haskell 在函数中将用户输入更改为用户定义的数据类型时出错

Question

handleStatements :: [Term] -> IO ()
handleStatements statements = do
    let (queries, clauses) = partition isQuery statements
    mapM_ (clausesEntailProof clauses) queries
    --apply clauses to queries and ignore result

handleArgs :: String-> IO ()
handleArgs args = do
    contents <- readFile $ args
    case parseInput contents of
      Left err -> print err
      Right statements -> handleStatements statements


main :: IO ()
main = do
  handleStatements(input)
    where input = getLine >>= (\str -> ((readIO str)::IO[Term]))

我出错了。无法将预期类型“[Term]”与实际类型“IO Term”匹配。我该如何解决这个问题？

Answer 1

如果你在where子句中定义了input = …，那么input的类型是IO [Term]，而不是[Term]，所以你不能传递input 到 handleStatements.

你可以这样写：

main :: IO ()
main = do
    <b>input</b> <- getLine >>= (\str -> ((readIO str)::IO[Term]))
    handleStatements <b>input</b>

但是你把事情搞得太复杂了，你可以用 readLn :: Read a => IO a 代替：

main :: IO ()
main = do
    input <- <b>readLn</b>
    handleStatements input

或者用 (>>=) :: Monad m => m a -> (a -> m b) -> m b function:

替换 do 块

main :: IO ()
main = readLn <b>>>=</b> handleStatements

haskell 在函数中将用户输入更改为用户定义的数据类型时出错

haskell error change user input to user defined data type in function

haskell

logic

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