您能否告诉我为什么会有错误“无法将预期的类型IO t与推断的类型String相匹配”-参见下文以查看错误的行:
data RectangleType = Rectangle Int Int Int deriving(Show)
menuRectangles :: [RectangleType] -> IO [RectangleType]
menuRectangles rs = do
putStrLn "Please choose option:"
putStrLn "3 - Show all rectangles"
putStrLn "4 - Quit"
putStr "Number: "
n <- getLine
case n of
"3" -> do { showRectangles rs; menuRectangles rs } -- this line is wrong
"4" -> do { putStrLn "Quitting"; return rs }
otherwise -> do { putStrLn "The End"; return rs }
showRectangles :: [RectangleType] -> String
showRectangles x = showingRectangles x
showingRectangles [] = "Empty"
showingRectangles (x:xs) = do printRectangle x
showingRectangles xs
printRectangle :: RectangleType -> String
printRectangle (Rectangle id width height) = "id: " ++ show id ++ "width: " ++ show width ++ "height: " ++ show height ++ "\n";
最佳答案
showingRectangles
是返回字符串的纯函数。为什么在这里使用do
表示法?将其更改为printRectangle x ++ showingRectangles xs
。
由于showRectangles
只是一个纯函数,因此不会打印到控制台。因此,“声明” showRectangles rs
无效。您需要putStrLn
它。
"3" -> do { putStrLn (showRectangles rs); menuRectangles rs }
(顺便说一句,使用此简单的修复程序,
showingRectangles
将始终在最后一行显示Empty
。您需要在函数中再添加一个定义以避免这种情况。)