dictionary - 为什么这个字典功能不起作用

Question

我正在尝试使用 here 中的代码创建和使用字典:

import Data.List (lookup)

insert :: Eq a => (a,b) -> [(a,b)] -> [(a,b)]
insert (a,b)  []           = [(a,b)]
insert (a,b) ((c,d):rest) = if a == c
    then (a,b) : rest
    else (c,d) : insert (a,b) rest

dict :: [(String, String)]
dict = [("", "")]

main = do 
    insert ("onekey", "onevalue") dict
    print dict
    print $ lookup "onekey" dict

但我收到以下错误:

$ runghc rndict.hs

rndict.hs:22:1: error:
    • Couldn't match expected type ‘IO t0’ with actual type ‘[()]’
    • In the expression: main
      When checking the type of the IO action ‘main’

rndict.hs:24:9: error:
    • Couldn't match type ‘IO’ with ‘[]’
      Expected type: [()]
        Actual type: IO ()
    • In a stmt of a 'do' block: print dict
      In the expression:
        do { insert ("onekey", "onevalue") dict;
             print dict;
             print $ lookup "onekey" dict }
      In an equation for ‘main’:
          main
            = do { insert ("onekey", "onevalue") dict;
                   print dict;
                   print $ lookup "onekey" dict }

rndict.hs:25:9: error:
    • Couldn't match type ‘IO’ with ‘[]’
      Expected type: [()]
        Actual type: IO ()
    • In a stmt of a 'do' block: print $ lookup "onekey" dict
      In the expression:
        do { insert ("onekey", "onevalue") dict;
             print dict;
             print $ lookup "onekey" dict }
      In an equation for ‘main’:
          main
            = do { insert ("onekey", "onevalue") dict;
                   print dict;
                   print $ lookup "onekey" dict }

有什么问题，在 Haskell 中使用字典的正确方法是什么？

Answer 1

您需要使用 let将新字典绑定(bind)到名称:

import Data.List (lookup)

insert :: Eq a => (a,b) -> [(a,b)] -> [(a,b)]
insert (a,b)  []           = [(a,b)]
insert (a,b) ((c,d):rest) = if a == c
    then (a,b) : rest
    else (c,d) : insert (a,b) rest

dict :: [(String, String)]
dict = [("", "")]

main = do
    let d = insert ("onekey", "onevalue") dict
    print d
    print $ lookup "onekey" d

您询问插入多个元素。为此，您可以使用折叠来编写一个名为 insertMany 的函数。 .您可能应该使用 foldl'但我会把它作为一个练习来找出原因。

import Data.List (lookup)

insert :: Eq a => (a,b) -> [(a,b)] -> [(a,b)]
insert (a,b)  []           = [(a,b)]
insert (a,b) ((c,d):rest) = if a == c
    then (a,b) : rest
    else (c,d) : insert (a,b) rest

insertMany :: Eq a => [(a,b)] -> [(a,b)] -> [(a,b)]
insertMany elements dict =
  foldl (flip insert) dict elements

dict :: [(String, String)]
dict = [("", "")]

main = do
    let d = insert ("onekey", "onevalue") dict
    print d
    print $ lookup "onekey" d
    print $ insertMany [("onekey", "newvalue"), ("anotherkey", "anothervalue")]
      dict