如果我可以使用RPAR完成所有工作？＆＃x27;
我有操作x*x + 2*y，我想同时评估+的左右操作数。
我的问题是以下两个实现之间有什么区别：
import Control.Parallel.Strategies
foo x y = runEval $ do
left <- rpar a
right <- rpar b -- <==
return $ left+right
where a = x*x
b = 2*y
import Control.Parallel.Strategies
foo x y = runEval $ do
left <- rpar a
right <- rseq b -- <==
return $ left+right
where a = x*x
b = 2*y
？ 我需要哪一个，为什么？
我可以对par和pseq ~~＆gt;我可以做到这一点：
import Control.Parallel
foo x y = a `par` b `par` a+b
where a=x*x; b=2*y
而不是这样：
import Control.Parallel
foo x y = a `par` b `pseq` a+b
where a=x*x; b=2*y
I have the operation x*x + 2*y and I want to evaluate the left and right operands of the + concurrently.
My question is what is the difference between the following two implementations:
import Control.Parallel.Strategies
foo x y = runEval $ do
left <- rpar a
right <- rpar b -- <==
return $ left+right
where a = x*x
b = 2*y
import Control.Parallel.Strategies
foo x y = runEval $ do
left <- rpar a
right <- rseq b -- <==
return $ left+right
where a = x*x
b = 2*y
?
Which one do I need, and why?
I could ask the same about par and pseq ~~> I could do this:
import Control.Parallel
foo x y = a `par` b `par` a+b
where a=x*x; b=2*y
instead of this:
import Control.Parallel
foo x y = a `par` b `pseq` a+b
where a=x*x; b=2*y
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a`par`B`B`Par` a+bsparksa，然后sparksb，然后anforcea+b在某些任意顺序上强制a和b。在理想的情况下，火花将已经强迫a和b，在这种情况下，a+b的强迫被暂停等待计算要完成。但是，当thunk即将被强迫时，还有一个微妙的种族条件，另一个线程开始同时强迫它，然后他们俩都会冗余地运行相同的计算，并互相比赛以在Thunk上写下他们的结果，由于纯度，这应该是相同的，因此为什么在GHC的运行时可以容忍这种比赛条件。a`PAR`B`B`PSEQ`A+Bsparksa，然后强制b，最后强迫a+b代码>。如果b运行足够长的时间，则为a的火花提供了机会，在a+b被迫之前，避免了上述竞赛健康）状况。a `par` b `par` a+bsparksa, then sparksb, then forcesa+bwhich forcesaandbin some arbitrary order. In an ideal scenario, the sparks will already be forcingaandb, in which case the forcing ofa+bis suspended waiting for the computations to finish. However there is a subtle race condition, when a thunk is just about to be forced, and another thread starts forcing it at the same time, then they will both run the same computation redundantly, and race each other to write their results over the thunk, which should be the same thanks to purity hence why this race condition is tolerated in GHC's runtime.a `par` b `pseq` a+bsparksa, then forcesb, and finally forcesa+b. Ifbruns for long enough, that gives the opportunity for the spark ofato fire beforea+bgets forced, avoiding the aforementioned race condition.