超时的Haskell函数

Question

import Math.NumberTheory.Primes (factorise)
import System.Timeout (timeout)
import Control.Monad (liftM)
type RetType = [(Integer, Int)] -- factorise's return type
-- proposed function
timedFact :: Integer -> Integer -> Either RetType Integer
timedFact u n = ?

试着理解如何为factorise编写一个包装器函数，该函数在使用后会超时。如果成功，则返回RetType，否则返回Integer(传入的内容)

我对Haskell有点陌生。我理解超时需要在IO Monad工作，但我很难收回适当的结果。(注:我没有和Either结婚。Maybe RetType也可以)。

感谢您的帮助

Answer 1

看看timeout :: Int -> IO a -> IO (Maybe a)这个类型，它可以被用作

import Math.NumberTheory.Primes (factorise)
import System.Timeout (timeout)
import Control.Exception (evaluate)
import Control.DeepSeq (force)
timedFact :: Int -> Integer -> IO (Maybe [(Integer, Int)])
timedFact u = 
      timeout u . evaluate . force . factorise 

测试:

 #> timedFact 3000000 1231231231223234234273434343469494949494499437141
Nothing
(3.04 secs, 2639142736 bytes)
 #> timedFact 4000000 1231231231223234234273434343469494949494499437141
Just [(1009,1),(47729236307,1),(125199345589541,1),(204202903382078984027,1)]
(3.07 secs, 2662489296 bytes)

update: as user2407038在评论中说(谢谢!)，

timedFact u n = timeout u (return $!! factorise n)

同样适用。($!!)也来自Control.DeepSeq。要引用文档，"在表达式f $!!