带有条件的 Haskell 中的 while 循环 [英] While loop in Haskell with a condition

问题描述

我在这里遇到了一些 Haskell 情况.我正在尝试用 monad 编写两个函数.只要函数的输入/输出条件为真，第一个就应该遍历一个函数.第二个应该使用第一个将数字作为输入并将其作为输出写入，直到您输入空格.

I'm having a little Haskell Situation over here. I'm trying to write two functions with monads. First one is supposed to iterate through a function as long as the condition is true for the input / output of the function. Second one is supposed to use the first one to take a number as input and write it as output until you enter a space.

我遇到了这个问题，有什么帮助吗?

I'm stuck with this, any help?

module Test where

while :: (a -> Bool) -> (a -> IO a) -> a -> IO a
while praed funktion x = do
                         f <- praed (funktion x)
                         if f == True then do
                                             y <- funktion x
                                             while praed funktion y
                         else return x



power2 :: IO ()
power2 = do putStr (Please enter a number.")
            i <- getChar
            while praed funktion
            where praed x = if x /= ' ' then False else True
                  funktion = i

推荐答案

import Control.Monad

while :: (a -> Bool) -> (a -> IO a) -> a -> IO a
while praed funktion x
    | praed x   = do
        y <- funktion x
        while praed funktion y
    | otherwise = return x


power2 :: IO ()
power2 = do
    putStr "Please enter a number."
    i <- getChar
    let praed x = x /= ' '
    let f x = do
        putChar x
        getChar
    while praed f '?'
    return ()

一些注意事项:

• 使用 if x then True else False 是多余的，它等同于 x.
• 类似地，if x == True ... 是多余的，等同于 if x ....

• 您需要区分 IO 操作及其结果.例如，如果你这样做

• Using if x then True else False is redundant, it's equivalent to just x.
• Similarly if x == True ... is redundant and equivalent to if x ....

• You need to distinguish between IO actions and their results. For example, if yo do

do
    i <- getChar
    ...

then in ... i 代表动作的结果，一个字符，所以i :: Char.但是 getChar :: IO Char 是动作本身.您可以将其视为执行时返回 Char 的配方.您可以将配方传递给函数等，并且只有在某处执行时才会执行.

then in ... i represents the result of the action, a character, so i :: Char. But getChar :: IO Char is the action itself. You can view it as a recipe that returns Char when performed. You can pass the recipe around to functions etc., and it is only performed when executed somewhere.

你的 while 调用了 funktion 两次，这可能不是你想要的 - 它会读取一个字符两次，检查第一个并返回第二个一.请记住，您的 funktion 是一个动作，因此每次您调用"该动作时(例如通过在 do<中使用 <- funktion .../code> 符号)，动作再次运行.所以它应该像

Your while called funktion twice, which probably isn't what you intend - it would read a character twice, check the first one and return the second one. Remember, your funktion is an action, so each time you "invoke" the action (for example by using <- funktion ... in the do notation), the action is run again. So it should rather be something like

do
    y <- funktion x
    f <- praed y
    -- ...

(我的代码有些不同，它检查传递给它的参数.)

(My code is somewhat different, it checks the argument that is passed to it.)

这篇关于带有条件的 Haskell 中的 while 循环的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！