 15, so the ideal size to fit that much data would be a short.

#include <stdio.h>  typedef struct {   char A:3;   char B:3;   char C:3;   char D:3;   char E:3; } col;    typedef struct {   short A:3;   short B:3;   short C:3;   short D:3;   short E:3; } col2;    int main(){    printf("size of col: %lu\n", sizeof(col));   printf("size of col2: %lu\n", sizeof(col2));  } 

The above code (for a 64-bit platform like mine) will indeed yield 2 for the second struct. For anything larger than a short, the struct will fill no more than one element of the used type, so - for that same platform - the struct will end up with size four for int , eight for long, etc.

Answer 2

Because you can't have a bit packet field that spans across the minimum alignment boundary (which is 1 byte) so they'll probably get packed like

byte 1   A : 3   B : 3   padding : 2 byte 2   C : 3   D : 3   padding : 2 byte 3   E : 3   padding : 5 

(the orders of field/padding inside the same byte is not intentional, it's just to give you the idea, since the compiler could laid them down how it prefers)