F# 为什么源按seq作为最后一个参数

当我在f#中查看seq type时，最后一个参数总是源代码类型

Seq.pick chooser source
Seq.concat sources

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为什么?？这是因为currying吗？

将源作为最后一个参数允许部分应用和使用如下Piping：

let source = seq { 1..2 }
let chooser x = Some x
let result = source |> Seq.pick chooser

let source = seq { 1..2 }
let chooser x = Some x
let result = source 
             |> Seq.map (fun x -> x*x )
             |> Seq.pick chooser

如果您打算通过以下一系列操作对序列进行管道化处理，这一点尤其有用：

let source = seq { 1..2 }
let chooser x = Some x
let result = source |> Seq.pick chooser

let source = seq { 1..2 }
let chooser x = Some x
let result = source 
             |> Seq.map (fun x -> x*x )
             |> Seq.pick chooser

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将源作为最后一个参数允许部分应用和如下所示的pipping使用：

let source = seq { 1..2 }
let chooser x = Some x
let result = source |> Seq.pick chooser

let source = seq { 1..2 }
let chooser x = Some x
let result = source 
             |> Seq.map (fun x -> x*x )
             |> Seq.pick chooser

如果您打算通过以下一系列操作对序列进行管道化处理，这一点尤其有用：

let source = seq { 1..2 }
let chooser x = Some x
let result = source |> Seq.pick chooser

let source = seq { 1..2 }
let chooser x = Some x
let result = source 
             |> Seq.map (fun x -> x*x )
             |> Seq.pick chooser

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是的，对函数参数进行排序以使部分应用程序更有用是很常见的。是的，对函数参数进行排序以使部分应用程序更有用是很常见的。