位域成员的大小？

Question

Would anyone know how to extract the size of a bit-field member. The below code naturally gives me the size of an integer, but how do I find out how many bits or bytes are in mybits.one ? I've tried sizeof(test.one) but which clearly won't work. I realize this is a measure of bits:

#include <iostream>

using namespace std;

int main()
{
    struct mybits {
        unsigned int one:15;
    };

    mybits test;
    test.one = 455;
    cout << test.one << endl;
    cout << "The size of test.one is:  " << sizeof(test) << endl;
}

Answer 1

The draft C++ standard says sizeof shall not be applied to a bit-field in section 5.3.3 Sizeof paragraph 1 . If you have control of the source then using an enum sounds much simpler and neater:

struct mybits
{
    enum bitFieldSizes
    {
        field1 = 15,
        field2 = 2,
        field3 = 4,
        field4 = 8,
        field5 = 31
    };

    unsigned int one : field1 ;  
    unsigned int two : field2 ;  
    unsigned int three : field3 ;
    unsigned int four : field4 ;
    unsigned int five : field5 ;
};

If you don't have control of the source it is possible to use bit hacks to obtain the size of your bit-field and std::bitset makes it easier:

#include <iostream>
#include <bitset>

struct mybits
{
    unsigned int one : 15 ;  
    unsigned int two : 2 ;  
    unsigned int three : 4 ;
    unsigned int four : 8 ;
    unsigned int five : 31 ;
};

int main()
{
    mybits mb1 ;

    mb1.one   =  ~0 ;
    mb1.two   =  ~0 ;
    mb1.three =  ~0 ;
    mb1.four  =  ~0 ;
    mb1.five  =  ~0 ;

    std::bitset<sizeof(unsigned int)*8> b1(mb1.one);
    std::bitset<sizeof(unsigned int)*8> b2(mb1.two);
    std::bitset<sizeof(unsigned int)*8> b3(mb1.three);
    std::bitset<sizeof(unsigned int)*8> b4(mb1.four);
    std::bitset<sizeof(unsigned int)*8> b5(mb1.five);

    std::cout << b1 << ":" << b1.count() << std::endl ;
    std::cout << b2 << ":" << b2.count() << std::endl ;
    std::cout << b3 << ":" << b3.count() << std::endl ;
    std::cout << b4 << ":" << b4.count() << std::endl ;
    std::cout << b5 << ":" << b5.count() << std::endl ;
}

which produces the following output:

00000000000000000111111111111111:15
00000000000000000000000000000011:2
00000000000000000000000000001111:4
00000000000000000000000011111111:8
01111111111111111111111111111111:31

Answer 2

Runtime solution, the idea from this discussion: http://social.msdn.microsoft.com/Forums/en-US/7e4f01b6-2e93-4acc-ac6a-b994702e7b66/finding-size-of-bitfield

#include <iostream>
using namespace std;

int BitCount(unsigned int value)
{
    int result = 0;

    while(value)
    {
        value &= (value - 1);
        ++result;
    }

    return result;
}

int main()
{
    struct mybits {
        unsigned int one:15;
    };

    mybits test;
    test.one = ~0;

    cout << BitCount(test.one) << endl;

    return 0;
}

Prints 15.

Answer 3

Because of padding it is not possible to see number of bits in a bit field using sizeof operator.

The only way is to open up the header where the structure is defined, and look it up.

Answer 4

There is no way to get this information (apart from reading the declaration yourself). As per the standard, [C++11]expr.sizeof§1 , it's illegal to call sizeof on a bit-field:

The sizeof operator shall not be applied to ... an lvalue that designates a bit-field.

Answer 5

A compile-time solution using constexpr :

struct S
{
    unsigned int a : 4;
    unsigned int b : 28;
};

#define GET_BIT_FIELD_WIDTH(T, f) \
    []() constexpr -> unsigned int \
    { \
        T t{}; \
        t.f = ~0; \
        unsigned int bitCount = 0; \
        while (t.f != 0) \
        { \
            t.f >>= 1; \
            ++bitCount; \
        } \
        return bitCount; \
    }()

int main()
{
    constexpr auto a = GET_BIT_FIELD_WIDTH(S, a);
    constexpr auto b = GET_BIT_FIELD_WIDTH(S, b);
    static_assert(a == 4);
    static_assert(b == 28);
}

I think it does not invoke any undefined behavior, but it does invoke some implementation-defined behavior:

1. Wrap-around of bit-fields is implementation-defined.
2. Above solution will not work for signed fields if right-shift of signed fields uses sign-extension (implemented-defined). The compiler will hit an infinite loop in that case.

Answer 6

Here is a little bit tricky generalized version:

#include <iostream>
#include <limits>
#include <bitset>
#include <cstring>
using namespace std;

template <class T>
T umaxof()
{
      T t;
      memset(&t, 0xFF, sizeof(T));
      return t;
}

template <class T>
size_t bitsof(const T& umax)
{
    return bitset<sizeof(T)*8>(umax).count();
}

int main() 
{
    struct A
    {
        uint32_t bf1:19;
        uint32_t bf2:1;
    };

    cout << bitsof(umaxof<A>().bf1) << "\n";
    cout << bitsof(umaxof<A>().bf2) << "\n";

    return 0;
}

It can be tried out at https://ideone.com/v4BiBH

Note: Works with unsigned bit fields only.