Exploring Typescript's conditional types and narrowing branches

Question

Exploring Typescript's conditional types and narrowing branches

When I use the following code snippet:

type Identity <T extends string> = T;

type MaybeString = string | undefined;

type StringOrNever = MaybeString extends undefined ? never : Identity<MaybeString>;

The compiler raises an error stating that 'undefined' is not assignable to type 'string' when using MaybeString as the Identity parameter.

I started wondering if this issue could be related to the distributive property of conditional types, so I decided to try the following:

type StringOrNever = [MaybeString] extends [undefined] ? [never] : Identity<MaybeString>;

Unfortunately, this approach did not resolve the error either.

I'm confused about what I might be doing wrong here. Can anyone provide some insights?

typescript conditional-types

Answer 1

Answer №1

This is not about the concept of distribution.

Distributive types work with a generic type. The example of MaybeString in StringOrNever is not considered generic.

The statement MaybeString extends undefined does not change it to just string. Even after this condition, MaybeString remains as string | undefined.

To address this issue, you need to add parameters to StringOrNever.

Take a look at this sample:

type Identity<T extends string> = T;

type MaybeString = string | undefined;

type StringOrNever<T extends string | undefined> = T extends string ? Identity<T> : never

type Result1 = StringOrNever<'Hello'> // hello
type Result2 = StringOrNever<undefined> // never
type Result3 = StringOrNever<undefined | 'some string'> // 'some string'

Play around with code

An interesting scenario arises with Result3. In reality, Result3 becomes a union of string and never: "some string" | never. This leads to a clash between never and string. How would you like to handle this union? One approach is to disable distributivity, similar to what you have implemented

type Identity<T extends string> = T;

type MaybeString = string | undefined;

type StringOrNever<T extends string | undefined> = [T] extends [string] ? Identity<T> : never

type Result1 = StringOrNever<'Hello'> // hello
type Result2 = StringOrNever<undefined> // never
type Result3 = StringOrNever<undefined | 'some string'> // never

Answer 2

This is not about the concept of distribution.

Distributive types work with a generic type. The example of MaybeString in StringOrNever is not considered generic.

The statement MaybeString extends undefined does not change it to just string. Even after this condition, MaybeString remains as string | undefined.

To address this issue, you need to add parameters to StringOrNever.

Take a look at this sample:

type Identity<T extends string> = T;

type MaybeString = string | undefined;

type StringOrNever<T extends string | undefined> = T extends string ? Identity<T> : never

type Result1 = StringOrNever<'Hello'> // hello
type Result2 = StringOrNever<undefined> // never
type Result3 = StringOrNever<undefined | 'some string'> // 'some string'

Play around with code

An interesting scenario arises with Result3. In reality, Result3 becomes a union of string and never: "some string" | never. This leads to a clash between never and string. How would you like to handle this union? One approach is to disable distributivity, similar to what you have implemented

type Identity<T extends string> = T;

type MaybeString = string | undefined;

type StringOrNever<T extends string | undefined> = [T] extends [string] ? Identity<T> : never

type Result1 = StringOrNever<'Hello'> // hello
type Result2 = StringOrNever<undefined> // never
type Result3 = StringOrNever<undefined | 'some string'> // never

Exploring Typescript's conditional types and narrowing branches

Answer №1

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