编写一个函数将十六进制转换为十进制 [英] Writing a function to convert hex to decimal

问题描述

我需要编写一个将十六进制转换为十进制的函数，并且我已经知道如何去做，但我还是停留在第一位.到目前为止，我正在接受用户输入并以十六进制返回它，但这仅在我一次输入 1 件事时才有效.代码是:

def hex(x):如果 x == "0":返回 0elif x == "1":返回 1elif x == "2":返回 2elif x == "3":返回 3elif x == "4":返回 4elif x == "5":返回 5elif x == "6":返回 6elif x == "7":返回 7elif x == "8":返回 8elif x == "9":返回 9elif x == "A":返回 10elif x == "B":返回 11elif x == "C":返回 12elif x == "D":返回 13elif x == "E":返回 14elif x == "F":返回 15打印十六进制(原始输入().上())

例如，如果我输入 C，它会返回 12，但如果我输入 8C，则它不起作用.我不明白这是为什么.

解决方案

您可以使用 int 使这更容易.功能:

def hex_to_dex(strng_of_hex):返回整数(strng_of_hex，16)

示例:

<预><代码>>>>int("0xff", 16)255

int 前面是否有 0x 并不重要，它会忽略字符串中的 0x.因此，以下内容也将起作用:

<预><代码>>>>int("a", 16)10

对于从头开始的原始十六进制代码，请尝试以下操作:

def 十六进制:_hexer = "0123456789ABCDEF"return sum([_hexer.find(var) * 16 ** i for i, var in enumerate(reversed(s.upper()))])

如果你想放一些守卫:

def 十六进制:_hexer = "0123456789ABCDEF"如果不是全部([s.upper() 中的 var 的 _hexer 中的变量]):打印无效字符串"返回无return sum([_hexer.find(var) * 16 ** i for i, var in enumerate(reversed(s.upper()))])

我在这里使用了很多函数，但为了快速参考，这里有一个附录:

1. str.find
2. 枚举
3. 反转 -> 反转字符串
4. sum

至于 [...]，它是一个列表推导式，你可以在上面找到大量资源.

I need to write a function that converts hex to decimal and I've got an idea of how to do it but I'm stuck at the first bit. So far I am taking the user input and returning it in hex, but this only works if I enter 1 thing at a time. The code is:

def hex(x):
    if x == "0":
        return 0
    elif x == "1":
        return 1
    elif x == "2":
        return 2
    elif x == "3":
        return 3
    elif x == "4":
        return 4    
    elif x == "5":
        return 5
    elif x == "6":
        return 6
    elif x == "7":
        return 7
    elif x == "8":
        return 8
    elif x == "9":
        return 9
    elif x == "A":
        return 10
    elif x == "B":
        return 11
    elif x == "C":
        return 12
    elif x == "D":
        return 13
    elif x == "E":
        return 14
    elif x == "F":
        return 15

print hex(raw_input().upper())

It works if I enter, for example, C then it returns 12, but if I enter 8C then it doesn't work. I can't figure out why this is.

解决方案

You can use int to make this a lot easier. Function:

def hex_to_dex(strng_of_hex):
    return int(strng_of_hex, 16)

Example:

>>> int("0xff", 16)
255

It does not matter if you have a 0x infront for int, it will ignore 0x in the string. So, the following will also work:

>>> int("a", 16)
10

As for a raw hex code from scratch, try the following:

def hex(s):
    _hexer = "0123456789ABCDEF"
    return sum([_hexer.find(var) * 16 ** i for i, var in enumerate(reversed(s.upper()))])

If you wanted to put some guards in:

def hex(s):
    _hexer = "0123456789ABCDEF"

    if not all([var in _hexer for var in s.upper()]):
        print "Invalid string"
        return None

    return sum([_hexer.find(var) * 16 ** i for i, var in enumerate(reversed(s.upper()))])

I've used a lot of functions here, but for quick reference, here's an appendix:

1. str.find
2. enumerate
3. reversed -> reverses the string
4. sum

As for the [...], its a list comprehension, you can find loads of resources on that.

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