It is true that variable is of type "function", however, it does not have a call signature as expected because of the unexpected type merging

Question

It is true that variable is of type "function", however, it does not have a call signature as expected because of the unexpected type merging

Similar to React, I am interested in working with states in a custom library. The current class I have is as follows:

export abstract class Room<State> {
    protected state: State;

    protected setState<Key extends keyof State>(
        state: ((previousState: Readonly<State>) => Pick<State, Key> | State)
             | (Pick<State, Key>)
    ) {
        if (typeof state === "function") {
            // TypeScript is complaining about the call signature of 'state', even though I expected to be able to distinguish between both types of the state argument. Removing the final | State at the end of the state type makes it work, but Intellisense in VS Code no longer offers key completion in lines like this.setState({ foo: 1 });
            const newState = state(this.state);
            // ...
        }
        // ...
    }
    // ...
}

In the comment, TypeScript is merging the second type of state to State & Function. Why is this happening? Is there a safe way to rewrite this?

typescript

Answer 1

Answer №1

It seems like there are a couple of things going on here. Firstly, the extra brackets don't serve any purpose since unions are commutative, meaning A | (B | C) is the same as (A | B) | C - essentially, the brackets are redundant. Secondly, to address the issue with the function, you can simply write `typeof state !== "object"` to eliminate 2/3 of the union and focus solely on the function. I hope this explanation clarifies things for you.

export abstract class Room<State extends object> {
    protected state: State = "" as any;

    protected setState<Key extends keyof State>(
        state: ((previousState: Readonly<State>) => Pick<State, Key> | State) | (Pick<State, Key>) | State
    ) {
        if (typeof state !== "object") {
            const newState = state(this.state);
        }
    }
}

Upon further review, the reason for the failure is the lack of a generic constraint on `State`. Without this constraint, TypeScript is unable to make assumptions about `State` when reasoning about it generically. This ambiguity could result in `typeof state === "function"` returning whatever type `State` represents.

Answer 2

It seems like there are a couple of things going on here. Firstly, the extra brackets don't serve any purpose since unions are commutative, meaning A | (B | C) is the same as (A | B) | C - essentially, the brackets are redundant. Secondly, to address the issue with the function, you can simply write `typeof state !== "object"` to eliminate 2/3 of the union and focus solely on the function. I hope this explanation clarifies things for you.

export abstract class Room<State extends object> {
    protected state: State = "" as any;

    protected setState<Key extends keyof State>(
        state: ((previousState: Readonly<State>) => Pick<State, Key> | State) | (Pick<State, Key>) | State
    ) {
        if (typeof state !== "object") {
            const newState = state(this.state);
        }
    }
}

Upon further review, the reason for the failure is the lack of a generic constraint on `State`. Without this constraint, TypeScript is unable to make assumptions about `State` when reasoning about it generically. This ambiguity could result in `typeof state === "function"` returning whatever type `State` represents.

Answer 3

Answer №2

To effectively narrow down a union type to a specific function type, implementing a type guard is essential.

function isFunctionType(value: any): value is Function {
  return typeof value === "function";
}

// Instead of using
if (typeof value === "function") { ... }

// Use
if (isFunctionType(value)) { ... }

Answer 4

To effectively narrow down a union type to a specific function type, implementing a type guard is essential.

function isFunctionType(value: any): value is Function {
  return typeof value === "function";
}

// Instead of using
if (typeof value === "function") { ... }

// Use
if (isFunctionType(value)) { ... }

It is true that variable is of type "function", however, it does not have a call signature as expected because of the unexpected type merging

Answer №1

Answer №2

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