Haskell中的输入类型

Question

我想问问题.我是Hakskell的biginner,我有一些非常简单的程序困难,应该告诉我divident%divider == 0.

我有这个代码:

f::Integer -> Integer -> Bool
f x y = if ((x `mod` y) == 0) then True
                              else False

main = do putStrLn "Set up dividend"
          x <- getLine
          putStrLn "Set Up divider"
          y <- getLine
          f read x::Int read y::Int

但是当我想运行它时,我遇到了一个错误:

 Couldn't match expected type `Int' with actual type `m0 b0'
    Expected type: m0 a0 -> m0 b0 -> Int
      Actual type: m0 a0 -> m0 b0 -> m0 b0
    In a stmt of a 'do' block: putStrLn "Set up dividend"
    In the expression:
        do { putStrLn "Set up dividend";
             x <- getLine;
             putStrLn "Set Up divider";
             y <- getLine;
             .... } ::
          Int

我真的不知道,有什么不对.我也试过f x y (not f read x::Int .....)没有任何结果.我必须做错事.我知道有很多关于这个问题的话题,但没有任何帮助.我错过了什么.

Answer 1

问题出在最后一行:

f read x::Int read y::Int

这个代码基本上是说f read x read y,类型Int和f read x类型也在哪里Int.您必须添加括号,以便f正确应用,并在正确的术语上使用类型注释.你得到:

f ((read x) :: Int) ((read y) :: Int)
-- or with fewer parentheses, but meaning the same thing:
f (read x :: Int) (read y :: Int)

你的定义中的if语句f也是不必要的,为什么不使用:

f x y = (x `mod` y) == 0

Answer 2

f read x::Int read y::Int

该应用功能f的参数read,x,read和y.它也说f read y应该是一个Int结果,整个事情的结果也应该是一个Int.那显然不是你想要的.你需要的是应用f到的结果read x和read y,所以你需要围绕这些括号.

另一个问题是f将Integers作为参数,但是你要告诉read你Int.您可以通过更改Int为修复它,Integer或者您可以完全删除类型注释,因为它们可以被推断.您还可以更改f接受任何类型的类型Integral,以便它可以同时使用Int和Integer.

Lastly the type of main needs to be IO (), but your definition evaluates to a Bool. Maybe you want to print the Bool?

The combination of getLine and read can be simplified to readLine by the way.

So you could do:

main = do putStrLn "Set up dividend"
          x <- readLine
          putStrLn "Set Up divider"
          y <- readLine
          print $ f x y