我需要一种快速的方法来获取 64 位整数中所有一位的位置。例如，给定 x = 123703 , 我想填充一个数组 idx[] = {0, 1, 2, 4, 5, 8, 9, 13, 14, 15, 16} .我们可以假设我们先验地知道位数。这将被称为 1012 - 1015 次，因此速度至关重要。到目前为止，我想出的最快的答案是以下怪物，它使用 64 位整数的每个字节作为表的索引，这些表给出了在该字节中设置的位数和这些位的位置:

int64_t x;            // this is the input
unsigned char idx[K]; // this is the array of K bits that are set
unsigned char *dst=idx, *src;
unsigned char zero, one, two, three, four, five;  // these hold the 0th-5th bytes
zero  =  x & 0x0000000000FFUL;
one   = (x & 0x00000000FF00UL) >> 8;
two   = (x & 0x000000FF0000UL) >> 16;
three = (x & 0x0000FF000000UL) >> 24;
four  = (x & 0x00FF00000000UL) >> 32;
five  = (x & 0xFF0000000000UL) >> 40;
src=tab0+tabofs[zero ]; COPY(dst, src, n[zero ]);
src=tab1+tabofs[one  ]; COPY(dst, src, n[one  ]);
src=tab2+tabofs[two  ]; COPY(dst, src, n[two  ]);
src=tab3+tabofs[three]; COPY(dst, src, n[three]);
src=tab4+tabofs[four ]; COPY(dst, src, n[four ]);
src=tab5+tabofs[five ]; COPY(dst, src, n[five ]);

double doTest(double *data, int64_t model) {
  int nidx, idx[];
  double submatrix[][];
  nidx = getIndices(model, idx);  // get the locations of ones in model
  // copy data into submatrix
  for(int i=0; i<nidx; i++) {
    for(int j=0; j<nidx; j++) {
      submatrix[i][j] = data[idx[i]][idx[j]];
    }
  }
  factorize(submatrix, nidx);
  return the_answer;
}

double *data;  // move data to GPU/Phi once into shared memory
template<unsigned int K> double doTestUnrolled(int *idx) {
  double submatrix[K][K];
  // copy data into submatrix
  #pragma unroll
  for(int i=0; i<K; i++) {
    #pragma unroll
    for(int j=0; j<K; j++) {
      submatrix[i][j] = data[idx[i]][idx[j]];
    }
  }
  factorizeUnrolled<K>(submatrix);
  return the_answer;
}

for(int i=0; i<K; i++) {
    idx[i] = __builtin_ctzll(tmp);
    lb = tmp & -tmp;    // get lowest bit
    tmp ^= lb;      // remove lowest bit from tmp
} 

for(int i=0; i<K; i++) {
    *dst = i;
    dst += x & 1;
    x >>= 1;
} 

#define COPY(d, s, n) \
switch(n) { \
case 8: *(d++) = *(s++); \
case 7: *(d++) = *(s++); \
case 6: *(d++) = *(s++); \
case 5: *(d++) = *(s++); \
case 4: *(d++) = *(s++); \
case 3: *(d++) = *(s++); \
case 2: *(d++) = *(s++); \
case 1: *(d++) = *(s++); \
case 0: break;        \
}

#define COPY2(d, s, n) { *((uint64_t *)d) = *((uint64_t *)s); d+=n; }

 K    Base    CTZ   Mark   Tab1   Tab2
001  4.97s  6.42s  6.66s 18.23s 12.77s
002  4.95s  8.49s  7.28s 19.50s 12.33s
004  4.95s  9.83s  8.68s 19.74s 11.92s
006  4.95s 16.86s  9.53s 20.48s 11.66s
008  4.95s 19.21s 13.87s 20.77s 11.92s
010  4.95s 21.53s 13.09s 21.02s 11.28s
015  4.95s 32.64s 17.75s 23.30s 10.98s
020  4.99s 42.00s 21.75s 27.15s 10.96s
030  5.00s 100.64s 35.48s 35.84s 11.07s
040  5.01s 131.96s 44.55s 44.51s 11.58s

最佳答案

我相信这里的性能关键是关注更大的问题，而不是从随机整数中提取位位置的微观优化。

根据您的示例代码和之前的 SO 问题判断，您正在枚举按顺序设置 K 位的所有单词，并从中提取位索引。这大大简化了事情。

如果是这样，那么每次迭代不是重建位位置，而是尝试直接增加位数组中的位置。一半的时间这将涉及单个循环迭代和增量。

沿着这些路线的东西:

// Walk through all len-bit words with num-bits set in order
void enumerate(size_t num, size_t len) {
    size_t i;
    unsigned int bitpos[64 + 1];

    // Seed with the lowest word plus a sentinel
    for(i = 0; i < num; ++i)
        bitpos[i] = i;
    bitpos[i] = 0;

    // Here goes the main loop
    do {
        // Do something with the resulting data
        process(bitpos, num);

        // Increment the least-significant series of consecutive bits
        for(i = 0; bitpos[i + 1] == bitpos[i] + 1; ++i)
            bitpos[i] = i;
    // Stop on reaching the top
    } while(++bitpos[i] != len);
}

// Test function
void process(const unsigned int *bits, size_t num) {
    do
        printf("%d ", bits[--num]);
    while(num);
    putchar('\n');
}

关于c - 返回 64 位整数中所有设置位的位置的最快方法是什么？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/20713017/