将 double 组转换为无符号短整型数组？

Question

  Prob = *((unsigned short*)(Prob));

  Prob = (unsigned short)(Prob);

这对我不起作用。如何正确转换？

Answer 1

您面临的最大问题是那些超过 USHRT_MAX 的 double 值。其次，您需要解决舍入问题。解决这两个问题并将舍入设置为 .5 的简单方法可能如下所示，其中函数 array_dbl2short 返回成功转换为 unsigned Short 的 double 值的数量:

size_t array_dbl2short (unsigned short *usa, double *da, size_t n)
{
    if (!usa || !da || !n) return 0;
    register size_t i, ncnvt = 0;

    for (i = 0; i < n; i++) {
        if (da[i] < 0 || (unsigned long long)(da[i] + 0.5) > USHRT_MAX)
            fprintf (stderr, "error: %lf outside range for conversion.\n",
                    da[i]);
        else
            usa[ncnvt++] = (unsigned short)(da[i] + 0.5);
    }

    return ncnvt;
}

其使用的一个简短示例是:

#include <stdio.h>
#include <limits.h>

size_t array_dbl2short (unsigned short *usa, double *da, size_t n);

int main (void) {

    double arr[] = { 1.49, 1.5, 65535.49, 65535.5, -1.9, 25671.8 };
    size_t i, n, nelem = sizeof arr/sizeof *arr;
    unsigned short usarr[nelem];

    if ((n = array_dbl2short (usarr, arr, nelem))) {
        printf ("\n unsigned short array:\n\n");
        for (i = 0; i < n; i++)
            printf ("  usarr[%zu] : %hu\n", i, usarr[i]);
    }

    return 0;
}

size_t array_dbl2short (unsigned short *usa, double *da, size_t n)
{
    if (!usa || !da || !n) return 0;
    register size_t i, ncnvt = 0;

    for (i = 0; i < n; i++) {
        if (da[i] < 0 || (unsigned long long)(da[i] + 0.5) > USHRT_MAX)
            fprintf (stderr, "error: %lf outside range for conversion.\n",
                    da[i]);
        else
            usa[ncnvt++] = (unsigned short)(da[i] + 0.5);
    }

    return ncnvt;
}

(注意:double 值的实部必须能够表示为unsigned long long 使转换验证正常工作)

示例使用/输出

$ ./bin/dbl2ushort
error: 65535.500000 outside range for conversion.
error: -1.900000 outside range for conversion.

 unsigned short array:

  usarr[0] : 1
  usarr[1] : 2
  usarr[2] : 65535
  usarr[3] : 25672

此解决方案无法解决所有可能的舍入和转换问题。您还可以使用 math.h 中的函数(例如 modf、round 等)来执行实部和舍入。如果您还有其他问题，请告诉我。