Question

我正在学习Haskell并遵循http://learnyouahaskell.com/starting-out上的指南。我正处于显示的位置：

ghci> let nouns = ["hobo","frog","pope"]  
ghci> let adjectives = ["lazy","grouchy","scheming"]  
ghci> [adjective ++ " " ++ noun | adjective <- adjectives, noun <- nouns]  
["lazy hobo","lazy frog","lazy pope","grouchy hobo","grouchy frog",  
"grouchy pope","scheming hobo","scheming frog","scheming pope"]

我想要实现的，它是类似的，但结合两个字符串中包含的字母，因为字符串基本上是Haskell中的char列表，这就是我尝试过的：

 [x ++ ' ' ++ y | x <- "ab", y <- "cd"]

但编译器抱怨：

Prelude> [y ++ ' ' ++ y | x <- "abd", y <- "bcd"]

<interactive>:50:2:
    Couldn't match expected type ‘[a]’ with actual type ‘Char’
    Relevant bindings include it :: [[a]] (bound at <interactive>:50:1)
    In the first argument of ‘(++)’, namely ‘y’
    In the expression: y ++ ' ' ++ y

<interactive>:50:7:
    Couldn't match expected type ‘[a]’ with actual type ‘Char’
    Relevant bindings include it :: [[a]] (bound at <interactive>:50:1)
    In the first argument of ‘(++)’, namely ‘' '’
    In the second argument of ‘(++)’, namely ‘' ' ++ y’
    In the expression: y ++ ' ' ++ y

<interactive>:50:14:
    Couldn't match expected type ‘[a]’ with actual type ‘Char’
    Relevant bindings include it :: [[a]] (bound at <interactive>:50:1)
    In the second argument of ‘(++)’, namely ‘y’
    In the second argument of ‘(++)’, namely ‘' ' ++ y’

我做了很多尝试，例如将表达式包装在括号中以获取列表，将空格更改为String而不是char ...我怎样才能使其工作？

由于

Answer 1

++仅适用于列表，但x和y仅适用于Char。毕竟，它们是来自String（= [Char]）的元素，而LYAH示例包含Char列表：[String] = [[Char]]：

-- [a] -> [a] -> [a]
-- vv     vv
[y ++ ' ' ++ y | x <- "abd", y <- "bcd"]
--           ^   ^           ^
--           Char           Char

-- vs

--                                        [String]          [String]
--                                       vvvvvvvvvv          vvvvv
[adjective ++ " " ++ noun | adjective <- adjectives, noun <- nouns]  
-- ^^^^^^^           ^^^^
-- String           String

相反，使用(:)将彼此的字符合并到空列表中：

[x : ' ' : y : [] | x <- "abd", y <- "bcd"]

Answer 2

x ++ ' ' ++ y

这里的实际问题是，您尝试连接三个字符，并且只为项目列表定义一个函数。 ++实际上会连接两个列表，而不是两个单独的项目并列出一个列表。

因此，您可以通过将所有字符转换为字符串来修复程序，例如
```
> [[x] ++ " " ++ [y] | x <- "ab", y <- "cd"]
["a c","a d","b c","b d"]
```
请注意" "，而不是' '。因为" "表示只包含空格字符的字符串，但' '仅表示空格字符。
或者，将y转换为字符串，将cons运算符与' '一起使用，并将其连接到x转换为字符串，就像这样< / p>
```
> [[x] ++ (' ' : [y]) | x <- "ab", y <- "cd"]
["a c","a d","b c","b d"]
```
或者，更简单直观的as suggested by chi，创建一个字符列表，就像这样
```
> [[x, ' ', y] | x <- "ab", y <- "cd"]
["a c","a d","b c","b d"]
```

注意：用[]包裹一个字符会使其成为一个只包含一个字符的字符列表。它基本上变成了一个String。

如何在haskell中组合两个字符串中的字母

2 个答案: