将64位浮点格式字符串转换为浮点数

Question

I have a bunch of 64-bit floating-point-format strings, and I have to convert them into floating-point numbers. I know what the format looks like, but I'm wondering whether there's a built-in function can fulfill this job directly, as in:

convertToFloat(C06FCA5E35000000) --> -254.324 
convertToFloat(405F7D70A4000000) --> +125.96

I know how to convert these bit patterns to numbers manually, but it needs a lot of bit shifting. Is there a better way?

I modified 'C06FCA5E35000000' to '\\xC0\\x6F\\xCA\\x5E\\x35\\x00\\x00\\x00', and stored it to myString.

>>>print (myString)
\xC0\x6F\xCA\x5E\x35\x00\x00\x00
>>>d = struct.unpack('>d', myString)
       d = struct.unpack('>d', myString)
 struct.error: unpack requires a string argument of length 8

Why did it happened?

I didn't use binascii.a2b_hex or binascii.b2a_hex, beacuse it converts 'A' to '41' or '41' to 'A'. Both are not what I want. Right?

>>>print len('\xC0\x6F\xCA\x5E\x35\x00\x00\x00')
8
>>>print len(myString)
32

Now I know why it happened, but I still don't know how to solve the situation.

Answer 1

You can use struct.unpack . >d is a big-endian double; see help(struct) for more!

import struct

d, = struct.unpack('>d', b'\xc0\x6f\xca\x5e\x35\x00\x00\x00')

print(d)
# -254.32399988174438

If your string is really 'C06FCA5E35000000' , you can convert it first into bytes using binascii.a2b_hex .