Question

我需要创建一个接受整数的函数，并返回一个编码为Hex Coded Decimal的整数的二进制字符串，以便以后打包成结构。

例如，我写了这个：

def convert_int(x):
    """
    Accepts an integer, outputs a hexadecimal string in HCD format
    Caution! Byte order is ALREADY little endian!
    """
    result = b''
    while x > 0:
        hcd = chr(int(str(divmod(x, 100)[1]), 16))
        result = result + hcd
        x = divmod(x, 100)[0]
    return result

因此convert_int(1234)将为3412h，依此类推。写这篇文章的Pythonic和优雅方式是什么？

upd：使函数输出小端字符串准备打包。

Answer 1

def convert_to_hcd(num):
    chars = []
    while num:
        num, ones = divmod(num, 10)
        num, tens = divmod(num, 10)
        chars.append(chr(tens * 16 + ones))
    chars.reverse()
    return "".join(chars)

convert_to_hcd(1234)    # => returns '\x124' (which is correct because '\x34' == '4')

Answer 2

所以，对我来说正确的代码如下，请注意字节顺序是反转的（小端）：

def convert_int(x):
"""
Accepts an integer, outputs a hexadecimal string in HCD format
Caution! Byte order is ALREADY little endian!
"""
result = b''
while x > 0:
    hcd = chr(int(str(divmod(x, 100)[1]), 16))
    result = result + hcd
    x = divmod(x, 100)[0]
return result

Answer 3

def intToHex(num):
    numBin = [int(n) for n in str(num)]
    result = 0
    for n in numBin:
        result = result*16 + n
    return(hex(result))

这很干净

Answer 4

这个小片段怎么样......：）

def convert_to_hcd(num): return "".join([ "\\x"+ (lambda i, k: ""+i+k if i !='m' else "0"+k)(i,k) for i,k in zip(( {NUM {1}} {NUM {1}} {NUM {1}} {NUM {1}} {NUM {1}} {NUM {1}}

Answer 5

>>> hex(1234)
'0x4d2'

谷歌不是你的敌人......

Python中的十六进制编码十进制

5 个答案: