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string -++ 字符串和 IO 字符串运算符

转载 作者:行者123 更新时间:2023-12-02 05:07:32 25 4
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explandDol :: String -> String -> [String] -> IO String
explandDol conclusion operators atoms =
let (ys,zs) = splitAt (head (take 1 replacement)) conclusion in ys ++ getConclusion operators atoms ++ (tail zs)
where replacement = elemIndices '$' conclusion



getConclusion :: String -> [String] -> IO String
getConclusion operators atoms =
runRVar (choice [atom1 ++ " " ++ [operator] ++ " " ++ atom2 | atom1 <- atoms, atom2 <- atoms, operator <- operators,checkAtoms atom1 atom2]) StdRandom

有什么好的方法可以解决这个问题吗?我收到此错误:

/home/joe/Documents/haskell/LAG/main/main.hs: line 73, column 69:
Couldn't match expected type `IO String' with actual type `[Char]'
In the expression: ys ++ getConclusion operators atoms ++ (tail zs)
In the expression:
let (ys, zs) = splitAt (head (take 1 replacement)) conclusion
in ys ++ getConclusion operators atoms ++ (tail zs)
In an equation for `explandDol':
explandDol conclusion operators atoms
= let (ys, zs) = splitAt (head (take 1 replacement)) conclusion
in ys ++ getConclusion operators atoms ++ (tail zs)
where
replacement = elemIndices '$' conclusion
/home/joe/Documents/haskell/LAG/main/main.hs: line 73, column 75:
Couldn't match expected type `[Char]' with actual type `IO String'
In the return type of a call of `getConclusion'
In the first argument of `(++)', namely
`getConclusion operators atoms'
In the second argument of `(++)', namely
`getConclusion operators atoms ++ (tail zs)'
/home/joe/Documents/haskell/LAG/main/main.hs: line 73, column 75:
Warning: Redundant bracket
Found:
getConclusion operators atoms ++ (tail zs)
Why not:
getConclusion operators atoms ++ tail zs

最佳答案

因为 getConclusion 返回的值是IO String您不能简单地将它与对未包装值进行操作的函数一起使用。首先使用 x <- getConclusion operators atom 展开值或者如果你想要函数组合使用 fmap .

关于string -++ 字符串和 IO 字符串运算符,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/16051951/

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