F#:向下播放序列至IEnumerator
为什么在F#中,我可以这样做F#:向下播放序列至IEnumerator,f#,ienumerator,seq,F#,Ienumerator,Seq,为什么在F#中,我可以这样做 let s = seq { for i in 0 .. 4095 do yield i } :?> IEnumerator 。。。但这会抛出一个系统。InvalidCastException let s = Seq.init 4095 (fun i -> i) :?> IEnumerator 序列表达式创建一个实现IEnumerable和IEnumerator 您可以重构代码以使用IEnumerable而不是IEnumerator,因为这两种构
let s = seq { for i in 0 .. 4095 do yield i } :?> IEnumerator
。。。但这会抛出一个系统。InvalidCastException
let s = Seq.init 4095 (fun i -> i) :?> IEnumerator
序列表达式创建一个实现
IEnumerable
和IEnumerator
您可以重构代码以使用IEnumerable
而不是IEnumerator
,因为这两种构造都会生成IEnumerable
或者,如果您确实需要一个IEnumerator
,只需调用GetEnumerator
,即可从可枚举对象返回枚举器
:
let s = (Seq.init 4095 (fun i -> i)).GetEnumerator()
printfn "%b" (s :? IEnumerable<int>) // false
printfn "%b" (s :? IEnumerator<int>) // true
let s=(Seq.init 4095(fun i->i)).GetEnumerator()
printfn“%b”(s:?IEnumerable)//false
printfn“%b”(s:?IEnumerator)//true
如果查看,序列表达式将转换为:
Seq.collect (fun pat -> Seq.singleton(pat)) (0 .. 4095)
如果您查看源代码以了解Seq.collect
的定义,它是:
let collect f sources = map f sources |> concat
let concat sources =
checkNonNull "sources" sources
mkConcatSeq sources
如果你看一下concat
的定义,它是:
let collect f sources = map f sources |> concat
let concat sources =
checkNonNull "sources" sources
mkConcatSeq sources
mkConcatSeq
定义为:
let mkConcatSeq (sources: seq<'U :> seq<'T>>) =
mkSeq (fun () -> new ConcatEnumerator<_,_>(sources) :> IEnumerator<'T>)
let init count f =
if count < 0 then invalidArg "count" (SR.GetString(SR.inputMustBeNonNegative))
mkSeq (fun () -> IEnumerator.upto (Some (count-1)) f)
let mkSeq f =
{ new IEnumerable<'U> with
member x.GetEnumerator() = f()
interface IEnumerable with
member x.GetEnumerator() = (f() :> IEnumerator) }
而mkSeq
定义为:
let mkConcatSeq (sources: seq<'U :> seq<'T>>) =
mkSeq (fun () -> new ConcatEnumerator<_,_>(sources) :> IEnumerator<'T>)
let init count f =
if count < 0 then invalidArg "count" (SR.GetString(SR.inputMustBeNonNegative))
mkSeq (fun () -> IEnumerator.upto (Some (count-1)) f)
let mkSeq f =
{ new IEnumerable<'U> with
member x.GetEnumerator() = f()
interface IEnumerable with
member x.GetEnumerator() = (f() :> IEnumerator) }
让mkSeq f=
{new IEnumerable
而不是IEnumerator
你确定你不想要IEnumerable而不是IEnumerator吗?我正在为Unity编写一个协同程序。必须是IEnumerable。Dang,偷了我的答案。嗯,部分原因是…不知道序列表达式产生的结果与Seq模块对应的结果不完全相同。