Question

我想让程序跳回选项部分，不要为变量“expression”赋值

我试着制作像嵌套的do函数：

main = do 
...
option = do

然后我会在某些情况下只调用“选项”，但它似乎没有这样工作，我找不到任何地方任何不同的解决方案。请问有人给出建议吗？

main = do
putStrLn "Enter expression:"
expression <- getLine

putStrLn "Enter a number from 1 to 5:"
putStrLn "1. Convert to Postfix"
putStrLn "2. Convert to Prefix"
putStrLn "3. Convert to Binary Expression Tree"
putStrLn "4. Substitution"
putStrLn "5. Evaluate Expression"
choice <- getLine

case (choice) of
    "1" -> putStrLn $ toPostfix expression
    "2" -> putStrLn $ toPrefix expression
    "3" -> print (parse expression)
    "4" -> do { putStrLn "Enter a variable: "
              ; variable <- getLine
              ; putStrLn "Enter a number: " 
              ; number <- getLine
              ; print (sub expression variable number)
              ; print (solvePostfix (toPostfix (sub expression variable number))) }
    "5" -> print (solvePostfix (toPostfix expression))
    otherwise -> putStrLn "This is not a number from 1 to 5!"

Answer 1

我不完全理解这个问题，但也许你正在寻找类似的东西。

main = do
  putStrLn "Enter expression:"
  expression <- getLine

  putStrLn "Enter a number from 1 to 5:"
  putStrLn "1. Convert to Postfix"
  putStrLn "2. Convert to Prefix"
  putStrLn "3. Convert to Binary Expression Tree"
  putStrLn "4. Substitution"
  putStrLn "5. Evaluate Expression"
  choice <- getLine

  let option = do 
        { putStrLn "Enter a variable: "
        ; variable <- getLine
        ; putStrLn "Enter a number: " 
        ; number <- getLine
        ; print (sub expression variable number)
        ; print (solvePostfix (toPostfix (sub expression variable number))) }

  case (choice) of
      "1" -> putStrLn $ toPostfix expression
      "2" -> putStrLn $ toPrefix expression
      "3" -> print (parse expression)
      "4" -> option
      "5" -> print (solvePostfix (toPostfix expression))
      otherwise -> putStrLn "This is not a number from 1 to 5!"

或者，如果你想要一个循环：

main = do
  putStrLn "Enter expression:"
  expression <- getLine

  let loop = do
        putStrLn "Enter a number from 1 to 5:"
        putStrLn "1. Convert to Postfix"
        putStrLn "2. Convert to Prefix"
        putStrLn "3. Convert to Binary Expression Tree"
        putStrLn "4. Substitution"
        putStrLn "5. Evaluate Expression"
        choice <- getLine

        case (choice) of
          "1" -> putStrLn $ toPostfix expression
          "2" -> putStrLn $ toPrefix expression
          "3" -> print (parse expression)
          "4" -> do
            { putStrLn "Enter a variable: "
            ; variable <- getLine
            ; putStrLn "Enter a number: " 
            ; number <- getLine
            ; print (sub expression variable number)
            ; print (solvePostfix (toPostfix (sub expression variable number))) }
          "5" -> print (solvePostfix (toPostfix expression))
          otherwise -> putStrLn "This is not a number from 1 to 5!"

        loop -- restart loop from start

  -- in main, start loop
  loop

可以使用库中的forever帮助程序实现此循环，但了解基本的递归解决方案非常重要。

Haskell - 单独的主要=做成块

1 个答案: