Question

我正在将Michael Herf's radix sort翻译成直接的C代码。对于正值，它可以正常工作，但负值会被破坏，如下所示：

Input:
-0.100000 0.100000 0.000000 0.700000 0.900000 -0.400000 -1.000000 0.200000 -0.400000 -0.700000 
Expected output:
-1.000000 -0.700000 -0.400000 -0.400000 -0.100000 0.000000 0.100000 0.200000 0.700000 0.900000
Actual output:
-44.799999 -11.200000 -11.200000 -6.400000 -4.000000 0.000000 0.100000 0.200000 0.700000 0.900000

如何解决这个问题？

代码（在Eclipse CDT上，Ubuntu 16.10，使用GCC 6.2编译）：

#include <inttypes.h>
#include <stdlib.h>
#include <string.h>
#include <xmmintrin.h>  // for prefetch

#define PFVAL   64
#define PFVAL2  128
#define PF(x, i)    _mm_prefetch((char *)((x)+(i)+PFVAL), 0)
#define PF2(x, i)   _mm_prefetch((char *)((x)+(i)+PFVAL2), 0)

#define HISTOGRAM   2048
#define HISTOGRAMS  6144    // 3*histogram

#define FLIP_FLOAT_A(x) ((x)^(((x) >> 31) | 0x80000000))
#define FLIP_FLOAT_B(x) ((x)^((((x) >> 31)-1) | 0x80000000))

// utils for accessing 11-bit quantities
#define _0(x)   (x & 0x7FF)
#define _1(x)   (x >> 11 & 0x7FF)
#define _2(x)   (x >> 22 )

void radixsort_float(float *a, size_t num) {
    uint32_t *base = (uint32_t *)a;
    size_t arraysize = num*sizeof(uint32_t);
    uint32_t *aux = malloc(arraysize);

    uint32_t byte0[HISTOGRAMS] = { 0 };
    uint32_t *byte1 = byte0+HISTOGRAM;
    uint32_t *byte2 = byte1+HISTOGRAM;

    // 1. parallel histogramming pass
    for(size_t i = 0; i < num; ++i) {
        PF(base, i);

        uint32_t bi = FLIP_FLOAT_A(base[i]);

        ++byte0[_0(bi)];
        ++byte1[_1(bi)];
        ++byte2[_2(bi)];
    }

    // 2. sum the histograms: each histogram entry
    // records the number of values preceding itself
    uint32_t sum0 = 0, sum1 = 0, sum2 = 0;
    uint32_t total;
    for(size_t i = 0; i < HISTOGRAM; i++) {
        total = byte0[i]+sum0;
        byte0[i] = sum0-1;
        sum0 = total;

        total = byte1[i]+sum1;
        byte1[i] = sum1-1;
        sum1 = total;

        total = byte2[i]+sum2;
        byte2[i] = sum2-1;
        sum2 = total;
    }

    // byte 0: floatflip entire value, read/write histogram, write out flipped
    for(size_t i = 0; i < num; ++i) {
        uint32_t bi = FLIP_FLOAT_A(base[i]);
        uint32_t pos = _0(bi);

        PF2(base, i);
        aux[++byte0[pos]] = bi;
    }

    // byte 1: read/write histogram, copy
    // aux -> array
    for(size_t i = 0; i < num; ++i) {
        uint32_t ai = aux[i];
        uint32_t pos = _1(ai);
        PF2(aux, i);
        base[++byte1[pos]] = ai;
    }

    // byte 2: read/write histogram, copy & flip out
    // base -> aux
    for(size_t i = 0; i < num; ++i) { 
        uint32_t bi = base[i];
        uint32_t pos = _2(bi);

        PF2(base, i);
        aux[++byte2[pos]] = FLIP_FLOAT_B(bi);
    }

    memcpy(base, aux, arraysize);
    free(aux);
}

Answer 1

问题是你的FLIP_FLOAT_A宏 - 这是错的。你需要它来翻转负数的所有位（不仅仅是底部位），而且宏需要相反（对FLIP_FLOAT_B(FLIP_FLOAT_A(X)) == X的所有值都是X）。尝试：

#define FLIP_FLOAT_A(x) ((x)^(((~(x) >> 31)-1) | 0x80000000))
#define FLIP_FLOAT_B(x) ((x)^((((x) >> 31)-1) | 0x80000000))

基数排序C中的浮点数 - 负值被破坏

1 个答案: