如何将十六进制ASCII字符串转换为有符号整数

Question

Input = 'FFFF' # 4 ASCII F's

desired result ... -1 as an integer

code tried:

hexstring = 'FFFF'
result = (int(hexstring,16))
print result #65535

Result: 65535

Nothing that I have tried seems to recognized that a 'FFFF' is a representation of a negative number.

Answer 1

Python converts FFFF at 'face value', to decimal 65535

input = 'FFFF'
val = int(input,16) # is 65535

You want it interpreted as a 16-bit signed number. The code below will take the lower 16 bits of any number, and 'sign-extend', ie interpret as a 16-bit signed value and deliver the corresponding integer

val16 = ((val+0x8000)&0xFFFF) - 0x8000

This is easily generalized

def sxtn( x, bits ):
     h= 1<<(bits-1)
     m = (1<<bits)-1
     return ((x+h) & m)-h

Answer 2

In a language like C, 'FFFF' can be interpreted as either a signed (-1) or unsigned (65535) value. You can use Python's struct module to force the interpretation that you're wanting.

Note that there may be endianness issues that the code below makes no attempt to deal with, and it doesn't handle data that's more than 16-bits long, so you'll need to adapt if either of those cases are in effect for you.

import struct

input = 'FFFF'

# first, convert to an integer. Python's going to treat it as an unsigned value.
unsignedVal = int(input, 16)
assert(65535 == unsignedVal)
# pack that value into a format that the struct module can work with, as an 
# unsigned short integer
packed = struct.pack('H', unsignedVal)
assert('\xff\xff' == packed)

# ..then UNpack it as a signed short integer
signedVal = struct.unpack('h', packed)[0]
assert(-1 == signedVal)