2字节解压为4字节浮点数

Question

I have some animation data (x,y,z), which is represented as 2 byte structures and written in Little Endian. I know that they should be a 4 byte floating point, so i have to unpack them. I collected a few sample values as precise as it was possible (they doesn't represent exactly packed values, but very close to them) and roughly divided packed values on few ranges.

Sample values (Little Endian):

• 0.048879981 - 0x0046
• 0.056879997 - 0x0047
• 0.253880024 - 0x0050
• 0.313879967 - 0x0051
• 0.623880029 - 0x0055
• 1.003879905 - 0x0058
• -0.066120029 - 0x00С8
• -0.1561199428 - 0x00СD
• -0.8691199871 - 0x00D7

Ranges:

• 0x0000 : zero
• [0x0000,0x0014] : invisible changes (increasing probably)
• [0x0014, ....] : increasing (visible)
• 0x0080 : zero , probably the point of sign change
• [0x0080,0x00B0] : invisible changes (decreasing probably)
• [0x00B0, ....] : decreasing (visible)

There are gaps (....) on the ends of ranges because it is hard to check them correctly, but i assume such big values which are lying close to these ends doesn't used in practice.

Also, it looks like a symmetry between positive and negative ranges, for example i tested 0x0058 which gave 1.003879905 and 0x00D8 which gave value close to -1.003879905 but not precise. Maybe it happened because of slightly offset observed after 0x0080 , when visible decreasing starts from 0x00B0 , but it should be about 0x0094 if entire range had equal symmetry. But slight measure inaccuracy might be as well.

So, how to get a function in C#, that will convert source data to 4 byte floating point?

Answer 1

Some initial comments based on the information in the question so far:

• byte[] buffer = new byte[4]; is a bad approach because it addresses bytes individually while the other code manipulates bits using shifts within words, and C# does not define endianness. Simply use an unsigned 32-bit integer for all the work. The code will actually be simpler.
• The code does not handle subnormal values properly. If num2 is zero and num3 is not zero, the significand ( num3 ) must be shifted and the exponent ( num2 ) must be adjusted.