I think the output of this program should 1. But program give 4, can anyone explain?

Question

#include <stdio.h>
struct bits{
unsigned int b1:1;
unsigned int b2:1;
unsigned int b3:1;
unsigned int b4:1;
unsigned int b5:1;
unsigned int b6:1;
unsigned int b7:1;
unsigned int b8:1;

}pr;
int main() {
printf("%ld",sizeof(struct bits));
return 0;
}

/*
Its output should be 1 byte because struct contains bitfields and when calculate overall size of struct should be 8 bit means 1 byte .
There is no case of structure padding because size of struct is 1 byte . So why its size become 4 byte
*/

Answer 1

But why you expected the size to be 1?

Answer 2

I'm not exactly an expert on this but basically your CPU has 32 or 64 physical wires that go to RAM ("bus width") so it will always read 4/8 bytes of memory and never less than that.

Say you have your 1-byte `bits` struct in memory location 0x0000 and a 4-byte struct called `cheese` directly after that in locations 0x0001 - 0x0004.

If your CPU wants to read `cheese` from RAM it will have to do two read operations: 0x0000 - 0x0003 and 0x0004 - 0x0007. That's because the CPU can only read memory in this 4 byte grid, and `cheese` is misaligned.

If `bits` is padded to 4 bytes, the problem disappears. It's a waste of space but makes the program faster.

I hope that made sense!

Answer 3

Structure padding *is* the answer! Your struct is 1 byte in size but will be padded to 4 bytes.

Answer 4

But why