问题
I've had to do this many times in the past, and I've never been satisfied with the results.
Can anyone suggest a fast way of copying a contiguous bit array from source to destination where both the source and destination's may not be aligned (right shifted) on convenient processor boundaries?
If both the source and destination's aren't aligned , the problem can quickly be changed into one where only either of them aren't aligned (after the first copy say).
As a starting point, my code inevitably ends up looking something like the following (untested, ignore side effects this is just an off the cuff example):
const char mask[8] = { 1, 3, 7, 15, 31, 63, 127, 255 };
/* Assume:
* - destination is already zeroed,
* - offsets are right shifts
* - bits to copy is big (> 32 say)
*/
int bitarray_copy(char * src, int src_bit_offset, int src_bit_len,
char * dst, int dst_bit_offset) {
if (src_bit_offset == dst_bit_offset) { /* Not very interesting */
} else {
int bit_diff_offset = src_bit_offset - dst_bit_offset; /* assume positive */
int loop_count;
char c;
char mask_val = mask[bit_diff_offset];
/* Get started, line up the destination. */
c = (*src++ << bit_diff_offset) | ((*src >> (8 - bit_diff_offset)) & mask_val);
c &= mask[8-dst_bit_offset];
*dst++ |= c;
src_bit_len -= 8 - dst_bit_offset;
loop_count = src_bit_len >> 3;
while (--loop_count >= 0)
* dst ++ = (*src++ << bit_diff_offset) | ((*src >> (8 - bit_diff_offset)) & mask_val);
/* Trailing tail copy etc ... */
if (src_bit_len % 8) /* ... */
}
}
(actually this is better than I've done before. It doesn't look too bad)
回答1:
Your inner loop takes pieces of two bytes and moves them to a destination byte. That's almost optimal. Here are a few more hints in no particular order:
- There's no need to limit yourself to a byte at a time. Use the largest integer size your platform will let you get away with. This of course will complicate your starting and trailing logic.
- If you use unsigned chars or integers, you may not need to mask the second piece of the source after it's shifted right. This will depend on your compiler.
- If you do need the mask, make sure your compiler is moving the table lookup outside of the loop. If it isn't, copy it to a temporary variable and use that.
回答2:
This is what I ended up doing. (EDIT Changed on 8/21/2014 for a single bit copy bug.)
#include <limits.h>
#include <string.h>
#include <stddef.h>
#define PREPARE_FIRST_COPY() \
do { \
if (src_len >= (CHAR_BIT - dst_offset_modulo)) { \
*dst &= reverse_mask[dst_offset_modulo]; \
src_len -= CHAR_BIT - dst_offset_modulo; \
} else { \
*dst &= reverse_mask[dst_offset_modulo] \
| reverse_mask_xor[dst_offset_modulo + src_len]; \
c &= reverse_mask[dst_offset_modulo + src_len]; \
src_len = 0; \
} } while (0)
static void
bitarray_copy(const unsigned char *src_org, int src_offset, int src_len,
unsigned char *dst_org, int dst_offset)
{
static const unsigned char mask[] =
{ 0x00, 0x01, 0x03, 0x07, 0x0f, 0x1f, 0x3f, 0x7f, 0xff };
static const unsigned char reverse_mask[] =
{ 0x00, 0x80, 0xc0, 0xe0, 0xf0, 0xf8, 0xfc, 0xfe, 0xff };
static const unsigned char reverse_mask_xor[] =
{ 0xff, 0x7f, 0x3f, 0x1f, 0x0f, 0x07, 0x03, 0x01, 0x00 };
if (src_len) {
const unsigned char *src;
unsigned char *dst;
int src_offset_modulo,
dst_offset_modulo;
src = src_org + (src_offset / CHAR_BIT);
dst = dst_org + (dst_offset / CHAR_BIT);
src_offset_modulo = src_offset % CHAR_BIT;
dst_offset_modulo = dst_offset % CHAR_BIT;
if (src_offset_modulo == dst_offset_modulo) {
int byte_len;
int src_len_modulo;
if (src_offset_modulo) {
unsigned char c;
c = reverse_mask_xor[dst_offset_modulo] & *src++;
PREPARE_FIRST_COPY();
*dst++ |= c;
}
byte_len = src_len / CHAR_BIT;
src_len_modulo = src_len % CHAR_BIT;
if (byte_len) {
memcpy(dst, src, byte_len);
src += byte_len;
dst += byte_len;
}
if (src_len_modulo) {
*dst &= reverse_mask_xor[src_len_modulo];
*dst |= reverse_mask[src_len_modulo] & *src;
}
} else {
int bit_diff_ls,
bit_diff_rs;
int byte_len;
int src_len_modulo;
unsigned char c;
/*
* Begin: Line things up on destination.
*/
if (src_offset_modulo > dst_offset_modulo) {
bit_diff_ls = src_offset_modulo - dst_offset_modulo;
bit_diff_rs = CHAR_BIT - bit_diff_ls;
c = *src++ << bit_diff_ls;
c |= *src >> bit_diff_rs;
c &= reverse_mask_xor[dst_offset_modulo];
} else {
bit_diff_rs = dst_offset_modulo - src_offset_modulo;
bit_diff_ls = CHAR_BIT - bit_diff_rs;
c = *src >> bit_diff_rs &
reverse_mask_xor[dst_offset_modulo];
}
PREPARE_FIRST_COPY();
*dst++ |= c;
/*
* Middle: copy with only shifting the source.
*/
byte_len = src_len / CHAR_BIT;
while (--byte_len >= 0) {
c = *src++ << bit_diff_ls;
c |= *src >> bit_diff_rs;
*dst++ = c;
}
/*
* End: copy the remaing bits;
*/
src_len_modulo = src_len % CHAR_BIT;
if (src_len_modulo) {
c = *src++ << bit_diff_ls;
c |= *src >> bit_diff_rs;
c &= reverse_mask[src_len_modulo];
*dst &= reverse_mask_xor[src_len_modulo];
*dst |= c;
}
}
}
}
回答3:
What is optimal will depend upon the target platform. On some platforms without barrel shifters, shifting the whole vector right or left one bit, n times, for n<3, will be the fastest approach (on the PIC18 platform, an 8x-unrolled byte loop to shift left one bit will cost 11 instruction cycles per eight bytes). Otherwise, I like the pattern (note src2 will have to be initialized depending upon what you want done with the end of your buffer)
src1 = *src++; src2 = (src1 shl shiftamount1) | (src2 shr shiftamount2); *dest++ = src2; src2 = *src++; src1 = (src2 shl shiftamount1) | (src1 shr shiftamount2); *dest++ = src1;
That should lend itself to very efficient implementation on an ARM (eight instructions every two words, if registers are available for src, dest, src1, src2, shiftamount1, and shiftamount2. Using more registers would allow faster operation via multi-word load/store instructions. Handling four words would be something like (one machine instruction per line, except the first four lines would together be one instruction, as would the last four lines ):
src0 = *src++; src1 = *src++; src2 = *src++; src3 = *src++; tmp = src0; src0 = src0 shr shiftamount1 src0 = src0 | src1 shl shiftamount2 src1 = src1 shr shiftamount1 src1 = src1 | src2 shl shiftamount2 src2 = src2 shr shiftamount1 src2 = src2 | src3 shl shiftamount2 src3 = src3 shr shiftamount1 src3 = src3 | tmp shl shiftamount2 *dest++ = src0; *dest++ = src1; *dest++ = src2; *dest++ = src3;
Eleven instructions per 16 bytes rotated.
回答4:
Your solution looks similar to most I've seen: basically do some unaligned work at the start and end, with the main loop in the middle using aligned accesses. If you really need efficiency and do this on very long bitstreams, I would suggest using something architecture-specific like SSE2 in the main loop.
来源：https://stackoverflow.com/questions/3534535/whats-a-time-efficient-algorithm-to-copy-unaligned-bit-arrays