F#，在不使用临时变量的情况下向前传递匹配项 [英] F#, Pipe forward a match case without using temp variable

问题描述

我想在不使用temp变量或lambda的情况下将变量通过管道传递给匹配项.想法:

I would like to pipe-forward a variable to a match case without using a temp variable or a lambda. The idea:

let temp =
    x 
    |> Function1
    |> Function2
    // ........ Many functions later.
    |> FunctionN

let result =
    match temp with
    | Case1 -> "Output 1"
    | Case2 -> "Output 2"
    | _ -> "Other Output"

---

我希望写类似以下内容的东西:

---

I hope to write something similar to the following:

// IDEAL CODE (with syntax error)
let result =
    x
    |> Function1
    |> Function2
    // ........ Many functions later.
    |> FunctionN
    |> match with   // Syntax error here! Should use "match something with"
        | Case1 -> "Output 1"
        | Case2 -> "Output 2"
        | _ -> "Other Output"

---

我最接近的东西是使用lambda.但是我认为下面的代码也不是那么好，因为我仍在命名"临时变量.

---

The closest thing that I have is the following by using a lambda. But I think the code below is not really that great either, because I am still "naming" the temp variable.

let result =
    x
    |> Function1
    |> Function2
    // ........ Many functions later.
    |> FunctionN
    |> fun temp -> 
        match temp with   
        | Case1 -> "Output 1"
        | Case2 -> "Output 2"
        | _ -> "Other Output"

---

另一方面，我可以用大量代码直接替换"temp"变量:

---

On the other hand, I can directly replace the "temp" variable with a big chunk of code:

let result =
    match x 
          |> Function1
          |> Function2
          // ........ Many functions later.
          |> FunctionN with
    | Case1 -> "Output 1"
    | Case2 -> "Output 2"
    | _ -> "Other Output"

是否可以编写类似于代码2的代码?还是我必须选择代码3或代码4?谢谢.

Is it possible to write a code similar to Code #2? Or do I have to choose either Code #3 or #4? Thank you.

推荐答案

let result =
    x
    |> Function1
    |> Function2
    // ........ Many functions later.
    |> FunctionN
    |> function  
        | Case1 -> "Output 1"
        | Case2 -> "Output 2"
        | _ -> "Other Output"

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