struct.error: unpack requires a buffer of 4 bytes

问题

I want to convert data from a device from bites to float I use the code from this answer

bytes to float

import struct

byte_file = b'+001.80\r'
print(type(byte_file))

y = struct.unpack('f' , byte_file)
print(y)

I get this struct.error: unpack requires a buffer of 4 bytes

The correct outcome should be 1.80 do I need to implement a buffer argument ?

回答1:

struct is used for binary packed data - data that is not human-readable. b'+001.80\r' is 8 bytes long: b'+', b'0', b'0', b'1', b'.', ....

You can just decode it and use float:

>>> b'+001.80\r'.decode()
'+001.80\r'
>>> float(_)
1.8
>>> import struct
>>> struct.pack('f', _)
b'ff\xe6?'  # doesn't look anything like your data!

However, because your data is 8 bytes long, you could treat it as a single double-precision floating-point value:

>>> struct.unpack('d', b'+001.80\r')
(3.711588247816385e-245,)

But that treats the data as binary-packed: +001.80\r, also known as 2b 30 30 31 2e 38 30 0d, is what 3.711588247816385e-245 looks like in memory.

来源：https://stackoverflow.com/questions/60376438/struct-error-unpack-requires-a-buffer-of-4-bytes