在位域结构上转换Endianess

Question

I need to convert a bit-field structure from little-endian to big-endia architecture. What is the best way to do that, as there will be issues in byte boundaries, if I simply swap the structure elements.

Ex Structure is:

struct {
    unsigned int    b1:1;
    unsigned int    b2:8;
    unsigned int    b3:7;
    unsigned int    b4:8;
    unsigned int    b5:7;
    unsigned int    b6:1;
}; 

Answer 1

You could use a 32 bit integer, and extract information out of it using and- and bitshift operators. With that in place, you could simply use htonl (host-to-network, long). Network byte order is big endian.

This won't be as elegant as a bit-field, but at least you'll know what you have and won't have to worry about the compiler padding your structures.

Answer 2

Processor endianness is unrelated to bit field ordering. It's quite possible to have two compilers on the same computer use opposite ordering for bitfields. So, given this:

union {
    unsigned char x;
    struct {
        unsigned char b1 : 1;
        unsigned char b2 : 7;
    };
} abc;
abc.x = 0;
abc.b1 = 1;
printf( "%02x\n", abc.x );

Unless you happen to have detailed documentation, the only way to know whether that will print out 01 or 80 is to try it.

Answer 3

In a project porting code from MIPS to Linux/x86 we did like this.

struct {

#ifdef __ONE_ENDIANESS__
    unsigned int    b1:1;
    unsigned int    b2:8;
    unsigned int    b3:7;
    unsigned int    b4:8;
    unsigned int    b5:7;
    unsigned int    b6:1;
#define _STRUCT_FILLED
#endif /* __ONE_ENDIANESS__ */

#ifdef __OTHER_ENDIANESS__
    unsigned int    b6:1;
    unsigned int    b5:7;
    unsigned int    b4:8;
    unsigned int    b3:7;
    unsigned int    b2:8;
    unsigned int    b1:1;
#define _STRUCT_FILLED
#endif /* __OTHER_ENDIANESS__ */

};

#ifndef _STRUCT_FILLED
#  error Endianess uncertain for struct
#else
#  undef _STRUCT_FILLED
#endif /* _STRUCT_FILLED */

The macros __ONE_ENDIANESS__ and __OTHER_ENDIANESS__ was the appropriate for the compiler we used so you might need to look into which is appropriate for you...

Answer 4

You have two 16 bit sections there (the first three fields and the last three fields are 16 bits).

That's only 65536 entries. So have a lookup table that holds the bit-reversed version of the fields. Wrap the struct in a union with another struct that has two 16 bit fields to make this easier?

Something like (untested, I'm not near a C compiler):

union u {
    struct {
        unsigned int    b1:1;
        unsigned int    b2:8;
        unsigned int    b3:7;
        unsigned int    b4:8;
        unsigned int    b5:7;
        unsigned int    b6:1;
     } bits;
     struct {
        uint16 first;
        uint16 second;
     } words
} ;

unit16 lookup[65536];

/* swap architectures */

void swapbits ( union u *p)
{
   p->words.first = lookup[p->words.first];
   p->words.second = lookup[p->words.second];
}

Population of the lookup table left as an exercise for the reader :)

However, read your compiler doc carefully. I'm not sure if the C standard requires that struct to fit in a word (although I'd expect most compilers to do that).

Answer 5

You want to do this between the channel (file, or network) and your structure. My preferred practice is to isolate file I/O from structures by write code that builds the file buffers in a known representation, and matching read code that reverses that transformation.

Your specific example is particularly difficult to guess at because the bitfields are defined to be unsigned int and sizeof(unsigned int) is particularly non-portable.

Assuming as a SWAG that sizeof(int)==4 then getting a pointer to the struct and reording the individual bytes probably gets you the answer you want.

The trick of defining the struct differently for different platforms might work, but in the example you cite there isn't a clean break at byte boundaries, so it is not likely to be possible to produce an equivalent of one platform in the other without splitting one or more of the fields into two pieces.

Answer 6

It should be enough to swap the bytes. Bit position within a byte is the same in big and little endian.
eg :

char* dest = (char*)&yourstruct;
unsigned int orig = yourstruct;
char* origbytes = (char*)&orig;
dest[0] = origbytes[3];
dest[1] = origbytes[2];
dest[2] = origbytes[1];
dest[3] = origbytes[0];

Answer 7

当物理布局很重要时，不应使用位字段，因为它是实现定义的，填充较大的字的顺序。

Answer 8

To get this going I finally got a solution (some what derived from epatel's solution above). This is if I convert from x86 to Solaris SPARC.

We need to first swap the incoming sturcture and then read the elements in reverse order. Basically after looking at how the structures are alligned I saw that the endianess changed both in byte ordering and bit ordering. Here is a pseudo code.

struct orig
{    
    unsigned int    b1:1;
    unsigned int    b2:8;
    unsigned int    b3:7;
    unsigned int    b4:8;
    unsigned int    b5:7;
    unsigned int    b6:1;
};

struct temp
{    
    unsigned int    b6:1;
    unsigned int    b5:7;
    unsigned int    b4:8;
    unsigned int    b3:7;
    unsigned int    b2:8;
    unsigned int    b1:1;
}temp;


func (struct orig *toconvert)
{
    struct temp temp_val;
    //Swap the bytes
    swap32byte((u32*)toconvert);
    //Now read the structure in reverse order - bytes have been swapped
    (u32*)&temp_val = (u32 *)toconvert;
    //Write it back to orignal structure
    toconvert->b6=temp_val.b6;
    toconvert->b5=temp_val.b5;
    toconvert->b4=temp_val.b4;
    toconvert->b3=temp_val.b3;
    toconvert->b2=temp_val.b2;
    toconvert->b1=temp_val.b1;

}

After some experimenting I found that this approach is only valid if the elements completely fill the structure, ie there are no unused bits.