Question

我有一个包含姓名，月，日的3元组的出生记录列表。我想添加新的元组并显示birthBook。但是，在我添加或删除值后，birthBook不会更改比如，我添加了tom 5 6但是打印birthBook只显示“dd”4 5，这意味着（tom，5,6）不会在birthBook中添加。我该怎么办？这是我的代码：

birthBook::[(String,String,String)]
birthBook=[("dd","4","5")]

prompt :: String -> IO ()
prompt todos = do
    putStrLn "Enter a command(add,remove,print)"
    cmd <- getLine
    interpret cmd todos
interpret :: String->String-> IO ()
interpret   "print"      todos= do
     print(birthBook)
     prompt todos
interpret   "add"        todos= do
    putStrLn "Enter a name"
    cmdname<-getLine
    putStrLn "Enter an integer month"
    cmdmonth<-getLine
    putStrLn "Enter an integer day"
    cmdday<-getLine
    print(add birthBook cmdname cmdmonth cmdday)
    prompt todos
add::(Eq key)=>[(key,v,z)]->key->v->z->[(key,v,z)]
add birthBook name month day = (name, month, day):birthBook

main= prompt []

Answer 1

这里的问题是你忘了Haskell纯粹是功能性的，并且默认情况下不会将值视为可变。

您需要将出生簿作为参数传递给prompt和interpret，以便保持任何更改。您还尝试打印birthBook的初始声明，它始终保持相同，因为Haskell是一种纯函数式编程语言，如果您想要更改任何内容，则需要创建数据的新副本以及您希望进行的所需更改 - 因此我在您的解释例程结束时放置了let newBirthBook表达式。

{-# LANGUAGE OverloadedStrings #-}
module Main where

birthBook::[(String,String,String)]
birthBook=[("dd","4","5")]

prompt :: [(String,String,String)] -> String -> IO ()
prompt bBook todos = do
    putStrLn "Enter a command(add,remove,print)"
    cmd <- getLine
    interpret bBook cmd todos

interpret :: [(String,String,String)] -> String -> String -> IO ()
interpret bBook "print" todos = do
  print bBook
  prompt bBook todos
interpret bBook "add" todos = do
  putStrLn "Enter a name"
  cmdname<-getLine
  putStrLn "Enter an integer month"
  cmdmonth<-getLine
  putStrLn "Enter an integer day"
  cmdday<-getLine
  let newBirthBook = add bBook cmdname cmdmonth cmdday
  prompt newBirthBook todos

add :: (Eq key) => [(key,v,z)] -> key -> v -> z -> [(key,v,z)]
add birthBook name month day = (name, month, day):birthBook

main= prompt birthBook []

如何在haskell中添加和删除元素后保存列表？

1 个答案: