Retrieving a nested type within the generic argument of a generic class in TypeScript

I'm currently working on a TypeScript project where I have a generic class called AbstractDAL. This class has a generic type parameter named T_DEFAULT_RETURN. Within the AbstractDAL class, my goal is to extract a nested type that is specified within the T_DEFAULT_RETURN generic type parameter. However, I am encountering some difficulties in achieving this.

Below is a simplified version of the code structure:

class A {
    alpha() { }
};

class B extends A {
    beta() { }
};

abstract class AbstractDAL<
    T_DEFAULT_RETURN extends BaseEntity = BaseEntity,       
    T_DATA = T_DEFAULT_RETURN extends BaseEntity<infer D> ? D : never
> {
    get result() {
        return {} as T_DATA
    }
}

class BaseEntity<       
    T_DATA extends A = A
> { }

class TestDAL extends AbstractDAL<TestEntity> {
    delta() {
        this.result.alpha // should also be beta, not just alpha
    }
}

class TestEntity extends BaseEntity<B> { }

In the above code snippet, the AbstractDAL class uses the T_DEFAULT_RETURN generic type parameter. The objective is to extract a nested type from this parameter. To accomplish this, I've utilized a conditional type with infer and created a helper type called T_DATA. Despite these efforts, the inferred type for T_DATA turns out to be A instead of the expected type B.

Could you provide suggestions or modifications needed in the code to accurately extract the nested type B from the T_DEFAULT_RETURN generic type parameter within the AbstractDAL class?

Answer №1

The issue lies in the fact that BaseEntity<T> does not have a structural dependency on T, leading to unreliable inference from BaseEntity<T> to T.


TypeScript employs a primarily structural type system rather than a nominal one. This means that types are compared based on their structure rather than their name or declaration location. If two object types, like X and Y, have the same members, they are considered the same type:

interface X { w: string; z: number; }
let x: X = { w: "abc", z: 123 };

interface Y { w: string; z: number; }
let y: Y = { w: "def", z: 456 };

x = y; // valid
y = x; // valid

In this scenario, using either X or Y as reference to the type makes no difference - they are interchangeable.

This principle extends to generics too. If a generic type does not utilize its type parameter within its definition, such as:

interface F<T> { w: string; z: number }

then any types created by specifying T with a type argument will be indistinguishable and compatible:

let fs: F<string> = x; // valid
let fn: F<number> = y; // valid
fs = fn; // valid
fn = fs; // valid

Thus, there is no distinction between F<string> and F<number>. The key aspect is the shape { w: string; z: number }, not the specific name F<string>.

This concept exposes an issue in coding practices regarding type inference, especially when inferring from structure rather than names.

To address this issue, ensure that your generic types have a structural dependence on their type parameters. For instance, linking properties to T in BaseEntity<T> can resolve these inconsistencies.

\Playground link for code

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