Question

我才刚刚开始学习Haskell，我对此确实有很多疑问。在我正在做的教程中，我需要开发一个功能，从列表和特定的String中，您将在列表中找到String的位置。在网上搜索时，我找到了此代码，但我真的不明白，有人可以向我解释。

TEST_F

Answer 1

lookUp :: [String] -> String -> Int
lookUp [] s = error "..."  -- An empty list does not contain _anything_, in
                           -- particular not the string you're looking for.
lookUp (x:xs) s     -- We've eliminated the empty case, which guarantees there's
                    -- at least one (head) element in the list if we get here.
                    -- Let's examine it!
   | not (x == s) -- If the element isn't the one we were looking for...
       = 1 + (lookUp xs s)   -- then we need to continue the search, i.e. search
                             -- through the _remaining_ elements `xs` and, if `s` is
                             -- found there, report a one higher position (because
                             -- we've already skipped one element).
   | otherwise  -- Else, the element _is_ the one we were looking for...
       = 0   -- IOW, it occurs at position 0 of the (part of) the list we're
             -- currently examining.

再加上几句话：

正如Willem Van Onsem所说，error在这里是一个坏主意：在现实情况下，列表中不包含您要查找的元素，即，这不仅仅是一个“哎呀，流星罢工打破了银行的债权人”之类的东西，但是您应该期待实际的风险。但是error将默认使整个程序崩溃。相反，您应该返回一个Maybe Int，它使您可以通过呼叫者可以轻松处理的方式来表示失败。
```
lookUp :: [String] -> String -> Maybe Int
lookUp [] _ = Nothing
lookUp (x:xs) s | not(x == s)  = fmap (1 +) (lookUp xs s)
                | otherwise    = Just 0
```
此函数中的所有内容实际上都不需要列表中的字符串。它与整数，单个字符，布尔值等同样适用。任何允许相等比较的东西。因此，您最好进行签名
```
lookUp :: Eq a => [a] -> a -> Maybe Int
```

Answer 2

lookUp :: [String] -> String -> Int

函数lookUp接受String和返回Int的String的列表

lookUp [] s = error "String no encontrado"

如果第一个参数为空字符串，则返回error ...

lookUp (x:xs) s | not(x == s) = 1 + (lookUp xs s)
                | otherwise = 0

有趣的部分(x:xs)从列表中获取第一个字符串，然后是字符串 |是守卫的，因此如果x中的字符串不等于s字符串return 1 + ( lookup xs s) .. ==>递归调用lookUp与xs-不带比较字符串x和字符串s作为参数的字符串列表

最后othervise返回0

手动：

lookUp [] "foo" ==>第一个模式[]，因此返回错误

lookUp ["foo"] "foo" ==>第二个模式并运行后卫==> not( "foo" == "foo") = 1 + ( lookUp [] "foo")，该命令以第二行othervise 0结尾，因此它返回正确的位置0

lookUp [ "bar", "foa", "foo", "fao" ] "foo" ==>第二种模式，并扩展为：not ( "bar" == "foo") return 1 + (lookUp ["foa", "foo", "fao"] "foo")然后not( "bar" == "foo") return 1 + (not ("foa" == "foo") = return 1 + (lookUp ["foo", "fao"] "foo"))然后not（“ bar” ==“ foo”）返回1 +（not（“ foa” = =“ foo”）=返回1 +（not（“ foo” ==“ foo”）=返回1）.. but because now test is *True* uses othervise = 0 so 1 + 1 = 2 and 2`是列表中正确的字符串位置。

最后一种可能性：

lookUp ["bar"] "foo" ==> not("bar" == "foo") = return 1 + (lookUp [] "foo")和lookUp空列表会引发错误

Haskell获取项目在列表中的位置

2 个答案: