这段Haskell代码使用-O运行很多，但-O应为non-dangerous。谁能告诉我发生了什么？如果重要，那就是尝试解决this problem，它使用二分搜索和持久段树：
import Control.Monad
import Data.Array
data Node =
Leaf Int -- value
| Branch Int Node Node -- sum, left child, right child
type NodeArray = Array Int Node
-- create an empty node with range [l, r)
create :: Int -> Int -> Node
create l r
| l + 1 == r = Leaf 0
| otherwise = Branch 0 (create l m) (create m r)
where m = (l + r) `div` 2
-- Get the sum in range [0, r). The range of the node is [nl, nr)
sumof :: Node -> Int -> Int -> Int -> Int
sumof (Leaf val) r nl nr
| nr <= r = val
| otherwise = 0
sumof (Branch sum lc rc) r nl nr
| nr <= r = sum
| r > nl = (sumof lc r nl m) + (sumof rc r m nr)
| otherwise = 0
where m = (nl + nr) `div` 2
-- Increase the value at x by 1. The range of the node is [nl, nr)
increase :: Node -> Int -> Int -> Int -> Node
increase (Leaf val) x nl nr = Leaf (val + 1)
increase (Branch sum lc rc) x nl nr
| x < m = Branch (sum + 1) (increase lc x nl m) rc
| otherwise = Branch (sum + 1) lc (increase rc x m nr)
where m = (nl + nr) `div` 2
-- signature said it all
tonodes :: Int -> [Int] -> [Node]
tonodes n = reverse . tonodes' . reverse
where
tonodes' :: [Int] -> [Node]
tonodes' (h:t) = increase h' h 0 n : s' where s'@(h':_) = tonodes' t
tonodes' _ = [create 0 n]
-- find the minimum m in [l, r] such that (predicate m) is True
binarysearch :: (Int -> Bool) -> Int -> Int -> Int
binarysearch predicate l r
| l == r = r
| predicate m = binarysearch predicate l m
| otherwise = binarysearch predicate (m+1) r
where m = (l + r) `div` 2
-- main, literally
main :: IO ()
main = do
[n, m] <- fmap (map read . words) getLine
nodes <- fmap (listArray (0, n) . tonodes n . map (subtract 1) . map read . words) getLine
replicateM_ m $ query n nodes
where
query :: Int -> NodeArray -> IO ()
query n nodes = do
[p, k] <- fmap (map read . words) getLine
print $ binarysearch (ok nodes n p k) 0 n
where
ok :: NodeArray -> Int -> Int -> Int -> Int -> Bool
ok nodes n p k s = (sumof (nodes ! min (p + s + 1) n) s 0 n) - (sumof (nodes ! max (p - s) 0) s 0 n) >= k
（这与code review的代码完全相同，但这个问题解决了另一个问题。）
这是我在C ++中的输入生成器：
#include <cstdio>
#include <cstdlib>
using namespace std;
int main (int argc, char * argv[]) {
srand(1827);
int n = 100000;
if(argc > 1)
sscanf(argv[1], "%d", &n);
printf("%d %d\n", n, n);
for(int i = 0; i < n; i++)
printf("%d%c", rand() % n + 1, i == n - 1 ? '\n' : ' ');
for(int i = 0; i < n; i++) {
int p = rand() % n;
int k = rand() % n + 1;
printf("%d %d\n", p, k);
}
}
如果您没有可用的C ++编译器，this is the result of ./gen.exe 1000。
这是我电脑上的执行结果：
$ ghc --version
The Glorious Glasgow Haskell Compilation System, version 7.8.3
$ ghc -fforce-recomp 1827.hs
[1 of 1] Compiling Main ( 1827.hs, 1827.o )
Linking 1827.exe ...
$ time ./gen.exe 1000 | ./1827.exe > /dev/null
real 0m0.088s
user 0m0.015s
sys 0m0.015s
$ ghc -fforce-recomp -O 1827.hs
[1 of 1] Compiling Main ( 1827.hs, 1827.o )
Linking 1827.exe ...
$ time ./gen.exe 1000 | ./1827.exe > /dev/null
real 0m2.969s
user 0m0.000s
sys 0m0.045s
这是堆配置文件摘要：
$ ghc -fforce-recomp -rtsopts ./1827.hs
[1 of 1] Compiling Main ( 1827.hs, 1827.o )
Linking 1827.exe ...
$ ./gen.exe 1000 | ./1827.exe +RTS -s > /dev/null
70,207,096 bytes allocated in the heap
2,112,416 bytes copied during GC
613,368 bytes maximum residency (3 sample(s))
28,816 bytes maximum slop
3 MB total memory in use (0 MB lost due to fragmentation)
Tot time (elapsed) Avg pause Max pause
Gen 0 132 colls, 0 par 0.00s 0.00s 0.0000s 0.0004s
Gen 1 3 colls, 0 par 0.00s 0.00s 0.0006s 0.0010s
INIT time 0.00s ( 0.00s elapsed)
MUT time 0.03s ( 0.03s elapsed)
GC time 0.00s ( 0.01s elapsed)
EXIT time 0.00s ( 0.00s elapsed)
Total time 0.03s ( 0.04s elapsed)
%GC time 0.0% (14.7% elapsed)
Alloc rate 2,250,213,011 bytes per MUT second
Productivity 100.0% of total user, 83.1% of total elapsed
$ ghc -fforce-recomp -O -rtsopts ./1827.hs
[1 of 1] Compiling Main ( 1827.hs, 1827.o )
Linking 1827.exe ...
$ ./gen.exe 1000 | ./1827.exe +RTS -s > /dev/null
6,009,233,608 bytes allocated in the heap
622,682,200 bytes copied during GC
443,240 bytes maximum residency (505 sample(s))
48,256 bytes maximum slop
3 MB total memory in use (0 MB lost due to fragmentation)
Tot time (elapsed) Avg pause Max pause
Gen 0 10945 colls, 0 par 0.72s 0.63s 0.0001s 0.0004s
Gen 1 505 colls, 0 par 0.16s 0.13s 0.0003s 0.0005s
INIT time 0.00s ( 0.00s elapsed)
MUT time 2.00s ( 2.13s elapsed)
GC time 0.87s ( 0.76s elapsed)
EXIT time 0.00s ( 0.00s elapsed)
Total time 2.89s ( 2.90s elapsed)
%GC time 30.3% (26.4% elapsed)
Alloc rate 3,009,412,603 bytes per MUT second
Productivity 69.7% of total user, 69.4% of total elapsed
答案 0 :(得分：41)
我想是时候这个问题得到了正确答案。
-O 让我放大你的主要功能，然后稍微重写一下：
main :: IO ()
main = do
[n, m] <- fmap (map read . words) getLine
line <- getLine
let nodes = listArray (0, n) . tonodes n . map (subtract 1) . map read . words $ line
replicateM_ m $ query n nodes
显然，此处的目的是NodeArray创建一次，然后在m的每个query次调用中使用。
不幸的是，GHC有效地将此代码转换为
main = do
[n, m] <- fmap (map read . words) getLine
line <- getLine
replicateM_ m $ do
let nodes = listArray (0, n) . tonodes n . map (subtract 1) . map read . words $ line
query n nodes
你可以立即在这里看到问题。
原因是状态黑客，它（粗略地）说：“当某些东西属于IO a类型时，假设它只被调用一次。” official documentation并不是更精细：
-fno-state-hack关闭＆＃34;状态黑客＆＃34;因此，任何带有State＃token作为参数的lambda都被认为是单项，因此可以内嵌其中的内容。这可以提高IO和ST monad代码的性能，但是存在降低共享的风险。
粗略地说，这个想法如下：如果你定义一个具有IO类型和where子句的函数，例如
foo x = do
putStrLn y
putStrLn y
where y = ...x...
IO a类型的某些内容可以被视为RealWord -> (a, RealWorld)类型的内容。在该观点中，上述变为（大致）
foo x =
let y = ...x... in
\world1 ->
let (world2, ()) = putStrLn y world1
let (world3, ()) = putStrLn y world2
in (world3, ())
对foo的调用（通常）会显示为此foo argument world。但是foo的定义只接受一个参数，而另一个只是稍后由本地lambda表达式使用！这对foo的调用非常缓慢。如果代码看起来像这样会快得多：
foo x world1 =
let y = ...x... in
let (world2, ()) = putStrLn y world1
let (world3, ()) = putStrLn y world2
in (world3, ())
这被称为eta-expansion，并且基于各种理由（例如analyzing the function’s definition，checking how it is being called，以及 - 在这种情况下 - 类型定向启发式）。
不幸的是，如果对foo的调用实际上是let fooArgument = foo argument形式，即带有参数，但没有通过world，则会降低性能。在原始代码中，如果多次使用fooArgument，y仍将只计算一次并共享。在修改后的代码中，每次都会重新计算y - 确切地说是nodes发生了什么。
可能。请参阅#9388以尝试这样做。修复它的问题在于，即使编译器不可能确切地知道这种情况， 会在很多情况下成本提高性能。并且可能存在技术上不合适的情况，即共享丢失，但它仍然是有益的，因为来自更快呼叫的加速超过了重新计算的额外成本。所以目前还不清楚从哪里开始。