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Home/ Questions/Q 3355022
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Editorial Team
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Editorial Team
Asked: May 18, 20262026-05-18T02:20:41+00:00 2026-05-18T02:20:41+00:00

#include <stdio.h> typedef struct size { unsigned int a:1; unsigned int b:31; unsigned int

  • 0
#include <stdio.h>

typedef struct size
{
        unsigned int a:1;
        unsigned int b:31;
        unsigned int c:1;
} mystruct;

int main()
{
        mystruct a;
        printf("%d", sizeof(a));
        return 0;
}
  • With int b:31, the output is 8.
  • With int b:1, the output is 4.
  • With int b:32, the output is 12.

Can somebody explain the reason for this?

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  1. Editorial Team
    Editorial Team
    2026-05-18T02:20:42+00:00Added an answer on May 18, 2026 at 2:20 am

    You don’t say whether you know what bitfields are, but I’ll assume you do.

    On your implementation, evidently unsigned int is a 32 bit integer, occupying 4 bytes. This accounts for the first and second examples. Clearly 3 bitfields totalling 33 bits don’t fit into a single unsigned int, hence the need for 8 bytes in the first example. 3 bitfields totalling 3 bits certainly do fit into an unsigned int, hence only 4 bytes in the second example.

    Furthermore, a bitfield cannot span multiple integers. This accounts for the third example. I can’t remember whether that’s a requirement of the standard, or just a detail of your implementation. Either way, since b is 32 bits, it fills a whole unsigned int on its own, forcing both of a and c to occupy their own unsigned int, before and after the middle one. Hence, 12 bytes.

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