haskell - 了解 Haskell 分析报告中的成本中心名称

Question

我试图在我的 Haskell 分析输出中解码各种成本中心名称的含义。这是 .prof 的示例文件

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 n_threads 内的绑定(bind) readBuffer )，但我不确定其他一些元素。和 .\.\.表示嵌套的 lambda 函数(例如来自 forM_ ... $ \arg -> do 之类的东西)。 但是，(...) 是什么意思？在 readBuffer.(...) ?

编辑:
作为第二个例子，我有:

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

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

Answer 1

(...) 表示可重复的操作，例如递归调用。在调查我的程序时，我有同样的问题。看看下面这个简单的例子，我正在评估 count和 mergeAndCount递归:

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)

将生成分析输出，如

    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    

在哪里 count.merged表示总体结果， count.a count.b功能模式匹配的成本中心。这个 (...)对 mergeAndCount 的每次内部调用都清晰可见.

如果您的函数包含许多不同的数据处理方法，您的分析输出将与您发送的数据不同且高度相关。