Haskell的主要功能 [英] Haskell main function

问题描述

 模块Main其中
 
 qsort :: Ord a => [a]  - > [a] 
 qsort [] = [] 
 qsort（x：xs）= qsort小++ [x] ++ qsort大
其中
小= [a | a< -xs，a< = x] 
 greater = [a | a< -xs，a> x] 
 
 main = do qsort [1,3,2] 
  

我得到以下错误：

 无法与实际类型为'[a0]'的预期类型'IO t0'匹配
在表达式中：main 
当检查函数`main'的类型时
  

我做错了吗？

解决方案

在 do 块中的所有函数都必须匹配返回的monadic值。您可以改为写
$ b $ pre $ $ $ c $ main $ =
print（qsort [1,3,2]）

因为 print 会返回 IO 值。同样，如果您使用的是 Maybe monad，则必须执行类似于

   -  lookup :: Eq k => k  - > [（k，v）]  - >也许v 
  -  listToMaybe :: [a]  - >也许是
 
 firstElementOf :: Eq q => k  - > [（k，[v]）]  - >也许v 
 firstElementOf键assocMap = do 
v<  -  lookup key assocMap 
 first<  -  listToMaybe v 
 return first 
  

这是可行的，因为 lookup 和 listToMaybe 都返回a Maybe ，它是<$ c $的类型签名所指定的整个 do 块的返回值看看 qsort 的类型，它只返回。 code> [a] ，而不是 IO something ，所以它不能直接在 main 的做块。您还可以使用 let 将其返回值赋值给名称：

  main = do 
 let result = qsort [1,3,2] 
打印结果
  

module Main where

qsort :: Ord a => [a] -> [a]
qsort [] = []
qsort (x : xs) = qsort smaller ++ [x] ++ qsort larger
                 where
                   smaller = [a | a <- xs , a <= x]
                   larger  = [a | a <- xs , a >  x]

main = do qsort [1,3,2]

I get the following error

Couldn't match expected type `IO t0' with actual type `[a0]'         
In the expression: main
When checking the type of the function `main'

What am I doing wrong?

解决方案

All functions within a do block must match the monadic value being returned. You could instead write

main = do
    print (qsort [1, 3, 2])

Because print returns an IO value. Similarly, if you were using the Maybe monad, you would have to do something like

-- lookup :: Eq k => k -> [(k, v)] -> Maybe v
-- listToMaybe :: [a] -> Maybe a

firstElementOf :: Eq q => k -> [(k, [v])] -> Maybe v
firstElementOf key assocMap = do
    v <- lookup key assocMap
    first <- listToMaybe v
    return first

This works because lookup and listToMaybe both return a Maybe, which is the return value of the overall do block as specified by the type signature of firstElementOf.

Looking at the type of qsort, it only returns [a], not IO something, so it can't be used directly inside main's do block. You could also assign it's returned value to a name using let:

main = do
    let result = qsort [1, 3, 2]
    print result

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