The parameter 'Barable' cannot be assigned to the parameter 'T' as they are not compatible

Could anyone help me understand why this code isn't functioning as intended:

class Fooable {
    foo: string;
}
class Barable extends Fooable { 
    bar: boolean;
}

function simplifiedExample<T extends Fooable>(): Array<T> {
    let list = new Array<T>();
        list.push(new Barable());
    return list;
}

Barable is extending Fooable. Why can't I add Barable to an Array when T must be Fooable? Check out the code in this playground

EDIT:

The issue stems from simplifiedExample() being an override from a base class, where the base class is solely a definition due to the project being a mix of JavaScript and TypeScript.

Refer to: new playground

I managed to find a workaround using casting, but it feels like more of a temporary fix than a proper solution:

class BarableService extends FooableService { 

    simplifiedExample<T extends Fooable>(): Array<T> { 
        let list = new Array<Fooable>();
            list.push(new Barable());
        return list as Array<T>;
    }
}
typescriptgenericsinheritance

Answer №1

Even though T is compatible with Fooable and Barable is also compatible with Fooable, it does not necessarily mean that Barable is compatible with T.

For instance:

class Bazable extends Fooable {
    baz: number;
}

const result: Bazable[] = simplifiedExample<Bazable>();

This situation can be likened to a logical fallacy, similar to a non sequitur:

If T (Bazable) is a Fooable
And Barable is a Fooable
Then T (Bazable) is a Barable

Alternatively...

If T (Lion) is a Cat
And Tiger is a cat
Then T (Lion) is a Tiger

     Cat
     / \
    /   \
Lion != Tiger

Generic vs Structural

In TypeScript, there are scenarios where problems can be resolved without using generics, thanks to the structural type system. It may suffice to accurately define your Barable:

function simplifiedExample(): Barable[] {
    let list: Barable[] = [];
    list.push(new Barable());
    return list;
}

All you need is a structurally compatible setup for this to pass type checking:

class Bazable {
    foo: string;
    bar: boolean;
}

const a: Bazable[] = simplifiedExample();

Alternatively:

class Bazable {
    bar: boolean;
}

const a: Bazable[] = simplifiedExample();

Or even:

const a: Fooable[] = simplifiedExample();

If this approach doesn't meet your needs, consider providing an example of when you believe a generic type introduction is necessary. Typically, this occurs when you want to create a new instance of T within the function rather than a predefined type.

Answer №2

The specific type for your array list is identified as Fooable, thus resulting in the return type being determined accordingly.

Answer №3

When considering the compatibility between T and Barable, it is important to note that this cannot be verified within the function itself. If there is a need to instantiate a new instance of T, it can be done by passing it into the constructor:

function example<T extends Fooable>(constructor: new ()=> T): Array<T> {
    let list = new Array<T>();
        list.push(new constructor());
    return list;
}

const output1: Bazable[] = example(Bazable);
const output2: Barable[] = example(Barable);

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