Haskell：递归转换十六进制字符串为整数？ [英] Haskell: recursively convert hex string to integer?

问题描述

对于我的家庭作业，我需要使用递归函数将十六进制字符串转换为基数为10的整数（必要时使用许多助手方法）。

For my homework assignment, I need to convert a hexadecimal string to a base-10 integer using a recursive function (with as many helper methods as necessary).

这是我到目前为止：

This is what I've got so far:

-- Question 1, part (c):
hexChar :: Char -> Integer
hexChar ch
    | ch == '0' = 0
    | ch == '1' = 1
    | ch == '2' = 2
    | ch == '3' = 3
    | ch == '4' = 4
    | ch == '5' = 5
    | ch == '6' = 6
    | ch == '7' = 7
    | ch == '8' = 8
    | ch == '9' = 9
    | ch == 'A' = 10
    | ch == 'B' = 11
    | ch == 'C' = 12
    | ch == 'D' = 13
    | ch == 'E' = 14
    | ch == 'F' = 15
    | otherwise     = 0

parseHex :: String -> Integer
parseHex hxStr 
    | length hxStr /= 0 = (hexChar(last(hxStr)))+(10*parseHex(init(hxStr)))
    | otherwise         = 0  

但是，这不会产生正确的结果。
有没有人知道这样做的正确方法？

However, this does not produce the correct results. Does anyone know of the correct way to do this?

推荐答案

你真的很接近。你的错误在这一行：

You are really close. your error is on this line:

    | length hxStr /= 0 = (hexChar(last(hxStr)))+(10*parseHex(init(hxStr)))

想想为什么你乘以10.记住......十六进制是16位。

Think about why you are multiplying by 10. Remember ... Hexadecimal is base 16.

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