Haskell mapm_不打印

发布于 2025-01-20 16:53:39 字数 838 浏览 3 评论 0原文

因此，我编写了一个程序来查询外汇API（外汇），它的工作方式就像魅力一样，但是当我想查询每个可用的货币对时，它会评估所有API调用，因为要执行时间很长，但没有打印什么。

import Data.Functor ((<&>))

supportedPairs :: IO (Maybe [(String, String)])
forex :: String -> String -> IO (Maybe (Scientific, UnixTime))


main :: IO ()
main = do
    x <- supportedPairs
    mapM_ (flip (<&>) print . uncurry forex) (fromJust x)
    -- this prints nothing at all

单个调用正常工作：

main = do
    x <- supportedPairs    
    u <- (uncurry forex . (flip (!!) 10 . fromJust)) x
    print u
    -- this prints "Just (438.685041,UnixTime {utSeconds = 1649588583, utMicroSeconds = 0})"

为什么对结果进行了评估，为什么mapm _不打印结果？如果我正确理解Haskell的懒惰，那么如果不打印结果，则不应首先对其进行评估？

原文

So I wrote a program to query a forex API (foreign exchange), and it works like a charm, but when I want to query every currency pair available, it evaluates all API calls as it takes a long time to execute but prints nothing.

import Data.Functor ((<&>))

supportedPairs :: IO (Maybe [(String, String)])
forex :: String -> String -> IO (Maybe (Scientific, UnixTime))


main :: IO ()
main = do
    x <- supportedPairs
    mapM_ (flip (<&>) print . uncurry forex) (fromJust x)
    -- this prints nothing at all

The single calls work just fine like this:

main = do
    x <- supportedPairs    
    u <- (uncurry forex . (flip (!!) 10 . fromJust)) x
    print u
    -- this prints "Just (438.685041,UnixTime {utSeconds = 1649588583, utMicroSeconds = 0})"

Why doesn't the mapM_ print the results although they are evaluated? If I understood Haskell's laziness correctly, then if the results are not to be printed they should not be evaluated in the first place?

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世界如花海般美丽 2025-01-27 16:53:39

检查类型：

print 是 ... -> IO()。

因此，... <&> print 是IO (IO ())。注意这里的双IO。

因此，对其进行映射将运行“最外部 IO”，但不会运行“最内部 IO”。更具体地说，比较一下：

main = do
  x <- print True >> return 5          -- x is 5
  y <- return (print True >> return 5) -- y is an IO action
  ...

这里只执行第一个 print True：第二个 IO 操作用于定义 y 但直到我们运行 y它不会被执行。

最后一点：在这里，您不需要 <&> 因为这会创建嵌套 IO。使用flip (>>=) print（或(=<<) print，或(>>= print) >) 而不是 flip <&>打印。

Check the types:

print is ... -> IO ().

Therefore, ... <&> print is IO (IO ()). Note the double IO here.

Hence, mapping over that, will run the "outermost IO" but not the "innermost IO". More concretely, compare this:

main = do
  x <- print True >> return 5          -- x is 5
  y <- return (print True >> return 5) -- y is an IO action
  ...

Only the first print True here is executed: the second IO action is used to define y but until we run y it won't be executed.

The final point: here, you do not need <&> since that creates the nested IO's. Use flip (>>=) print (or (=<<) print, or (>>= print)) instead of flip <&> print.

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