我试图理解为什么我用do-block编写的函数不能被重写为在列表上fmap类似的lambda表达式。
我有以下内容:
-- This works
test1 x = do
let m = T.pack $ show x
T.putStrLn m
test1 1
产生
1
但
-- This fails
fmap (\x -> do
let m = T.pack $ show x
T.putStrLn m
) [1..10]
-- And this also fails
fmap (\x -> do
T.putStrLn $ T.pack $ show x
) [1..10]
有错误:
<interactive>:1:1: error:
• No instance for (Show (IO ())) arising from a use of ‘print’
• In a stmt of an interactive GHCi command: print it
我的putStrLn在工作和非工作之间是一致的。进口是一样的。我需要打印的show-pack-putstrln舞蹈在工作和非工作之间也是一致的。
印刷品的使用在工作和非工作之间发生了什么变化?
-- I was also surprised that this fails
fmap (T.putStrLn $ T.pack $ show) [1..10]
-- it seemed as similar as possible to the test1 function but mapped.
<interactive>:1:7: error:
• Couldn't match expected type ‘Integer -> b’ with actual type ‘IO ()’
• In the first argument of ‘fmap’, namely ‘(T.putStrLn $ pack $ show)’
In the expression: fmap (T.putStrLn $ pack $ show) [1 .. 10]
In an equation for ‘it’: it = fmap (T.putStrLn $ pack $ show) [1 .. 10]
• Relevant bindings include it :: [b] (bound at <interactive>:1:1)
<interactive>:1:29: error:
• Couldn't match type ‘() -> String’ with ‘String’
Expected type: String
Actual type: () -> String
• Probable cause: ‘show’ is applied to too few arguments
In the second argument of ‘($)’, namely ‘show’
In the second argument of ‘($)’, namely ‘pack $ show’
In the first argument of ‘fmap’, namely ‘(T.putStrLn $ pack $ show)’
-- This lambda returns x of the same type as \x
-- even while incidentally printing along the way
fmap (\x -> do
let m = T.pack $ show x
T.putStrLn $ m
return x
) [1..10]
但也失败了:
<interactive>:1:1: error:
• No instance for (Show (IO Integer)) arising from a use of ‘print’
• In a stmt of an interactive GHCi command: print it
fmap f [1..10]
的类型是[T]
,其中T
是f
的返回类型。
在你的情况下,T = IO ()
,所以完整表达式的类型是[IO ()]
。
无法打印IO操作,因此当您尝试打印该列表时,GHCi会抱怨。您可能想要使用像sequence_ (fmap f [1..10])
之类的东西来运行这些操作而不是打印它们。
或者,考虑放弃fmap
,而是使用类似的东西
import Data.Foldable (for_)
main = do
putStrLn "hello"
for_ [1..10] $ \i -> do
putStrLn "in the loop"
print (i*2)
putStrLn "out of the loop"
你写了:
但是当我将lambda的返回类型更改为与更新2中的x相同的x时...
不,不。你没有。 lambda函数返回其最后一个表达式的值。你的lambda函数只有一个表达式 - 整个do { ... }
块定义了一个值,即lambda函数的返回值。不是x
。 return
属于do
,而不是lambda表达。我们更容易看到我们是否使用显式分隔符来编写它,如
fmap (\x -> do {
let m = T.pack $ show x ;
T.putStrLn $ m ;
return x
} ) [1..10]
do
块作为一个整体具有与其每个行语句相同的monadic类型。
其中一个是putStrLn ...
,其类型是IO ()
。所以你的lambda函数为某些IO t
返回t
。
由于return x
,t
是x
的类型。我们有return :: Monad m => t -> m t
,所以与m ~ IO
它是return :: t -> IO t
。
x
来自参数列表Num t => [t]
,所以总的来说你有
Num t => fmap (fx :: t -> IO t) (xs :: [t]) :: [IO t]
要么
xs :: [t]
fx :: t -> IO t
----------------------------
fmap fx xs :: [IO t]