我正在阅读为了一个很好的东西,你学习了一个haskell,但是当我读到输入和输出章节中的异常部分时,我遇到了一个不在范围内的错误:`catch'
这是我的代码:
import System.Environment
import System.IO
import System.IO.Error
main = toTry `catch` handler
toTry :: IO ()
toTry = do (fileName:_) <- getArgs
contents <- readFile fileName
putStrLn $ "The file has " ++ show (length (lines contents)) ++ " lines!"
handler :: IOError -> IO ()
handler e = putStrLn "Whoops, had some trouble!"
我收到此错误消息:
Not in scope: `catch'
我是Haskell的新手.主题来自Learn your Haskell书籍"递归数据结构"
这是我的代码
data List a = Empty | Cons a (List a) deriving (Show, Read, Eq, Ord)
main = do
print $ Empty
print $ 5 `Cons` Empty
print $ 4 (Cons 5 Empty)
print $ 3 `Cons` (4 `Cons` (5 `Cons` Empty))
这是我收到的错误消息
No instance for (Show a0) arising from a use of `print'
The type variable `a0' is ambiguous
Possible fix: add a type signature that fixes these type variable(s)
Note: there are several potential instances: …我有一个例子来自于了解你是一个很好的Haskell
class Eq1 a where
(===), (=/=) :: a -> a -> Bool
x === y = not $ x =/= y
x =/= y = not $ x === y
data TrafficLight = Red | Yellow | Green
instance Eq1 TrafficLight where
Red === Red = True
Green === Green = True
Yellow === Yellow = True
_ === _ = False
instance Show TrafficLight where
show Red = "Red light"
show Yellow = "Yellow light"
show Green = …我已经开始学习Haskell,并阅读Learn You Haskell.第8章处理"制作我们自己的类型和类型类",我有一个错误信息,这对我来说是一个问题.可能解决方案是一个小问题,但我找不到它,所以请揭示提示,或者帮我解释一下.
data Person = Person { firstName :: String
, lastName :: String
, age :: Int
} deriving (Eq, Show, Read)
mikeD = Person {firstName = "Michael", lastName = "Diamond", age = 43}
main = print $ read "Person {firstName =\"Michael\", lastName =\"Diamond\", age = 43}" :: Person
这是我收到的错误消息
Couldn't match expected type `Person' with actual type `IO ()'
In a stmt of a 'do' block:
print
$ read
"Person {firstName =\"Michael\", lastName =\"Diamond\", age = 43}" …我尝试实现一个基本的 lambda 函数，但遇到了一些错误，并且在此处的问题之间搜索后无法找出解决方案。我的代码是：
myMap :: (a -> b) -> [a] -> [b]
myMap addSomething [] = []
myMap addSomething (x:xs) = addSomething x : myMap addSomething xs
-- instance Show (a -> b) where
-- show a = "funcion"
list = [0..4]
listTwo = [(5 :: Int)..9]
addSomething :: Int -> Int
addSomething x = x + 1
addSomethingTwo :: Num a => a -> a-> a
addSomethingTwo x = (\x->x+1)
main = do
print $ myMap addSomething list
print …与您一起学习Haskell函子，应用函子和monoids章节，并面临一个新问题，即使我试图解决无法解决的问题，我的代码也在这里，我的代码在这里：
class Monoid m where
mempty :: m
mappend :: m -> m -> m
instance Monoid a => Monoid (Maybe a) where
mempty = Nothing
Nothing `mappend` m = m
m `mappend` Nothing = m
Just m1 `mappend` Just m2 = Just (m1 `mappend` m2)
main = print $ Nothing `mappend` Just "andy"
这是我收到的错误消息：
No instance for (Monoid [Char]) arising from a use of `mappend'
Possible fix: add an instance declaration for (Monoid [Char])
In the second argument of …我是一个真正的一级忍者,想要解决我的作业,但我不能这样做:
所以任务是:
说一个介于1-100之间的数字:
我说15个
Okey,这很好
说1-100之间的数字:
我说231
Grrr,不好回答,说另一个数字
我说-58
Grrr,不好回答,说另一个数字
78
Okey,那是好在
1-100之间说一个数字:
这是我的问题,因为我可以在第一行编写代码,但我无法前进
我的代码是:
x=input("Saying a number between 1-100: ")
print(x)