Issue arises while extracting a conditional type from a generic function

I've come across a TypeScript error while trying to implement a conditional type in a MessagesStore object. Here is the code snippet:

type Message = { id: number; text: string };

type MessagesStore = {
    queue: Array<Message>;
    read: <T extends boolean>(assertiveMode?: T) => T extends true ? Message : Message | undefined;
}

const store: MessagesStore = {
    queue: [],

    read(assertiveMode: boolean = false) {
        const message = this.queue.shift();

        if (assertiveMode && message === undefined) {
            throw new Error('The messages queue is empty');
        }

        return message;
    }
}

An error is being thrown by TypeScript with the following message:

Type '<T extends boolean>(assertiveMode?: boolean) => Message | undefined' is not assignable to type '<T extends boolean>(assertiveMode?: T | undefined) => T extends true ? Message : Message | undefined'.
  Type 'Message | undefined' is not assignable to type 'T extends true ? Message : Message | undefined'.
    Type 'undefined' is not assignable to type 'T extends true ? Message : Message | undefined'.

I'm having trouble figuring out why this error is occurring and how to fix it.

Answer №1

At this moment in time, with TypeScript 5.3, there isn't a compiler-verified method to safely return a conditional type that relies on a generic type parameter. The community has been requesting support for this through microsoft/TypeScript#33912, but it's not part of the language yet.

The issue lies in the fact that when you check assertiveMode, it can narrow the type from T to either true or false, but it doesn't impact the generic type parameter T itself. As a result, the compiler struggles to determine if message is a valid value of type

T extends true ? Message : Message | undefined
.

There are hopes that this might be addressed soon; a pull request exists at microsoft/TypeScript#56941 that could potentially be merged, and addressing these types of functions is mentioned in the TypeScript 5.5 roadmap at microsoft/TypeScript#57475. Your code might just start working flawlessly without any errors.

Until then, you may have to relax the type checking to compile your code. The simplest approach is by utilizing the any type as a way of indicating that the return type shouldn't be checked at all:

const store: MessagesStore = {
  queue: [],

  read(assertiveMode: boolean = false) {
    const message = this.queue.shift();

    if (assertiveMode && message === undefined) {
      throw new Error('The messages queue is empty');
    }

    return message as any; // 🤷‍♀️ ms/TS#33912
  }
}

Unfortunately, at present, generic conditional return types lean more towards being useful for function callers rather than function implementers.

Link to Playground for testing

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