Question

我想在不使用临时变量或lambda的情况下将变量传递给匹配大小写。想法：

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

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

我希望写下类似于以下内容的内容：

// 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"

我最接近的是使用lambda。但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。

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，我可以用大块代码直接替换“temp”变量：

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Code＃2的代码？或者我必须选择Code＃3还是＃4？谢谢。

Answer 1

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

F＃，在不使用临时变量

1 个答案: