CODEWITHSUNDEEP
typescriptdiscriminated-union
James B
James B

Reputation: 9595

Typescript Tagged Union Types

I want to discriminate some logic according to what interface my function receives. To do this I'm attempting to use tagged union types, e.g.,

someFunction(arg: TypeA | TypeB): void {
    if (arg.kind === "TypeA")
    {
        // do this
    }
    else
    {
        // do that
    }
}

where

interface TypeA {
    kind: "TypeA";
    propertyA: string;
}

interface TypeB {
    kind: "TypeB";
    propertyB: string;
}

But if I want to call this function, Typescript complains if I don't provide a value for kind, i.e.,

let typeA: TypeA;
typeA = {propertyA: ""}
someFunction(typeA);

with

TS2322: Type '{ propertyA: string; }' is not assignable to type 'TypeA'.
  Property 'kind' is missing in type '{ propertyA: string; }'.

So I don't understand how tagged types work if I have to implement the tag (kind in the example above) every time I want to discriminate. I can only assume I'm using them wrong?

Upvotes: 4

Views: 2719

Answers (1)

toskv
toskv

Reputation: 31600

You can define a type guard to do this. They allow you to tell the type of an argument by the value or existence of one of it's properties.

function isTypeA(arg: TypeA | TypeB): arg is TypeA {
    return (<TypeA>arg).propertyA !== undefined;
}

function someFunction(arg: TypeA | TypeB): void {
    if (isTypeA(arg))
    {
      arg.propertyA;
    }
    else
    {
        arg.propertyB
    }
}

You can read more about them here and check a working example here.

Upvotes: 5

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