CODEWITHSUNDEEP
c++c++11bit-manipulation
yat
yat

Reputation: 35

Create a custom bit fields using a struct

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 ?

Upvotes: 1

Views: 1145

Answers (1)

Sergey Kolesnik
Sergey Kolesnik

Reputation: 3678

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.

Upvotes: 1

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