Scala 是否有任何函数式语言编译器/运行库可以优化链式迭代?
任何函数式语言编译器/运行时是否会在可应用时将所有链式迭代减少为一个?从程序员的角度来看,我们可以使用lazyness和streams等结构优化函数代码,但我有兴趣了解故事的另一面。 我的函数示例是用Scala编写的,但请不要将答案局限于该语言 功能方式: 我希望编译器优化到与以下命令等效的命令:Scala 是否有任何函数式语言编译器/运行库可以优化链式迭代?,scala,optimization,functional-programming,compiler-optimization,Scala,Optimization,Functional Programming,Compiler Optimization,任何函数式语言编译器/运行时是否会在可应用时将所有链式迭代减少为一个?从程序员的角度来看,我们可以使用lazyness和streams等结构优化函数代码,但我有兴趣了解故事的另一面。 我的函数示例是用Scala编写的,但请不要将答案局限于该语言 功能方式: 我希望编译器优化到与以下命令等效的命令: /*仅一次迭代*/ 长和,我; 对于(i=1L,sum=0L;i,理论上正如一位评论者所写,编译器可以在编译时将其还原为结果。这在某些宏中是不可想象的,但在一般情况下不太可能 如果插入.view调
/*仅一次迭代*/
长和,我;
对于(i=1L,sum=0L;i,理论上正如一位评论者所写,编译器可以在编译时将其还原为结果。这在某些宏中是不可想象的,但在一般情况下不太可能
如果插入.view
调用,则Scala中会出现惰性语义,因此只会执行一次迭代,尽管不像命令式代码那么简单:
val lz = (1L to 1000000L).view.filter(_ % 2 == 0) // SeqView (lazy)!
lz.sum
另外,您的假设是错误的,否则会有三次迭代。(1L到1000000L)
创建了一个numeriRange
,它不涉及对元素的任何迭代。因此.view
为您节省了一次迭代。Haskell从定义上讲是一种非严格语言,我所知道的所有实现都使用延迟求值来提供非严格语义
类似的代码(具有用于开始和结束的参数,因此不可能进行编译时计算)
只需一次遍历并在恒定的小内存中计算总和。[low..high]
是enumFromTo low-high
的语法糖,而enumFromTo
对于Int
的定义基本上是
enumFromTo x y
| y < x = []
| otherwise = go x
where
go k = k : if k == y then [] else go (k+1)
和
总和:
sum l = sum' l 0
where
sum' [] a = a
sum' (x:xs) a = sum' xs (a+x)
即使没有任何优化,评估仍将继续进行
sum' (filter even (enumFromTo 1 6)) 0
-- Now it must be determined whether the first argument of sum' is [] or not
-- For that, the application of filter must be evaluated
-- For that, enumFromTo must be evaluated
~> sum' (filter even (1 : go 2)) 0
-- Now filter knows which equation to use, unfortunately, `even 1` is False
~> sum' (filter even (go 2)) 0
~> sum' (filter even (2 : go 3)) 0
-- 2 is even, so
~> sum' (2 : filter even (go 3)) 0
~> sum' (filter even (go 3)) (0+2)
-- Once again, sum asks whether filter is done or not, so filter demands another value or []
-- from go
~> sum' (filter even (3 : go 4)) 2
~> sum' (filter even (go 4)) 2
~> sum' (filter even (4 : go 5)) 2
~> sum' (4 : filter even (go 5)) 2
~> sum' (filter even (go 5)) (2+4)
~> sum' (filter even (5 : go 6)) 6
~> sum' (filter even (go 6)) 6
~> sum' (filter even (6 : [])) 6
~> sum' (6 : filter even []) 6
~> sum' (filter even []) (6+6)
~> sum' [] 12
~> 12
这当然比循环效率低,因为对于枚举的每个元素,都必须生成一个列表单元格,然后对于通过过滤器的每个元素,都必须生成一个列表单元格,而列表单元格只能立即被总和消耗
让我们检查一下内存使用情况是否确实很小:
module Main (main) where
import System.Environment (getArgs)
main :: IO ()
main = do
args <- getArgs
let (low, high) = case args of
(a:b:_) -> (read a, read b)
_ -> error "Want two args"
print $ sum $ filter even [low :: Int .. high]
它为50万个列表单元格分配了大约40MB的内存(1)和一些更改,但最大驻留空间约为44KB。以1000万的上限运行,总体分配(和运行时间)增长了10倍(减去常量),但最大驻留空间保持不变
(1) GHC将枚举和过滤器融合在一起,只生成类型为Int
的范围内的偶数。不幸的是,它无法融合掉sum
