What does 'unsigned temp:3' in a struct or union mean?

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What does this C++ code mean?

I'm trying to map a C structure to Java using JNA. I came across something that I've never seen.

The struct definition is as follows:

struct op 
{
    unsigned op_type:9;  //---> what does this mean? 
    unsigned op_opt:1; 
    unsigned op_latefree:1; 
    unsigned op_latefreed:1; 
    unsigned op_attached:1; 
    unsigned op_spare:3; 
    U8 op_flags; 
    U8 op_private;
};

You can see some variable being defined like unsigned op_attached:1 and I'm unsure what would that mean. Would that effect the number of bytes to be allocated for this particular variable?

The best thing you can do with bit fields is to avoid them. Better use masking & shifting to access groups of bits in integers in a well-defined way. — starblue, Jul 22 '10 at 05:41

score 47 · Accepted Answer · answered Jun 01 '10 at 14:43

47

This construct specifies the length in bits for each field.

The advantage of this is that you can control the sizeof(op), if you're careful. the size of the structure will be the sum of the sizes of the fields inside.

In your case, size of op is 32 bits (that is, sizeof(op) is 4).

The size always gets rounded up to the next multiple of 8 for every group of unsigned xxx:yy; construct.

That means:

struct A
{
    unsigned a: 4;    //  4 bits
    unsigned b: 4;    // +4 bits, same group, (4+4 is rounded to 8 bits)
    unsigned char c;  // +8 bits
};
//                    sizeof(A) = 2 (16 bits)

struct B
{
    unsigned a: 4;    //  4 bits
    unsigned b: 1;    // +1 bit, same group, (4+1 is rounded to 8 bits)
    unsigned char c;  // +8 bits
    unsigned d: 7;    // + 7 bits
};
//                    sizeof(B) = 3 (4+1 rounded to 8 + 8 + 7 = 23, rounded to 24)

I'm not sure I remember this correctly, but I think I got it right.

answered Jun 01 '10 at 14:43

utnapistim

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Remember kids! bit fields are compiler dependent, there is no C/C++ standard that specfies that the first 4 bits must be used on the a:4 above. – Gregor Brandt Jun 04 '10 at 22:02
1

Assuming 8 bits per byte is safe. Assuming the compiler won't use padding bits or has one specific byte order, however, isn't. – starturtle Dec 07 '16 at 09:55
C and C++ don't define a byte to be 8 bits. – user904963 Feb 01 '22 at 15:20
It is not multiple of 8，it is multiple a word，a word is related to the machine， such as 8, 16 or 32 bits – Jamishon Aug 18 '22 at 12:26

score 20 · Answer 2 · answered Jun 01 '10 at 13:17

20

It declares a bit field; the number after the colon gives the length of the field in bits (i.e., how many bits are used to represent it).

answered Jun 01 '10 at 13:17

James McNellis

348,265
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score 6 · Answer 3 · answered Jun 01 '10 at 13:18

6

unsigned op_type:9;

Means op_type is an integer variable with 9 bits.

answered Jun 01 '10 at 13:18

goedson

643
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score 4 · Answer 4 · answered Jun 01 '10 at 13:17

4

The colon modifier on integral types specifies how many bits the int should take up.

answered Jun 01 '10 at 13:17

plinth

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What does 'unsigned temp:3' in a struct or union mean?

4 Answers4

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