F＃ - 我可以从函数返回一个有区别的联合

Question

I have the following types:

type GoodResource = {
    Id:int;
    Field1:string }


type ErrorResource = {
    StatusCode:int;
    Description:string }

I have the following discriminated union:

type ProcessingResult = 
    | Good of GoodResource
    | Error of ErrorResource

Then want to have a function that will have a return type of the discriminated union ProcessingResult:

let SampleProcessingFunction value =
    match value with
    | "GoodScenario" -> { Id = 123; Field1 = "field1data" }
    | _ -> { StatusCode = 456; Description = "desc" }

Is what I am trying to do possible. The compiler is giving out stating that it expects GoodResource to be the return type. What am I missing or am I completely going about this the wrong way?

Answer 1

As it stands, SampleProcessingFunction returns two different types for each branch.

To return the same type, you need to create a DU (which you did) but also specify the case of the DU explicitly, like this:

let SampleProcessingFunction value =
    match value with
    | "GoodScenario" -> Good { Id = 123; Field1 = "field1data" }
    | _ -> Error { StatusCode = 456; Description = "desc" }

You might ask "why can't the compiler figure out the correct case automatically", but what happens if your DU has two cases of the same type? For example:

type GoodOrError = 
    | Good of string
    | Error of string

In the example below, the compiler cannot determine which case you mean:

let ReturnGoodOrError value =
    match value with
    | "GoodScenario" -> "Goodness"
    | _ -> "Badness"

So again you need to use the constructor for the case you want:

let ReturnGoodOrError value =
    match value with
    | "GoodScenario" -> Good "Goodness"
    | _ -> Error "Badness"

Answer 2

You have to state the case of the union type you want to return in either branch.

let SampleProcessingFunction value =
    match value with
    | "GoodScenario" -> { Id = 123; Field1 = "field1data" } |> Good
    | _ -> { StatusCode = 456; Description = "desc" } |> Error

I suggest to read this excellent articles by Scott Wlaschin Railway Oriented Programming

Answer 3

{ Id = 123; Field1 = "field1data" } { Id = 123; Field1 = "field1data" } is a value of type GoodResource , not of type ProcessingResult . To create a value of type ProcessingResult , you need to use one of its two constructors: Good or Error .

So your function can be written like this:

let SampleProcessingFunction value =
    match value with
    | "GoodScenario" -> Good { Id = 123; Field1 = "field1data" }
    | _ -> Error { StatusCode = 456; Description = "desc" }