Old*_*vec 3 performance f# lazy-evaluation
为什么Lazy类型的创建如此之慢?
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
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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.懒惰慢了13倍!!
我试图以这种方式优化我的代码,最终我的模拟代码速度慢了6倍.跟踪发生减速的地方很有趣...
懒惰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
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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
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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一个返回.
由于lazy版本有一个额外的调用,我希望它会明显变慢.
更新:
所以我想知道我们是否可以创建一个lazy不需要方法调用的自定义版本- 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";;
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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 = ()
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这里的lazy版本只比直接版本慢2倍,而假冒懒惰版本甚至比直接版本略快 - 这可能是由于在基准测试期间发生了GC.