什么是复制未对齐位数组的高效算法? [英] What's a time efficient algorithm to copy unaligned bit arrays?

问题描述

过去我不得不这样做很多次，但我从未对结果感到满意.

I've had to do this many times in the past, and I've never been satisfied with the results.

谁能建议一种将连续位数组从源复制到目标的快速方法，其中源和目标可能未在方便的处理器边界上对齐(右移)?

如果源和目标都没有对齐，问题可以很快变成只有其中一个没有对齐的问题(在第一个副本之后).

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).

作为一个起点，我的代码最终不可避免地看起来类似于以下内容(未经测试，忽略副作用，这只是一个即兴例子):

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) /* ... */
    }
}

(其实这个比我之前做的好，看起来还不错)

(actually this is better than I've done before. It doesn't look too bad)

推荐答案

你的内部循环需要两个字节的片段并将它们移动到目标字节.这几乎是最佳的.这里还有一些提示，不分先后:

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:

• 您无需一次只限于一个字节.使用您的平台允许的最大整数大小.这当然会使您的起始和尾随逻辑复杂化.
• 如果您使用无符号字符或整数，您可能不需要在源代码的第二部分右移后对其进行屏蔽.这取决于您的编译器.
• 如果您确实需要掩码，请确保您的编译器将表查找移到循环之外.如果不是，请将其复制到一个临时变量并使用它.

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