Can one specify a conditional return type in TypeScript without resorting to typecasting within the function's implementation?

Question

Can one specify a conditional return type in TypeScript without resorting to typecasting within the function's implementation?

I'm currently working on a function where the return type is determined by the input values, but I'm finding that typescript is requiring me to use the as keyword to address the issue I'm facing. Is there a more graceful way to handle this and avoid using as?

type TestReturnType<T> = T extends string ? number : string;

const test = <T,>(value: T): TestReturnType<T> => {
    if (typeof value === 'string') {
        return 1 as TestReturnType<T>; // ?
    } else {
        return '1' as TestReturnType<T>; // ?
    } 
};

const res1 = test('String'); 
const res2 = test(13123);

typescript

Answer 1

Answer №1

One way to achieve this is by utilizing function overloading:

type TestReturnType<T> = T extends string ? number : string;

function test(value: string): TestReturnType<string>;
function test(value: number): TestReturnType<number>;
function test<T>(value: T): TestReturnType<string | number> {
    if (typeof value === 'string') {
        return 1;
    } else {
        return '1';
    }
}

const res1 = test('String');   // res1: number
const res2 = test(13123);      // res2: string

Keep in mind that,

If the overloads are not included, both res1 and res2 will be of type string | number
You have the option to remove the type parameter of "test" and define it as function test(value: any)

Answer 2

One way to achieve this is by utilizing function overloading:

type TestReturnType<T> = T extends string ? number : string;

function test(value: string): TestReturnType<string>;
function test(value: number): TestReturnType<number>;
function test<T>(value: T): TestReturnType<string | number> {
    if (typeof value === 'string') {
        return 1;
    } else {
        return '1';
    }
}

const res1 = test('String');   // res1: number
const res2 = test(13123);      // res2: string

Keep in mind that,

If the overloads are not included, both res1 and res2 will be of type string | number
You have the option to remove the type parameter of "test" and define it as function test(value: any)

Answer 3

Answer №2

Unfortunately, at the current time, if a function returns a conditional type that relies on a generic type parameter, you must use a type assertion (referred to as "typecasting") in the return statement, or opt for a less strictly typed approach (such as overloading the function). This means there is no way to achieve compiler-verified type safety in this scenario. The feature request related to this issue can be found here: microsoft/TypeScript#33912.

The challenge lies in the fact that while control flow analysis can narrow down the types of actual values belonging to generic types, it does not impact the generic type parameter. For example, within your test() function, typeof value === "string" can narrow down value from T to T & string; however, T itself remains unaffected and is not constrained to be of type string. Therefore, the type

T extends string ? number : string

cannot be solely evaluated as number; it will still be

T extends string ? number : string

.

Answer 4

Unfortunately, at the current time, if a function returns a conditional type that relies on a generic type parameter, you must use a type assertion (referred to as "typecasting") in the return statement, or opt for a less strictly typed approach (such as overloading the function). This means there is no way to achieve compiler-verified type safety in this scenario. The feature request related to this issue can be found here: microsoft/TypeScript#33912.

The challenge lies in the fact that while control flow analysis can narrow down the types of actual values belonging to generic types, it does not impact the generic type parameter. For example, within your test() function, typeof value === "string" can narrow down value from T to T & string; however, T itself remains unaffected and is not constrained to be of type string. Therefore, the type

T extends string ? number : string

cannot be solely evaluated as number; it will still be

T extends string ? number : string

.

Can one specify a conditional return type in TypeScript without resorting to typecasting within the function's implementation?

Answer №1

Answer №2

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