Getting the size of an indiviual field from a c++ struct field

Question

The short version is: How do I learn the size (in bits) of an individual field of a c++ field?

To clarify, an example of the field I am talking about:

struct Test {
    unsigned field1 : 4;  // takes up 4 bits
    unsigned field2 : 8;  // 8 bits
    unsigned field3 : 1;  // 1 bit
    unsigned field4 : 3;  // 3 bits
    unsigned field5 : 16; // 16 more to make it a 32 bit struct

    int normal_member; // normal struct variable member, 4 bytes on my system
};

Test t;
t.field1 = 1;
t.field2 = 5;
// etc.

To get the size of the entire Test object is easy, we just say

sizeof(Test); // returns 8, for 8 bytes total size

We can get a normal struct member through

sizeof(((Test*)0)->normal_member); // returns 4 (on my system)

I would like to know how to get the size of an individual field, say Test::field4. The above example for a normal struct member does not work. Any ideas? Or does someone know a reason why it cannot work? I am fairly convinced that sizeof will not be of help since it only returns size in bytes, but if anyone knows otherwise I'm all ears.

Thanks!

Answer 1

You can calculate the size at run time, fwiw, e.g.:

//instantiate
Test t;
//fill all bits in the field
t.field1 = ~0;
//extract to unsigned integer
unsigned int i = t.field1;
... TODO use contents of i to calculate the bit-width of the field ...

Answer 2

You cannot take the sizeof a bitfield and get the number of bits.

Your best bet would be use #defines or enums:

struct Test {
    enum Sizes {
        sizeof_field1 = 4,
        sizeof_field2 = 8,
        sizeof_field3 = 1,
        sizeof_field4 = 3,
        sizeof_field5 = 16,
    };

    unsigned field1 : sizeof_field1;  // takes up 4 bits
    unsigned field2 : sizeof_field2;  // 8 bits
    unsigned field3 : sizeof_field3;  // 1 bit
    unsigned field4 : sizeof_field4;  // 3 bits
    unsigned field5 : sizeof_field5;  // 16 more to make it a 32 bit struct

    int normal_member; // normal struct variable member, 4 bytes on my system
};

printf("%d\n", Test::sizeof_field1); // prints 4

For the sake of consistency, I believe you can move normal_member up to the top and add an entry in Sizes using sizeof(normal_member). This messes with the order of your data, though.

Answer 3

Seems unlikely, since sizeof() is in bytes, and you want bits.

http://en.wikipedia.org/wiki/Sizeof

building on the bit counting answer, you can use.

http://www-graphics.stanford.edu/~seander/bithacks.html

Answer 4

Using ChrisW's idea (nice, by the way), you can create a helper macro:

#define SIZEOF_BITFIELD(class,member,out) { \
    class tmp_;                             \
    tmp_.member = ~0;                       \
    unsigned int tmp2_ = tmp_.member;       \
    ++tmp2_;                                \
    out = log2(tmp2_);                      \
}

unsigned int log2(unsigned int x) {
    // Overflow occured.
    if(!x) {
        return sizeof(unsigned int) * CHAR_BIT;
    }

    // Some bit twiddling...  Exploiting the fact that floats use base 2 and store the exponent.  Assumes 32-bit IEEE.
    float f = (float)x;
    return (*(unsigned int *)&f >> 23) - 0x7f;
}

Usage:

size_t size;
SIZEOF_BITFIELD(Test, field1, size);  // Class of the field, field itself, output variable.

printf("%d\n", size);  // Prints 4.

My attempts to use templated functions have failed. I'm not an expert on templates, however, so it may still be possible to have a clean method (e.g. sizeof_bitfield(Test::field1)).

Answer 5

I don't think you can do it. If you really need the size, I suggest you use a #define (or, better yet, if possible a const variable -- I'm not sure if that's legal) as so:

#define TEST_FIELD1_SIZE 4
struct Test {
    unsigned field1 : TEST_FIELD1_SIZE;
    ...
}

Answer 6

This is not possible

Answer to comment: Because the type is just an int, there is no 'bit' type. The bit field assignment syntax is just short hand for performing the bitwise code for reads and writes.