Typescript type generation issue in switch statement

Question

I have this map function that is rendering the type of an object.

function mapStuffs(fields: any) {
    switch (fields.type) {
        case "NewStuffs":
            return {
                type: "NewStuffs" as const,
                stuff: fields.stuff
            };
        case "StuffFile":
            const fileName = "jadflkjs";
            const stuffId = "adlkajsf";
            return {
                type: "StuffFile" as const,
                fileName,
                stuffId
            };
        default:
            return {
                type: "ExistingStuff" as const
            };
    }
}

And when I build, the generated type is:

mainstuff:
        | {
              stuff: string | undefined;
              type: "NewStuffs";
              fileName?: undefined;
              stuffId?: undefined;
          }
        | {
              type: "StuffFile";
              fileName: string;
              stuffId: string;
          }
        | {
              type: "ExistingStuff";
              fileName?: undefined;
              stuffId?: undefined;
          };

I don't understand how to remove fileName and stuffId as optional parameters when they're not required. Is it a typescript bug? What I would like to have is this type:

mainstuff:
        | {
              stuff: string | undefined;
              type: "NewStuffs";
          }
        | {
              type: "StuffFile";
              fileName: string;
              stuffId: string;
          }
        | {
              type: "ExistingStuff";
          };

jcalz · Accepted Answer · 2021-05-31 16:53:46Z

2

It's not a TypeScript bug. This is the intended behavior since TypeScript 2.7 introduced improved type inference for object literals. The fact that fileName and stuffId is not being included for some of the return values makes the compiler think that absence should be reflected in the return type.

The type inference algorithm in TypeScript is a set of heuristics, where the language designers try to guess what kinds of things most people want to see in certain circumstances. If you don't like the inferred type, you are free to annotate the return type of your mapStuffs() function with whatever type you prefer, and the compiler will be happy with this, assuming your function implementation conforms to it:

type MappedStuff = {
    stuff: string | undefined;
    type: "NewStuffs";
} | {
    type: "StuffFile";
    fileName: string;
    stuffId: string;
} | {
    type: "ExistingStuff";
};

function mapStuffs(fields: any): MappedStuff {

    switch (fields.type) {
        case "NewStuffs":
            return {
                type: "NewStuffs" as const,
                stuff: fields.stuff
            };
        case "StuffFile":
            const fileName = "jadflkjs";
            const stuffId = "adlkajsf";
            return {
                type: "StuffFile" as const,
                fileName,
                stuffId
            };
        default:
            return {
                type: "ExistingStuff" as const
            };
    }
}

Playground link to code

answered May 31, 2021 at 16:53

jcalz

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7 Comments

sharkbait Over a year ago

thanks a lot! now it's really clear! very helpful

sharkbait Over a year ago

but now my question is: why I don't see "stuff" attribute in the other objects?

jcalz Over a year ago

Sorry, I’m not sure what you’re asking. Follow up questions may deserve their own posts, in any case.

sharkbait Over a year ago

sure, sorry, I was just asking why the "issue" I had was happening only with 'fileName' and 'stuffId' (attributes for type="StuffFile") and not for 'stuff' (attribute for type="NewStuffs"). But no worries, I'll open a new question

jcalz Over a year ago

I’m not sure because the question has any for fields, but also string | undefined for fields.stuff somehow. I think I’ll need a true minimal reproducible example before I can answer (also on mobile right now so I can’t check easily at the moment)

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