,因为这是一个左折叠,而GHC的融合框架只融合右折叠
现在,为了融合和
,我们必须做大量的工作来教GHC如何使用重写规则。幸运的是,向量
包中的许多算法都这样做了,如果我们使用它
module Main where
import qualified Data.Vector.Unboxed as U
import System.Environment (getArgs)
val :: Int -> Int -> Int
val low high = U.sum . U.filter even $ U.enumFromN low (high - low + 1)
main :: IO ()
main = do
args <- getArgs
let (low, high) = case args of
(a:b:_) -> (read a, read b)
_ -> error "Want two args"
print $ val low high
以下是GHC为(val的工作人员)制作的核心,如果有人感兴趣:
Rec {
Main.main_$s$wfoldlM'_loop [Occ=LoopBreaker]
:: GHC.Prim.Int# -> GHC.Prim.Int# -> GHC.Prim.Int# -> GHC.Prim.Int#
[GblId, Arity=3, Caf=NoCafRefs, Str=DmdType LLL]
Main.main_$s$wfoldlM'_loop =
\ (sc_s303 :: GHC.Prim.Int#)
(sc1_s304 :: GHC.Prim.Int#)
(sc2_s305 :: GHC.Prim.Int#) ->
case GHC.Prim.># sc1_s304 0 of _ {
GHC.Types.False -> sc_s303;
GHC.Types.True ->
case GHC.Prim.remInt# sc2_s305 2 of _ {
__DEFAULT ->
Main.main_$s$wfoldlM'_loop
sc_s303 (GHC.Prim.-# sc1_s304 1) (GHC.Prim.+# sc2_s305 1);
0 ->
Main.main_$s$wfoldlM'_loop
(GHC.Prim.+# sc_s303 sc2_s305)
(GHC.Prim.-# sc1_s304 1)
(GHC.Prim.+# sc2_s305 1)
}
}
end Rec }
几年前,我发表了两篇关于这个话题的博客文章:
请注意,Scala编译器完成的专门化和优化从那时起有了很大的改进(可能在Hotspot中也有改进),所以今天的结果可能会更好。我们甚至可以对其进行优化以进行编译时评估:
val sum=250000500000
。也许有些编译器会这样做?@leems这是正确的,但我对编译时不知道值的情况感兴趣。您能给出您希望编译器生成的伪代码吗?@tsenart Since你似乎很感兴趣,我已经写了一个部分详细的答案。我希望即使没有太多Haskell知识也可以访问。你在寻找“毁林”或“融合”优化。这些已经在Haskell的上下文中进行了广泛的研究,但在其他FP-ish编译器中也有一些更受限制的形式。就最终结果而言,您的解决方案确实是优化的。我的问题,如上文所述,是指编译器对任何函数中第一个代码段的等效优化nal语言。关于你的P.S:我还没有反编译生成的Java字节码,所以我不能确定,但列表必须以某种方式构造和初始化。我相信在较低的级别必须通过迭代来完成。如果我错了,请纠正我。表示(1L到100000ml)
您需要存储1L
和1000000ML
的类型。为什么要迭代?当您尝试打印该对象(toString
)时可能会发生这种情况当然。是的,但是严格语言呢?嗯,我认为严格语言的编译器也有可能将其重写到循环中。但我不知道是否有严格函数语言的编译器被教导过这一点。问题特别提到了懒惰,OP对此不感兴趣;示例在scala中。我了解到OP对手动插入的惰性不感兴趣,因为在这种情况下,同样可以编写循环并完成它,特别是在OP要求详细说明我的评论之后。循环融合已经在OCaml和至少一个专有的strict Haskell编译器中实现。
sum' (filter even (enumFromTo 1 6)) 0
-- Now it must be determined whether the first argument of sum' is [] or not
-- For that, the application of filter must be evaluated
-- For that, enumFromTo must be evaluated
~> sum' (filter even (1 : go 2)) 0
-- Now filter knows which equation to use, unfortunately, `even 1` is False
~> sum' (filter even (go 2)) 0
~> sum' (filter even (2 : go 3)) 0
-- 2 is even, so
~> sum' (2 : filter even (go 3)) 0
~> sum' (filter even (go 3)) (0+2)
-- Once again, sum asks whether filter is done or not, so filter demands another value or []
-- from go
~> sum' (filter even (3 : go 4)) 2
~> sum' (filter even (go 4)) 2
~> sum' (filter even (4 : go 5)) 2
~> sum' (4 : filter even (go 5)) 2
~> sum' (filter even (go 5)) (2+4)
~> sum' (filter even (5 : go 6)) 6
~> sum' (filter even (go 6)) 6
~> sum' (filter even (6 : [])) 6
~> sum' (6 : filter even []) 6
~> sum' (filter even []) (6+6)
~> sum' [] 12
~> 12
module Main (main) where
import System.Environment (getArgs)
main :: IO ()
main = do
args <- getArgs
let (low, high) = case args of
(a:b:_) -> (read a, read b)
_ -> error "Want two args"
print $ sum $ filter even [low :: Int .. high]
$ ./sumEvens +RTS -s -RTS 1 1000000
250000500000
40,071,856 bytes allocated in the heap
12,504 bytes copied during GC
44,416 bytes maximum residency (2 sample(s))
21,120 bytes maximum slop
1 MB total memory in use (0 MB lost due to fragmentation)
Tot time (elapsed) Avg pause Max pause
Gen 0 75 colls, 0 par 0.00s 0.00s 0.0000s 0.0000s
Gen 1 2 colls, 0 par 0.00s 0.00s 0.0002s 0.0003s
INIT time 0.00s ( 0.00s elapsed)
MUT time 0.01s ( 0.01s elapsed)
GC time 0.00s ( 0.00s elapsed)
EXIT time 0.00s ( 0.00s elapsed)
Total time 0.01s ( 0.01s elapsed)
%GC time 6.1% (7.6% elapsed)
Alloc rate 4,367,976,530 bytes per MUT second
Productivity 91.8% of total user, 115.8% of total elapsed
module Main where
import qualified Data.Vector.Unboxed as U
import System.Environment (getArgs)
val :: Int -> Int -> Int
val low high = U.sum . U.filter even $ U.enumFromN low (high - low + 1)
main :: IO ()
main = do
args <- getArgs
let (low, high) = case args of
(a:b:_) -> (read a, read b)
_ -> error "Want two args"
print $ val low high
$ ./sumFilter +RTS -s -RTS 1 10000000
25000005000000
72,640 bytes allocated in the heap
3,512 bytes copied during GC
44,416 bytes maximum residency (1 sample(s))
17,024 bytes maximum slop
1 MB total memory in use (0 MB lost due to fragmentation)
Tot time (elapsed) Avg pause Max pause
Gen 0 0 colls, 0 par 0.00s 0.00s 0.0000s 0.0000s
Gen 1 1 colls, 0 par 0.00s 0.00s 0.0001s 0.0001s
INIT time 0.00s ( 0.00s elapsed)
MUT time 0.01s ( 0.01s elapsed)
GC time 0.00s ( 0.00s elapsed)
EXIT time 0.00s ( 0.00s elapsed)
Total time 0.01s ( 0.01s elapsed)
%GC time 1.0% (1.2% elapsed)
Alloc rate 7,361,805 bytes per MUT second
Productivity 97.7% of total user, 111.5% of total elapsed
Rec {
Main.main_$s$wfoldlM'_loop [Occ=LoopBreaker]
:: GHC.Prim.Int# -> GHC.Prim.Int# -> GHC.Prim.Int# -> GHC.Prim.Int#
[GblId, Arity=3, Caf=NoCafRefs, Str=DmdType LLL]
Main.main_$s$wfoldlM'_loop =
\ (sc_s303 :: GHC.Prim.Int#)
(sc1_s304 :: GHC.Prim.Int#)
(sc2_s305 :: GHC.Prim.Int#) ->
case GHC.Prim.># sc1_s304 0 of _ {
GHC.Types.False -> sc_s303;
GHC.Types.True ->
case GHC.Prim.remInt# sc2_s305 2 of _ {
__DEFAULT ->
Main.main_$s$wfoldlM'_loop
sc_s303 (GHC.Prim.-# sc1_s304 1) (GHC.Prim.+# sc2_s305 1);
0 ->
Main.main_$s$wfoldlM'_loop
(GHC.Prim.+# sc_s303 sc2_s305)
(GHC.Prim.-# sc1_s304 1)
(GHC.Prim.+# sc2_s305 1)
}
}
end Rec }