Create a custom bit fields using a struct

Question

I am trying to get a custom bit field I tried this method:

struct foo
{   
    unsigned z : 10;
    unsigned y : 16;
    unsigned x : 1;
    unsigned w : 16;
};


int main()
{
    foo test({0x345, 0x1234, 0x1 ,0x1234});
    char bytes[8] = {0};

    std::cout << sizeof(test) << std::endl;

    memcpy(bytes, &test, 8);

    std::cout << sizeof(bool)  << std::endl;

    for (int i = 0; i < sizeof(bytes) / sizeof(char); i++)
    {
        std::cout << std::bitset<8>(bytes[sizeof(bytes) / sizeof(char) - i - 1]);
    }
    std::cout << "" << std::endl;

    return 0;
}

With the test I am trying it returns me:

0000000000000000000100100011010000000100010010001101001101000101

(00000000000000000 | 0010 010 0011 0100 | 000001 | 0001 0010 0011 0100 |11 0100 0101 should correspond to: 0x1234 |0x1 | 0x1234 | 0x345)

I am reading it from right to left, in the right side I have the 10 first bits ( 11 0100 0101), then I have next 16 bits (0001 0010 0011 0100). After that field I am expecting just one bit for the next data, but I have 6 bits (000001) instead of (1) before the last 16 bits (0001 0010 0011 0100).

Do you have any insight for this please ?

Compilers are free to order, pad, and align bitfields however they like. In this case it seems that you compiler decided to add 5 bits of padding to x so that the overall structure would be 32-bit aligned. — Brian61354270
– Brian61354270, Commented Dec 7, 2021 at 19:59
How can I solve that ? It's very odd situation, specially since I want to have a certain definition in my bits because I am willing to use it to define a hardware message. — yat
– yat, Commented Dec 7, 2021 at 20:04
Little Endian might also skew the "expected" view of the bytes. But what problem are you trying to solve? If you are trying to guarantee a specific bit order (say for a network protocol or bus communication), write your own serialization (bit packing) code. — selbie
– selbie, Commented Dec 7, 2021 at 20:08
You literally use the bit manipulation operators like <<, >>, | or & to pack a binary message into a byte array instead of relying on the compiler to do the work for you. — selbie
– selbie, Commented Dec 7, 2021 at 20:11

Sergey Kolesnik · Accepted Answer · 2021-12-08 00:23:23Z

1

You have 5 spare bits, because the next bitfield occupies too much space to fit inside the remaining space (unsigned is 8 bits)

#include <cstdint> // types with fixed bit sizes

// force to remove padding
#pragma pack(push, 1) 

struct foo
{   
    // make bitsets occupy one address space
    uint32_t z : 10;
    uint32_t y : 16;
    uint32_t x : 1;
    // until now you have 27 bits, another 16 will not fit, 
// thus adding another 5 bits for padding. Nothing you can do.
    uint32_t w : 16; // or you can have uint16_t
    // 
};

#pragma pack(pop)

Also, bitsets can't share address space of different types of neighboring members.

edited Dec 8, 2021 at 0:23

answered Dec 8, 2021 at 0:18

Sergey Kolesnik

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7 Comments

yat Over a year ago

But there is no way to make that w "breaks" automatically in 5 bits to complete the first part from 27 to 32 instead of having 5 bits of padding and then the 11 following bits of w complete the rest ? I could do a hard coding with a union or another struct but I don't think it will be a clean code.

Sergey Kolesnik Over a year ago

@yacth you can fit several bitsets within one byte (for padding == 1 byte). But you can't fit one bitset (smaller than byte) across two bytes. Can you add a message scheme (lsb to msb) to your question? Also, you can add empty bits explicitly within one byte via uint32_t : 2 - this will add 2 bits between your bitsets.

yat Over a year ago

I want the following scheme (MSB on left side, LSB on right side): W= 16 bits | X = 1 bit | Y = 16 bits | Z = 10 bits If I understood what you told me, the compiler groups the bitset by 32 bits and since X,Y and Z adds up to 27 bits then it completes to 32 bits by adding 5 bits and then writes W. I wanted to know if I can write create a structure struct sW{uint8_t firstFiveBits : 5; uint16_t remainingElevelBits : 11;} and declare W as sW would that solve my issue here ?

yat Over a year ago

Actually this is an example that I was working on to know better about bitsets but my real data is devided by 1 bit, 32 bits, 6 bits, 64 bits, 16 bits, 2 bits and 7 bits (MSB → LSB), which will be even more trickier with these padding issues.

yat Over a year ago

Ok strangely if I define every bitfield as uint64_t it won't do a padding.

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