在Haskell中调用嵌套函数 [英] Invoking nested functions in Haskell

问题描述

我在一个名为 person 的函数内定义了一个名为 initials 的函数，但我无法弄清楚在 person 之外呼叫首字母：

  main = 
  - 我试图在这里打印initials函数的输出。 
（putStrLn（（personfirstNamelastName）。initials）） - 不在范围内：`首字母'
  - 如果此函数调用正常工作，则输出为f.l.。 
 
 person firstName lastName = 
 firstName ++ [。] ++ lastName 
其中
 fullName = firstName ++++ lastName 
 firstInitial = firstName !! 0 
 lastInitial =姓氏！ 0 
 initials = [firstInitial] ++。 ++ [lastInitial] ++。 
  

是否可以调用首字母函数在 main 函数中，即使它是在 person 函数中定义的

解决方案

您误解了其中的作用：它只是定义了定义，这意味着它们只能用于在这种情况下，首字母只能用于 person 。

换句话说，您正在定义一个函数 person :: String - >字符串 - > [String] （取两个字符串并返回它们的列表），其中此函数的内部实现细节包含一些名称为 initials 。它不像其他语言中的结构或类，其中首字母是某些值，所有 person s可以访问。 （无论如何 person 只是返回一个字符串列表，而不是一个新类型。）另外，。不是一个字段访问器，它只是一个方便的帮助函数组合。

看起来你可能想要制作一个 Person 数据类型，类似于

  data Person = Person {firstName :: String，lastName :: String} 
  

这是通过 Person //learnyouahaskell.com/making-our-own-types-and-typeclasses#record-syntaxrel =nofollow>record syntax。

<现在，您可以编写一个函数，它接受 Person 并返回它们的首字母：

   - 使用模式匹配
首字母（Person {firstName = f，lastName = l}）= 
 [f！ 0] ++。 ++ [l !! 0] ++。 
 
  - 或者，使用通过记录语法隐式定义的访问器
 initials p = [firstName p !! 0] ++。 ++ [lastName p !! 0] ++。 
  

这个用法就像

  putStrLn（首字母（Person {firstName =first，lastName =last}））
 
  - 或者只是使用位置参数（它仍然适用于记录语法）
 putStrLn（首字母（Personfirstlast））
  

I have defined a function called initials inside a function called person, but I can't figure out how to call initials outside of person:

main =
    --I attempted to print the output of the initials function here.
    (putStrLn ((person "firstName" "lastName") . initials)) --Not in scope: `initials'
    --If this function call worked correctly, the output would be "f.l.".

person firstName lastName =
    firstName ++ ["."] ++ lastName
    where
        fullName = firstName ++ " " ++ lastName
        firstInitial = firstName !! 0
        lastInitial = lastName !! 0
        initials = [firstInitial] ++ "." ++ [lastInitial] ++ "."

Is it possible to invoke the initials function inside the main function, even though it is defined inside the person function?

解决方案

You're misunderstanding what where does: it just scopes definitions, meaning they can only be used in the item they are attached to, in this case, initials can only be used in person.

In other words, you are defining a function person :: String -> String -> [String] (taking two strings and returning a list of them), where the internal implementation details of this function include some (unused) value called initials. It is not like a struct or class in other languages, where initials is some value that all persons can access. (And anyway person is just returning a list of strings, not a new type.) Also, . is not a field accessor, it is just a convenient helper for function composition.

It seems like you might be wanting to make a Person data type, something like

data Person = Person { firstName :: String, lastName :: String }

This is defining Person via "record syntax".

Now, you can write a function that takes a Person and returns their initials:

-- using pattern matching
initials (Person {firstName = f, lastName = l}) =
    [f !! 0] ++ "." ++ [l !! 0] ++ "."

-- or, using the accessors implicitly defined via record syntax
initials p = [firstName p !! 0] ++ "." ++ [lastName p !! 0] ++ "."

This is used like

putStrLn (initials (Person { firstName = "first", lastName = "last" }))

-- or, just using positional arguments (which still work with record syntax)
putStrLn (initials (Person "first" "last"))

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