了解Haskell分析报告中的成本中心名称
我试图解码Haskell分析输出中各种成本中心名称的含义。下面是了解Haskell分析报告中的成本中心名称,haskell,Haskell,我试图解码Haskell分析输出中各种成本中心名称的含义。下面是.prof文件的示例 COST CENTRE MODULE no. entries %time %alloc %time %alloc ... runSiT.\.\.readBufResults SiT.SiT 3487 0 0.0 46.3 51.9 ...
.prof
文件的示例
COST CENTRE MODULE no. entries %time %alloc %time %alloc
...
runSiT.\.\.readBufResults SiT.SiT 3487 0 0.0 46.3 51.9
...
...
readBuffer.(...) SiT.SiT 3540 1 0.0 0.2 0.0 0.2
readBuffer.tm0_vals SiT.SiT 3539 1 0.0 0.0 0.0 0.0
readBuffer.\ SiT.SiT 3499 0 18.4 12.8 31.0 27.7
...
似乎一个点分隔了嵌套的成本中心(例如,readBuffer.n_threads
表示readBuffer
中的绑定n_threads
),但我不确定其他一些元素。\.\.\.
表示嵌套的lambda函数(例如,来自表单\…$\arg->do
)但是,读取缓冲区中的(…)
是什么意思。(…
编辑:
第二个例子是:
statsFields.mkStr.\ Main 3801 4 0.0 0.0 0.0 0.0
statsFields.fmtModePct Main 3811 2 0.0 0.0 0.0 0.0
statsFields.fmtModePct.pct_str Main 3815 2 0.0 0.0 0.0 0.0
ssN SiT.SiT 3817 2 0.0 0.0 0.0 0.0
statsFields.fmtPctI Main 3816 2 0.0 0.0 0.0 0.0
statsFields.fmtModePct.(...) Main 3813 2 0.0 0.0 0.0 0.0
ssMode SiT.SiT 3814 2 0.0 0.0 0.0 0.0
statsFields.fmtModePct.m_fq Main 3812 2 0.0 0.0 0.0 0.0
这方面的资料来源是:
where ...
fmtModePct :: SiTStats -> String
fmtModePct ss = fmtI64 m_fq ++ " (" ++ pct_str ++ ")"
where (m_val,m_fq) = ssMode ss
pct_str = fmtPctI m_fq (ssN ss)
fmtF64 :: Double -> String
fmtF64 = commafy . printf "%.1f"
-- turns 1000 -> 1,000
commafy :: String -> String
commafy str
| head str == '-' = '-':commafy (tail str)
| otherwise = reverse (go (reverse sig)) ++ frac
where (sig,frac) = span (/='.') str
go (a:b:c:cs@(_:_)) = a : b : c : ',' : go cs
go str = str
(…)表示可重复的操作,如递归调用。我在调查我的程序时也有同样的问题。看看下面的简单示例,我递归地计算count
和mergeAndCount
:
count :: [Int] -> (Int, [Int])
count [] = (0, [])
count (x:[]) = (0, [x])
count xs =
let halves = splitAt (length xs `div` 2) xs
(ac, a) = count $ fst halves
(bc, b) = count $ snd halves
(mc, merged) = mergeAndCount a b
in
(ac + bc + mc, merged)
mergeAndCount :: [Int] -> [Int] -> (Int, [Int])
mergeAndCount [] [] = (0, [])
mergeAndCount xs [] = (0, xs)
mergeAndCount [] ys = (0, ys)
mergeAndCount xs@(x:xs') ys@(y:ys') =
let (larger, thisCount, (counted, merged))
= if x < y
then (x, 0, mergeAndCount xs' ys)
else (y, length xs, mergeAndCount xs ys')
in
(thisCount + counted, larger : merged)
其中,count.merged
表示总体结果,count.a
count.b
功能模式匹配的成本中心。此(…)
在每次对mergeAndCount
的内部调用中都清晰可见
如果您的函数包含许多不同的数据处理方法,那么您的分析输出将不同,并且与您发送的数据高度相关。谢谢您的回复。就我而言,我很难看到任何明显的循环。我到处都在使用递归的高阶函数(例如map)。内联代码是否可能出现这种情况?我添加了第二个示例。
commafy
函数确实会递归,但在其他地方显示为statsFields.commafy。(……
)。我想它可能是在这种情况下内联的,而不是在另一种情况下内联的?是否存在对这种分析格式的引用?
count Invariant 103 199999 0.1 4.3 99.2 37.5
count.merged Invariant 118 99998 0.0 0.0 0.0 0.0
count.a Invariant 113 99999 0.0 0.0 0.0 0.0
count.b Invariant 112 99999 0.0 0.0 0.0 0.0
count.(...) Invariant 110 99999 0.0 0.0 99.0 25.2
mergeAndCount Invariant 111 1636301 98.9 25.2 99.0 25.2
mergeAndCount.merged Invariant 122 726644 0.0 0.0 0.0 0.0
mergeAndCount.merged Invariant 121 709659 0.0 0.0 0.0 0.0
mergeAndCount.(...) Invariant 120 776644 0.0 0.0 0.0 0.0
mergeAndCount.cnt Invariant 119 776644 0.0 0.0 0.0 0.0
mergeAndCount.(...) Invariant 117 759658 0.0 0.0 0.0 0.0
mergeAndCount.cnt Invariant 116 759658 0.0 0.0 0.0