如何将浮点数保存为 2 个字节?

问题描述

是的，我知道 IEEE-754 半精度标准，是的，我知道在该领域所做的工作.简而言之，我试图将一个简单的浮点数(如 52.1 或 1.25)保存在 2 个字节中.

Yes I'm aware of the IEEE-754 half-precision standard, and yes I'm aware of the work done in the field. Put very simply, I'm trying to save a simple floating point number (like 52.1, or 1.25) in just 2 bytes.

我在 Java 和 C# 但它们通过解码不同的数字来破坏输入值.您输入 32.1 并在编码解码后得到 32.0985.

I've tried some implementations in Java and in C# but they ruin the input value by decoding a different number. You feed in 32.1 and after encode-decode you get 32.0985.

有什么方法可以在不破坏输入值的情况下以 16 位存储浮点数?

Is there ANY way I can store floating point numbers in just 16-bits without ruining the input value?

非常感谢.

推荐答案

您可以将三位数字存储为 BCD 并使用剩余的四位作为小数点位置:

You could store three digits in BCD and use the remaining four bits for the decimal point position:

52.1 = 521 * 10 ^ -1 => 0x1521
1.25 = 125 * 10 ^ -2 => 0x2125

这将为您提供从 0.0000000000000001 到 999 的范围.当然，您可以为小数点添加偏移量，例如范围为 0.0000000001 到 999000000.

This would give you a range from 0.0000000000000001 to 999. You can of course add an offset for the decimal point to get for example the range 0.0000000001 to 999000000.

四位用于小数点放置的简单实现，其余用于值.没有任何错误检查，也没有彻底检查.(使用 != 比较双精度时，某些值可能存在精度问题.)

Simple implementation of four bit used for decimal point placement, and the rest for the value. Without any error check, and not thoroughly checked. (May have precision issues with some values when using != to compare doubles.)

public static short Encode(double value) {
  int cnt = 0;
  while (value != Math.Floor(value)) {
    value *= 10.0;
    cnt++;
  }
  return (short)((cnt << 12) + (int)value);
}

public static double Decode(short value) {
  int cnt = value >> 12;
  double result = value & 0xfff;
  while (cnt > 0) {
    result /= 10.0;
    cnt--;
  }
  return result;
}

例子:

Console.WriteLine(Encode(52.1));
Console.WriteLine(Decode(4617));

输出:

4617
52.1

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