C 40bit byte swap (endian) - c

Question

I'm reading/writing a binary file in little-endian format from big-endian using C and bswap_{16,32,64} macros from byteswap.h for byte-swapping.
All values are read and written correctly, except a bit-field of 40 bits.
The bswap_40 macro doesn't exist and I don't know how do it or if a better solution is possible.
Here is a small code showing this problem:
#include <stdio.h>
#include <inttypes.h>
#include <byteswap.h>
#define bswap_40(x) bswap_64(x)
struct tIndex {
uint64_t val_64;
uint64_t val_40:40;
} s1 = { 5294967296, 5294967296 };
int main(void)
{
// write swapped values
struct tIndex s2 = { bswap_64(s1.val_64), bswap_40(s1.val_40) };
FILE *fp = fopen("index.bin", "w");
fwrite(&s2, sizeof(s2), 1, fp);
fclose(fp);
// read swapped values
struct tIndex s3;
fp = fopen("index.bin", "r");
fread(&s3, sizeof(s3), 1, fp);
fclose(fp);
s3.val_64 = bswap_64(s3.val_64);
s3.val_40 = bswap_40(s3.val_40);
printf("val_64: %" PRIu64 " -> %s\n", s3.val_64, (s1.val_64 == s3.val_64 ? "OK" : "Error"));
printf("val_40: %" PRIu64 " -> %s\n", s3.val_40, (s1.val_40 == s3.val_40 ? "OK" : "Error"));
return 0;
}
That code is compiled with:
gcc -D_FILE_OFFSET_BITS=64 -D_LARGEFILE_SOURCE -D_LARGEFILE64_SOURCE
swap_40.c -o swap_40
How can I define bswap_40 macro for read and write these values of 40 bits doing byte-swap?

Answer 1

By defining bswap_40 to be the same as bswap_64, you're swapping 8 bytes instead of 5. So if you start with this:
00 00 00 01 02 03 04 05
You end up with this:
05 04 03 02 01 00 00 00
Instead of this:
00 00 00 05 04 03 02 01
The simplest way to handle this is to take the result of bswap_64 and right shift it by 24:
#define bswap_40(x) (bswap_64(x) >> 24)

Answer 2

EDIT
I got better performance writing this macro (comparing with my initial code, this produced less assembly instructions):
#define bswap40(s) \
((((s)&0xFF) << 32) | (((s)&0xFF00) << 16) | (((s)&0xFF0000)) | \
(((s)&0xFF000000) >> 16) | (((s)&0xFF00000000) >> 32))
use:
s3.val_40 = bswap40(s3.val_40);
... but it might be an optimizer issue. I thinks they should be optimized to the same thing.
Original Post
I love dbush's answer better... I was about to write this:
static inline void bswap40(void* s) {
uint8_t* bytes = s;
bytes[0] ^= bytes[3];
bytes[1] ^= bytes[2];
bytes[3] ^= bytes[0];
bytes[2] ^= bytes[1];
bytes[0] ^= bytes[3];
bytes[1] ^= bytes[2];
}
It's a destructive inline function for switching the bytes...

Answer 3

I'm reading/writing a binary file in little-endian format from big-endian using C and bswap_{16,32,64} macros from byteswap.h for byte-swapping.
Suggest a different way of approaching this problem: Far more often, code needs to read a file in a known endian format and then convert to the code's endian. This may involve a byte swap, it may not the trick is to write code that works under all conditions.
unsigned char file_data[5];
// file data is in big endidan
fread(file_data, sizeof file_data, 1, fp);
uint64_t y = 0;
for (i=0; i<sizeof file_data; i++) {
y <<= 8;
y |= file_data[i];
}
printf("val_64: %" PRIu64 "\n", y);
uint64_t val_40:40; is not portable. Bit ranges on types other that int, signed int, unsigned are not portable and have implementation specified behavior.
BTW: Open the file in binary mode:
// FILE *fp = fopen("index.bin", "w");
FILE *fp = fopen("index.bin", "wb");