haskell -skipping getLine

Question

嘿 - 伟大的程序员和haskellers,我是一名哈斯克尔大学新生并且有一个程序问题归结为以下情况

main :: IO ()
main = do
    putStrLn "\nplease give me some input"
    input1 <- getLine
    putStrLn "\nplease give me another input"
    input2 <-getLine
    putStrLn ("\nyour inputs were "++show(input1)++" and "++ show(input2)")
    putStrLn "restart ?? yY or nN"
    c <- getChar
    restart c
    where 
    restart c
        |elem c "yY" = do
            main
        |elem c "nN" = putStrLn "\nExample Over"
        |otherwise = do
            putStrLn "\nyou must type one of Yy to confirm or nN to abort"
            c'<- getChar
            restart c'

除了第一次执行main之外的任何一个

input1 <- getLine

被跳过,我找不到它的理由,如下所示

input2 <- getLine

按预期执行,我对任何建议持开放态度,并提前感谢ε/ 2

Answer 1

修复:设置NoBuffering在程序的开头:

hSetBuffering stdin NoBuffering

为什么这会解决问题？当你不使用NoBuffering时,看看你正在输入什么!您输入并getLine消耗:

first input[enter]

然后键入,getLine#2消耗:

second input[enter]

然后键入:

 y[enter]

但是getChar只消耗了y并且离开了[enter]缓冲区,这是您的第一个getLine呼叫读取的!你为什么打字[enter]？因为你必须这样做,只是点击'y'不会导致main循环,因为终端是行缓冲的.