Haskell函数超时

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 = ?

试图理解如何编写因使用过后超时的因子的包装函数.如果成功则返回,RetType否则返回Integer(传入的内容)

我是Haskell的新手.我知道超时需要在IOMonad 工作,但我无法撤回相应的结果.(注意:我没有结婚Either. Maybe RetType也没关系.)

谢谢你的帮助

Answer 1

查看类型,timeout :: Int -> IO a -> IO (Maybe a)它可以用作

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)

更新:正如user2407038 在评论中所说(谢谢!),

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

也有效.($!!)也来自Control.DeepSeq.引用文档,"在表达式f $ !! x中,x在函数f应用于它之前完全评估".