如何在Python中将256位大端整数转换为小端？

不太复杂,或者我希望如此.我有一个256位十六进制整数编码为大端,我需要转换为小端. Python的struct模块通常就足够了,但the official documentation没有列出的格式,其大小甚至接近我需要的格式.

使用struct的非长度特定类型(虽然我可能这样做错了)似乎不起作用：

>> x = 'ffffffffffffffffffffffffffffffffffffffffffffffffffffffff00000000'
>> y = struct.unpack('>64s', x)[0] # Unpacking as big-endian
>> z = struct.pack('<64s', y) # Repacking as little-endian
>> print z
'ffffffffffffffffffffffffffffffffffffffffffffffffffffffff00000000'

示例代码(应该发生什么)：

>> x = 'ffffffffffffffffffffffffffffffffffffffffffffffffffffffff00000000'
>> y = endianSwap(x)
>> print y
'00000000ffffffffffffffffffffffffffffffffffffffffffffffffffffffff'

解决方法:

struct模块无法处理256位数.所以你必须手动编码.

首先,您应该将其转换为字节：

x = 'ffffffffffffffffffffffffffffffffffffffffffffffffffffffff00000000'
a = x # for having more successive variables
b = a.decode('hex')
print repr(b)
# -> '\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\x00\x00\x00\x00'

这样你可以逆转它using @Lennart’s method：

c = b[::-1]
# -> '\x00\x00\x00\x00\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff'

d = c.encode('hex')
z = d
print z
# -> 00000000ffffffffffffffffffffffffffffffffffffffffffffffffffffffff

标签：python,struct,endianness
来源： https://codeday.me/bug/20190729/1568236.html