Is there a way to retrieve a specific type from a union-based type by using a generic function in Typescript?

When working with my API, I have noticed a consistent pattern where it returns either a specific type or a TypeScript type called BadResult, as shown below:

type Result1 = CreatedPersonResult | BadResult;
type Result2 = CreatedRoleResult | BadResult;

To streamline my code, I am looking to create a generic wrapper function that can accept either Result1 or Result2 as input and return only the successful result, such as CreatedPersonResult.

I believe leveraging the Exclude option in TypeScript might be the solution.

In cases where the result is a BadResult, throwing an error would be acceptable.

const resolveActualType = (result: T): Exclude<T, BadResult> => {    
    // if(... is BadResult) {
    //   throw Error();
    // }
    
    return result;
}
typescriptgenerics

Answer №1

To effectively constrain the generic parameter type in TypeScript and ensure that the result does not match a specific type after error handling, we can utilize union types directly in the function parameter for inference.

Instead of considering the entire union as the generic parameter type T, we can let TypeScript infer it as the part without the specified type (BadResult):

// T is inferred as the type excluding BadResult directly
const resolveActualType = <T,>(result: T | BadResult): T => {
    result
    //^? BadResult | T

    if (isBadResult(result)) {
        result
        //^? BadResult
        throw Error();
    }

    return result;
    //     ^? T
}

declare const r1b: Result1
const r1 = resolveActualType(r1b)
//    ^? CreatedPersonResult

declare const r2b: Result2
const r2 = resolveActualType(r2b)
//    ^? CreatedRoleResult
 | BadResult;

// The actual implementation of the type guard predicate depends on the specific BadResult type
declare function isBadResult(result: unknown): result is BadResult

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