What is the use of declaring different datatypes inside bitfields?

Question

A typical use of bitfield is to declare a space efficient variable smaller than 8 bits. What i don't understand is the value of declaring those bits as short, int , long , bool etc. For example

typedef struct{
    int first:3,
    short second:3,
    char third:3
    } somestruct;

In above case, all 3 variables, i.e. first, second and third are 3 bit long. What is the value of declaring the variable first as int, second as short and third as char?

Or, why is even a data type required? I should be able to declare the above as

typedef struct{
    first:3,
    second:3,
    third:3
    } modifiedstruct;

The modifiedstruct assumes no datatype for the variables first, second and third. The responsibility of interpreting the 3 bits as character, numeric or floating should be responsibility of something else.

Both gcc and g++ on linux allow the above behavior.

score 5 · Accepted Answer · answered Feb 09 '12 at 17:25

Actually, the C standard only allows bitfields to be of type signed int or unsigned int (and _Bool in C99). If you can throw a short, long or char in there, that's a compiler extension.

As to why, the main reason is signedness. Consider:

struct {
   int s: 3;
   unsigned u: 3;
} bf;

bf.s = 7;
bf.u = 7;

Both of these bitfields are all ones. However, C preserves sign, so:

(int)bf.s == -1    // Because signed conversions preserve the sign bit
bf.s >> 1 == -1    // So do right shifts on signed values

while:

(int)bf.u == 7     // Because the source is unsigned and so just a series of bits
bf.u >> 1 == 3     // Unsigned right shifts are just moving bits around as well

For compilers that allow char, it's probably the same sort of thinking. The default signedness of char is implementation-defined so if you want a bitfield's signedness to match your compiler's char's signedness, you can define it as char.

My whole question is why introduce different datatypes in bitfields such as char(Bool) and int and second, what is difference between int x:2 and char y:2 ? — Jimm, Feb 10 '12 at 18:37

score 2 · Answer 2 · answered Feb 09 '12 at 15:34

2

The size is for the storage (or transportation) of the data. The data type is how you intend to use and process that information.

answered Feb 09 '12 at 15:34

Ed Heal

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score 0 · Answer 3 · answered Feb 09 '12 at 15:46

Using any other type than int for a bit-field is implementation-defined behavior. So without knowing what specific compiler you are using, there is no telling what that code does, or if it even compiles.

Bit-fields in general are very poorly defined by the standard, and therefore completely unportable. Since bit-fields are also pretty much a superfluous feature of the C language, it is most likely wiser to do bit manipulations using the bit-wise operators.

score 0 · Answer 4 · answered Feb 09 '12 at 16:46

0

The only time that types inside a bitfield matter to me is signed vs. unsigned types. For fields wider than 1 , this can make a difference in how the data is interpreted during assignment.

answered Feb 09 '12 at 16:46

AShelly

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What is the use of declaring different datatypes inside bitfields?

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