CODEWITHSUNDEEP
cbit-manipulationbitbit-shift
Kinyugo
Kinyugo

Reputation: 471

How can I concatenate integer bit fields in c into a uint64_t integer?

I have a struct in c like this

struct RegisterStruct
{
    uint64_t b_0 : 64;
    uint64_t b_1 : 64;
    uint64_t c_0 : 64;
    uint64_t c_1 : 64;
    uint64_t c_2 : 64;
    uint64_t d_0 : 64;
    uint64_t d_1 : 64;
};

I would like to concatenate the fields into an uint64_t integer. Each of the fields should occupy a given number of bits defined as follows:

b_0: 4bits 
b_1: 4bits
c_0: 8bits
c_1: 8bits
c_2: 8bits
d_1: 16bits
d_2: 16bits

The result should be an uint64_t integer containing the concatenated bit fields(from b_0 to d_2) each occupying the given number of bits.

Here is what I have tried but I don't think this solution is correct:

struct RegisterStruct Register;

   Register.b_0 = 8;
   Register.b_1 = 8;
   Register.c_0 = 128;
   Register.c_1 = 128;
   Register.c_2 = 128;
   Register.d_0 = 32768;
   Register.d_1 = 32768;

   uint64_t reg_frame =Register.b_0<<60|Register.b_1<<56|Register.c_0<<48|Register.c_1<<40|Register.c_2<<32|Register.d_0<<16|Register.d_1;

Upvotes: 0

Views: 452

Answers (2)

Ian Abbott
Ian Abbott

Reputation: 17403

You can put the structure containing the bit fields in a union with the full 64-bit unsigned integer like this:

union RegisterUnion
    struct
    {
        uint64_t b_0 : 4;
        uint64_t b_1 : 4;
        uint64_t c_0 : 8;
        uint64_t c_1 : 8;
        uint64_t c_2 : 8;
        uint64_t d_0 : 16;
        uint64_t d_1 : 16;
    };
    uint64_t val;
};

The main problem with the above is that it is not portable. The C standard leaves the order in which bit fields are packed into their underlying storage unit type (uint64_t in this case) as an implementation defined decision. This is entirely separate from the ordering of bytes within a multi-byte integer, i.e. the little endian versus big endian byte ordering.

In addition, using uint64_t as the base type of a bit-field might not be supported. An implementation is only required to support bit-field members of types _Bool, signed int and unsigned int (or qualified versions thereof). According to the C11 draft 6.7.2.1 paragraph 5:

  1. A bit-field shall have a type that is a qualified or unqualified version of _Bool, signed int, unsigned int, or some other implementation-defined type. It is implementation-defined whether atomic types are permitted.

Upvotes: 1

0___________
0___________

Reputation: 67475

typedef union
{ 
    struct
    {
        uint64_t b_0 : 4;
        uint64_t b_1 : 4;
        uint64_t c_0 : 8;
        uint64_t c_1 : 8;
        uint64_t c_2 : 8;
        uint64_t d_0 : 16;
        uint64_t d_1 : 16;
    };
    uint64_t u64;
}R_t;


int main()
{

    R_t Register;

    Register.b_0 = 8;
    Register.b_1 = 8;
    Register.c_0 = 128;
    Register.c_1 = 128;
    Register.c_2 = 128;
    Register.d_0 = 32768;
    Register.d_1 = 32768;

    printf("%llx\n", (long long unsigned)Register.u64);
}

https://godbolt.org/z/_dYuz2

Upvotes: 0

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