我有一个data类型:
data BuildException a = KillBuild JobID a Stage
| FailBuild JobID a Stage
| CancelBuild JobID a Stage
| StopBuild JobID a Stage
deriving Typeable
其中a必须有一个Foo类的实例.我记得读过(在RWH中,也许)虽然可能在data定义中有类约束,但这是不可取的.那么这样做的正确方法是什么？
这是我得到的类型错误,其次是相关代码.这可能是因为函数组合的使用不正确,如果是这样,我想解释一下如何解决这个问题.如果这是别的什么,我也想谈谈这个.
frameworkDaemon.lhs:125:5:
Couldn't match expected type `IO ()' with actual type `a0 -> c0'
In the expression: popJobState . popProcessConfig . readJobFile
In an equation for `makeJobState':
makeJobState world
= popJobState . popProcessConfig . readJobFile
where
readJobFile
= do { let ...;
.... }
where
getFileContents (x : xs) = readFile (ixiaRoot ++ "/" ++ (show x))
popProcessConfig = undefined
popJobState = undefined
失败,模块加载:无.
makeJobState :: JobState -> IO ()
makeJobState world =
popJobState . popProcessConfig . readJobFile -- f …这是代码,我试着让类型推断找出函数的类型.代码编译时,它在运行时失败.
Ambiguous type variables `b0', `m0' in the constraint:
(PersistBackend b0 m0) arising from a use of `isFree'
Probable fix: add a type signature that fixes these type variable(s)
In the expression: isFree testDay
In an equation for `it': it = isFree testDay
:t isFree
isFree :: PersistBackend b m => C.Day -> b m Bool
>isFree day = do
match <- selectList [TestStartDate ==. day,
TestStatus !=. Passed,
TestStatus !=. Failed] []
if (L.null match) then (liftIO $ return …I want to write a function which takes a list of integers and returns a list where every element is negative.
negate :: [Int] -> [Int]
negate xs = foldl (\x xs -> (abs x * (-1)) : xs) [] xs
This function negate all the array objects but also reverse the locations of all variables in the array. What make this function reverse the locations?
我一直在尝试采用一些简单的功能，并将其转换为无点样式进行练习。我开始是这样的：
zipSorted x y = (zip . sort) y $ sort x --zipSorted(x, y) = zip(sort(y), sort(x))
并最终将其转换为
zipSorted = flip (zip . sort) . sort
（我不确定这是否是最好的方法，但确实可行）
现在，我试图不让它依赖于进一步降低这种表达zip，并sort在所有。换句话说，我正在寻找此功能：（如果我的词汇没有误，我认为它是一个组合器）
P(f, g, x, y) = f(g(y), g(x))
sort出现两次但只传递一次的事实提示我应该使用应用函子运算符，<*>但由于某种原因我不知道怎么做。
根据我的理解，(f <*> g)(x) = f(x, g(x))因此，我尝试以这种形式重新编写第一个无点表达式：
flip (zip . sort) . sort
(.) (flip $ zip . sort) sort
(flip (.)) sort $ flip (zip . sort)
(flip (.)) sort $ flip $ (zip .) …我试图做一个带有3个参数的函数“ tokenize”。main String，应该在自己的String中的字符字符串，以及要从字符串中删除的chars字符串。
tokenize :: String -> String -> String -> [String]
tokenize [] imp remm = []
tokenize str imp remm = let chr = (head str) in
if elem chr imp then ([chr] : (tokenize (tail str) imp remm))
else if (elem chr remm ) then (tokenize (tail str) imp remm)
else chr: (tokenize (tail str) imp remm)
我收到此错误消息：
Occurs check:
cannot construct the infinite type: a ~ [a]
Expected type: [a]
Actual type: [[a]]
我想得到给定列表的总距离，其中包含Floats元组。我必须保证少于2个元素的列表将输出0.0
我到目前为止所做的是：
distancia :: [(Float,Float)] -> Float
distancia [] = 0.0
distancia [(_,_)] = 0.0
distancia (x:y:xs) = foldl(\(xa,ya)(xb,yb) -> sqrt(((xa-xb)**2)+((ya-yb)**2))) 0 xs
所以我期望的输出是
ghci> distancia [(0,0), (0,0), (1,0), (1,10)]
11.0
ghci> distancia [(1,1), (3,4)]
3.6055512
但是我收到以下错误：
t3_fc42035.hs:9:22: error:
* Couldn't match expected type `Float'
with actual type `(Float, Float)'
* In the expression:
foldl
(\ (xa, ya) (xb, yb) -> sqrt (((xa - xb) ** 2) + ((ya - yb) ** 2)))
0
xs
In an equation for `distancia': …我必须创建一个包含两个数字（n和k）并返回其二项式系数的递归函数。
我必须使用pascal :: Int-> Int-> Int
老实说，我不明白哪里出了问题，谢谢您的帮助！
pascal :: Int -> Int -> Int
pascal n k
| k == 0 = 1
| n == n = 1
| k > n = 0
| otherwise = pascal(n-1)(k-1) + pascal(n-1) + k
以下错误是：
main.hs:7:40: error:
• Couldn't match expected type ‘Int’ with actual type ‘Int -> Int’
• Probable cause: ‘pascal’ is applied to too few arguments
In the second argument of ‘(+)’, namely ‘pascal (n - 1)’
In …我需要实现在列表头部插入两个元素的函数，但我得到
Exception: <interactive>:7:5-41: Non-exhaustive patterns in function addTwoElements
该函数的代码如下
addTwoElements a b [xs]= a : b : [xs]
提前致谢
haskell ×10
types ×3
fold ×2
combinators ×1
persistent ×1
pointfree ×1
recursion ×1
tuples ×1
yesod ×1