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cbit-fields
Shahar Gvirtz
Shahar Gvirtz

Reputation: 2438

Converting Bit Field to int

I have bit field declared this way:

typedef struct morder {
    unsigned int targetRegister : 3;
    unsigned int targetMethodOfAddressing : 3;
    unsigned int originRegister : 3;
    unsigned int originMethodOfAddressing : 3;
    unsigned int oCode : 4;
} bitset;

I also have int array, and I want to get int value from this array, that represents the actual value of this bit field (which is actually some kind of machine word that I have the parts of it, and I want the int representation of the whole word).

Upvotes: 16

Views: 32502

Answers (4)

0xNIC
0xNIC

Reputation: 151

You can simply do

typedef struct morder {
  unsigned int targetRegister : 3;
  unsigned int targetMethodOfAddressing : 3;
  unsigned int originRegister : 3;
  unsigned int originMethodOfAddressing : 3;
  unsigned int oCode : 4;
} bitset;

short result;
std::memcpy(&result, &bitset, sizeof(short));

In this way, bitset's memory area will be copied inside a memory area and interpreted as a short.

result is a short because your bitset has size of 16bits (2bytes).

If you want a better way to set the memory size of the new buffer you can do (with c++11 or more)

std::memcpy(&result, &bitset, sizeof(decltype(result));

Another solution (very ugly and dangerous) is to reinterpret memory area:

short result = *reinterpret_cast<short*>(&bitset);

Upvotes: 0

Paul R
Paul R

Reputation: 212969

Just use a union. You can then access your data either as a 16 bit int or as individual bit-fields, e.g.

#include <stdio.h>
#include <stdint.h>

typedef struct {
    unsigned int targetRegister : 3;
    unsigned int targetMethodOfAddressing : 3;
    unsigned int originRegister : 3;
    unsigned int originMethodOfAddressing : 3;
    unsigned int oCode : 4;
} bitset;

typedef union {
    bitset b;
    uint16_t i;
} u_bitset;

int main(void)
{
    u_bitset u = {{0}};
    
    u.b.originRegister = 1;
    printf("u.i = %#x\n", u.i); 

    return 0;
}

Upvotes: 3

strager
strager

Reputation: 90012

You can use a union:

typedef union bitsetConvertor {
    bitset bs;
    uint16_t i;
} bitsetConvertor;

bitsetConvertor convertor;
convertor.i = myInt;
bitset bs = convertor.bs;

Or you can use a cast:

bitset bs = *(bitset *)&myInt;

Or you can use an anonymous struct within a union:

typedef union morder {
    struct {
        unsigned int targetRegister : 3;
        unsigned int targetMethodOfAddressing : 3;
        unsigned int originRegister : 3;
        unsigned int originMethodOfAddressing : 3;
        unsigned int oCode : 4;
    };

    uint16_t intRepresentation;
} bitset;

bitset bs;
bs.intRepresentation = myInt;

Upvotes: 15

JXG
JXG

Reputation: 7403

Please, please, do not use a union. Or, rather, understand what you're doing by using a union--preferably before you use one.

As you can see in this answer, do not rely on bitfields to be portable. Specifically for your case, the ordering of the bitfields within a struct is implementation-dependent.

Now, if your question was, how can you print out the bitfield struct as an int, for occasional private review, sure, unions are great. But you seem to want the "actual value" of your bitfields.

So: if you only work on this one machine/compiler combination, and you don't need to rely on the mathematical value of the int, so long as it makes sense, you can use unions. But if you might port your code, or if you need the "actual value" of the int, you need to write bit-manipulation code to get the bit fields into the right int bits.

Upvotes: 29

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