Is there a TypeScript equivalent to C#'s generic type constraint that allows for extending a class?

Question

Is there a TypeScript equivalent to C#'s generic type constraint that allows for extending a class?

I'm working on creating a protected abstract class that allows for the subclass type to be passed as a type argument in the constructor of the superclass.

What I need is something similar to C#'s Generic Type Constraint (using the where keyword) so that I can specify a child type in the parameter list.

//                    where T : <base class name>
BaseAuthController<T> where T : BaseAuthController

Existing superclass

export abstract class BaseAuthController {
    protected constructor(
        protected dialogRef:
            //This class should not have knowledge of child classes
            MatDialogRef<Child1DialogComponent> |
            MatDialogRef<Child2DialogComponent> |
            MatDialogRef<Child3DialogComponent>
    ) {

    }
}

Existing subclass

export class Child1DialogComponent extends BaseAuthController {
    constructor(dialogRef: MatDialogRef<Child1DialogComponent>) {
        super(dialogRef);
    }
}

Preferred superclass structure

export abstract class BaseAuthController<T> {
    protected constructor(protected dialogRef: MatDialogRef<T>) {

    }
}

References

c#angular typescript generics typescript-generics

Answer 1

Answer №1

Consider utilizing self-bounded generics for your scenario:

export abstract class AuthBaseController<T extends AuthBaseController<T>> {
  protected constructor(protected dialogReference: MatDialogRef<T>) {}
}

In TypeScript, achieving this behavior is commonly done using polymorphic this types. However, referencing the this type within the constructor is currently not possible due to an existing issue. The workaround is similar to how Java handles it, ensuring that your subclass functions as intended:

export class CustomDialogComponent extends AuthBaseController<CustomDialogComponent> {
  constructor(dialogReference: MatDialogRef<CustomDialogComponent>) {
    super(dialogReference);
  }
}

I hope this information proves helpful to you in your endeavors!

Answer 2

Consider utilizing self-bounded generics for your scenario:

export abstract class AuthBaseController<T extends AuthBaseController<T>> {
  protected constructor(protected dialogReference: MatDialogRef<T>) {}
}

In TypeScript, achieving this behavior is commonly done using polymorphic this types. However, referencing the this type within the constructor is currently not possible due to an existing issue. The workaround is similar to how Java handles it, ensuring that your subclass functions as intended:

export class CustomDialogComponent extends AuthBaseController<CustomDialogComponent> {
  constructor(dialogReference: MatDialogRef<CustomDialogComponent>) {
    super(dialogReference);
  }
}

I hope this information proves helpful to you in your endeavors!

Answer 3

Answer №2

Although it operates differently at its core, there is a method to achieve the equivalent outcome:

export abstract class BaseAuthController<T extends ChildClass> {
    protected constructor(protected dialogReference: MatDialogRef<T>) {

    }
}

By incorporating unified types, we can specify multiple subclasses:

export abstract class BaseAuthController<T extends SubClassA | SubClassB> {
    protected constructor(protected dialogReference: MatDialogRef<T>) {

    }
}

Answer 4

Although it operates differently at its core, there is a method to achieve the equivalent outcome:

export abstract class BaseAuthController<T extends ChildClass> {
    protected constructor(protected dialogReference: MatDialogRef<T>) {

    }
}

By incorporating unified types, we can specify multiple subclasses:

export abstract class BaseAuthController<T extends SubClassA | SubClassB> {
    protected constructor(protected dialogReference: MatDialogRef<T>) {

    }
}

Is there a TypeScript equivalent to C#'s generic type constraint that allows for extending a class?

Answer №1

Answer №2

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