Why is a union-string generic type considered as a distinct literal value when passed to a conditional type?

Typescript version: 5.6.2

Can someone explain why the ApprovalType is being modified into specific values of ApprovalType even when the function parameter type is a conditional type? This happens even if I explicitly pass in the generic parameter <ApprovalType> when calling the function.

Playground

type ApprovalType = "PENDING" | "APPROVED" | "REJECTED";

type A<T> = {
    options: T[];
};

type B<T> = {
    options: T[];
};

function conditionalTypeFn<T>(props: T extends string ? B<T> : A<T>) {
    return props;
}
conditionalTypeFn<ApprovalType>({
    /**
     * ❌ Type '("PENDING" | "APPROVED" | "REJECTED")[]' is not assignable
     * to type '"PENDING"[] | "APPROVED"[] | "REJECTED"[]'
     */
    options: ["PENDING", "APPROVED", "REJECTED"],
});

function unionTypeFn<T>(props: A<T> | B<T>) {
    return props;
}
unionTypeFn<ApprovalType>({
    /* ✅ no error */
    options: ["PENDING", "APPROVED", "REJECTED"],
});

Answer №1

As per the default behavior described in the official documentation, conditional types exhibit a distinctive distributive approach when applied to a union type.

Here's a simplified example to showcase this concept:

type ApprovalType = "PENDING" | "APPROVED" | "REJECTED";

type Distributed<T> = T extends string ? T[] : never;

type ApprovalTypes = Distributed<ApprovalType>;
// type ApprovalTypes = "PENDING"[] | "APPROVED"[] | "REJECTED"[]

To prevent the distributive nature, you can utilize square brackets within your conditional type. Using the same example, it would result in:

type ApprovalType = "PENDING" | "APPROVED" | "REJECTED";

type NonDistributed<T> = [T] extends [string] ? T[] : never;

type ApprovalTypes = NonDistributed<ApprovalType>;
// type ApprovalTypes = ("PENDING" | "APPROVED" | "REJECTED")[]

Alternatively, when implementing the code in your query:

function conditionalTypeFn<T>(props: [T] extends [string] ? B<T> : A<T>) {
    return props;
}

Playground link

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