Question

我正在尝试编写一个固定点转换例程。我需要将两个值加上一些标志放入一个8字节的数据包中，因此每个值只有3个字节。

我的代码如下所示：

typedef struct
{
    signed short m;
    unsigned char f;
} Q16_8;

Q16_8 toQ(double d)
{
    Q16_8 q;
    q.m = (signed short)d;
    q.f = (unsigned char)(d * 256.0);
    return q;
}

double toDouble(const Q16_8 q)
{
    return q.m + q.f / 256.0;
}

我的测试结果如下所示。第1列是浮点数，2是固定点，3是差值。

-2.000000 -2.000000 0.000000
-1.750000 -0.750000 -1.000000
-1.500000 -0.500000 -1.000000
-1.250000 -0.250000 -1.000000
-1.000000 -1.000000 0.000000
-0.750000 0.250000 -1.000000
-0.500000 0.500000 -1.000000
-0.250000 0.750000 -1.000000
0.000000 0.000000 0.000000
0.250000 0.250000 0.000000
0.500000 0.500000 0.000000
0.750000 0.750000 0.000000
1.000000 1.000000 0.000000
1.250000 1.250000 0.000000
1.500000 1.500000 0.000000
1.750000 1.750000 0.000000

我做错了什么？

Answer 1

请改为尝试：

#include <stdio.h>

typedef struct
{
    signed short m;
    unsigned char f;
} Q16_8;

Q16_8 toQ(double d)
{
    Q16_8 q;
    long x = d * 256;
    q.m = x >> 8; // this assumes >> to be an arithmetic shift
    q.f = x & 0xFF; // this assumes signed ints to be 2's complement
    return q;
}

double toDouble(Q16_8 q)
{
    long x = ((long)q.m << 8) + q.f;
    return x / 256.0;
}

int main(void)
{
  int i;
  for (i = -2*4; i <= +2*4; i++)
  {
    double d = i / 4.0;
    Q16_8 q = toQ(d);
    double d2 = toDouble(q);
    printf("toDouble(toQ(%f)) = %f\n", d, d2);
  }
  return 0;
}

输出（ideone）：

toDouble(toQ(-2.000000)) = -2.000000
toDouble(toQ(-1.750000)) = -1.750000
toDouble(toQ(-1.500000)) = -1.500000
toDouble(toQ(-1.250000)) = -1.250000
toDouble(toQ(-1.000000)) = -1.000000
toDouble(toQ(-0.750000)) = -0.750000
toDouble(toQ(-0.500000)) = -0.500000
toDouble(toQ(-0.250000)) = -0.250000
toDouble(toQ(0.000000)) = 0.000000
toDouble(toQ(0.250000)) = 0.250000
toDouble(toQ(0.500000)) = 0.500000
toDouble(toQ(0.750000)) = 0.750000
toDouble(toQ(1.000000)) = 1.000000
toDouble(toQ(1.250000)) = 1.250000
toDouble(toQ(1.500000)) = 1.500000
toDouble(toQ(1.750000)) = 1.750000
toDouble(toQ(2.000000)) = 2.000000

Answer 2

以d = -1.75为例，考虑以下两个事实的后果：

m将为-1，
f只能是积极的......

对于负非整数，为什么我的定点转换偏差为1？

2 个答案: