Question

我正在做一个小型练习来帮助我做一个更大的练习。现在，我正在尝试编写一个名为＆＃34; digitStr＆＃34;的函数。它具有（我认为）digitStr :: String -> String的签名。

我在这里尝试做的是，当我输入如下内容时，我应该得到以下输出：

> digitStr "23"
"twenty three"

目前，假设数字最多只能达到25个。

我在自己的身上做过什么：

我的第一个想法是在数据库中添加数字。

wrdtxt :: txt
wrdtxt = " one"," two"," three"," four"," five","six"," seven"," eight"," nine"," ten"," eleven"," twelve","thirteen"," fourteen","fifteen"," sixteen"," seventeen"," eighteen"," nineteen", " twenty"

下一步是创建一个带字符串并将其输出到字符串列表中的函数：

numWords :: String -> [String]
numWords "" = []
numWords x = words x

下一步是获取函数的长度。如果它的长度为2，我们会查看第一个数字，以确定它是在10s还是20s。如果它是＆＃34; 25＆＃34;，不知何故，做＆＃34; 20＆＃34; ++＆＃34;五＆＃34;。

正如你所知，我对如何做有点困惑。在过去的两天里，我一直在考虑这个问题，这就是我能想到的。

Answer 1

以下是如何模式匹配字符串：

numWords :: String -> [String]
numWords []  =  ...    -- matches only the empty string
numWords [a] =  ...    -- matches when argument is just  a single char;
                          a = the digit
numWords [a,b] = ...   -- matches when there are exactly two chars;
                          a = first chars, b = second char
numWords ['1',b] = ... -- matches when there are exactly two characters and
                          the first char is '1'; b is set to the second char
numWords (a:as) = ...  -- matches when there is at least one character;
                          a = first character, as = second through last characters

但您可能还想考虑更改numWords的签名以获取Int：

numWords :: Int -> [String]
numWords x
  | x < 10  = ...     -- x is a single digit
  | x < 20  = ...     -- x is between 11 and 19
  | x < 30  = ...     -- x is between 20 and 29
  | ...

String Number to String Word Representation

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