Question

为什么Lazy类型的创建速度如此之慢？

假设以下代码：

type T() =
  let v = lazy (0.0)
  member o.a = v.Value

type T2() =
  member o.a = 0.0

#time "on"

for i in 0 .. 10000000 do
  T() |> ignore

#time "on"

for i in 0 .. 10000000 do
  T2() |> ignore

第一个循环给了我：Real: 00:00:00.647而第二个循环给了我Real: 00:00:00.051。懒惰是慢了13倍!!

我试图以这种方式优化我的代码，最终模拟代码速度慢了6倍。跟踪发生减速的地方很有趣......

Answer 1

Lazy版本有一些重要的开销代码 -

 60     .method public specialname
 61            instance default float64 get_a ()  cil managed
 62     {
 63         // Method begins at RVA 0x2078
 64     // Code size 14 (0xe)
 65     .maxstack 3
 66     IL_0000:  ldarg.0
 67     IL_0001:  ldfld class [FSharp.Core]System.Lazy`1<float64> Test/T::v
 68     IL_0006:  tail.
 69     IL_0008:  call instance !0 class [FSharp.Core]System.Lazy`1<float64>::get_Value()
 70     IL_000d:  ret
 71     } // end of method T::get_a

将此与直接版本

进行比较

 .method public specialname
130            instance default float64 get_a ()  cil managed
131     {
132         // Method begins at RVA 0x20cc
133     // Code size 10 (0xa)
134     .maxstack 3
135     IL_0000:  ldc.r8 0.
136     IL_0009:  ret
137     } // end of method T2::get_a

所以直接版本有一个加载然后返回，而间接版本有一个加载然后一个调用然后返回。

由于lazy版本有一个额外的调用，我预计它会明显变慢。

<强>更新所以我想知道我们是否可以创建一个不需要方法调用的lazy自定义版本 - 我还将测试更新为实际调用方法而不是仅创建对象。这是代码：

type T() =
  let v = lazy (0.0)
  member o.a() = v.Value

type T2() =
  member o.a() = 0.0

type T3() = 
  let mutable calculated = true
  let mutable value = 0.0
  member o.a() = if calculated then value else failwith "not done";;

#time "on"
let lazy_ = 
  for i in 0 .. 1000000 do
    T().a() |> ignore
  printfn "lazy"
#time "on"
let fakelazy = 
  for i in 0 .. 1000000 do
    T3().a() |> ignore
  printfn "fake lazy"

#time "on"
let direct = 
  for i in 0 .. 1000000 do
    T2().a() |> ignore
  printfn "direct";;

这给出了以下结果：

lazy
Real: 00:00:03.786, CPU: 00:00:06.443, GC gen0: 7

val lazy_ : unit = ()


--> Timing now on

fake lazy
Real: 00:00:01.627, CPU: 00:00:02.858, GC gen0: 2

val fakelazy : unit = ()


--> Timing now on

direct
Real: 00:00:01.759, CPU: 00:00:02.935, GC gen0: 2

val direct : unit = ()

此处lazy版本仅比直接版本慢2倍，而假冒懒惰版本甚至比直接版本略快 - 这可能是由于基准测试期间发生了GC。

Answer 2

更新.net核心世界

为Lazy添加了一个新的构造函数来处理常量，例如你的情况。不幸的是F＃＆＃39; s lazy＆＃34;伪关键字＆＃34;总是（此刻！）将常量包装为函数。

无论如何，如果你改变：

let v = lazy (0.0)

为：

let v = Lazy<_> 0.0 // NB. Only .net core at the moment

然后你会发现你的T()课程只需要T2的3倍。

（具有延迟常量的重点是什么？这意味着当你有混合的常量和真正的懒惰项时，你可以使用Lazy作为抽象而只需很少的开销......）

...和...

如果您实际多次使用创建的值，则开销会进一步缩小。例如：

open System.Diagnostics

type T() =
  let v = Lazy<_> 0.1
  member o.a () = v.Value

type T2() =
  member o.a () = 0.1

let withLazyType () =
    let mutable sum = 0.0
    for i in 0 .. 10000000 do
        let t = T()
        for __ = 1 to 10 do
            sum <- sum + t.a()
    sum

let withoutLazyType () =
    let mutable sum = 0.0
    for i in 0 .. 10000000 do
        let t = T2()
        for __ = 1 to 10 do
            sum <- sum + t.a()
    sum

let runtest name count f =
    let mutable checksum = 0.
    let mutable totaltime = 0L
    for i = 0 to count do
        if i = 0 then
            f () |> ignore // warm up
        else
            let sw = Stopwatch.StartNew ()
            checksum <- checksum + f ()
            totaltime <- totaltime + sw.ElapsedMilliseconds
    printfn "%s: %4d (checksum=%f for %d runs)" name (totaltime/int64 count) checksum count

[<EntryPoint>]
let main _ =
    runtest "w/o  lazy" 10 withoutLazyType
    runtest "with lazy" 10 withLazyType

    0

带来的时间差异为＆lt; 2次。

NB。我参与了new lazy implementation ......

F＃中懒惰的表现

2 个答案: