Get the dynamic type as the return value in a TypeScript abstract class method with generic type T

Within my codebase, I have created an abstract class:

export abstract class Filters<P> {
  protected field: string;

  protected constructor(field: string) {
    this.field = field;
  }

  protected abstract getFilter(): P;
}

Additionally, there is an interface defined as follows:

interface IRagne<T> {
  field: string;
  from: T;
  to: T;
}

To implement the abstract Filters class, I have a specific class called RangeFilter:

class RangeFilter<T, P = IRagne<T>> extends Filters<P> {
  private from: T;
  private to: T;

  constructor(field: string, from: T, to: T) {
    super(field);
    this.from = from;
    this.to = to;
  }

  getFilter(): P {
    return {
      field: super.field,
      from: this.from,
      to: this.to
    };
  }
}

However, my integrated development environment displays an error in the getFilter implementation. It suggests using getFilter(): IRange<T>, but this solution does not resolve the issue.

Answer №1

The issue here arises from attempting to calculate type P in the declaration of generic types, which is not allowed. The equal sign in a generic type declaration signifies a default value. Therefore, in the statement

class RangeFilter<T, P = IRange<T>>
, it indicates that the RangeFilter class requires 2 types to function. If the second type is not provided, it defaults to IRange<T>, but you have the flexibility to specify a different type.

As per your current declaration, using RangeFilter<xxx, string> or RangeFilter<xxx, number> will be acceptable, although the output of getFilter may be incorrect in these scenarios.

A more effective solution would be as follows:

class RangeFilter<T> extends Filters<IRange<T>> {
  private from: T;
  private to: T;

  constructor(field: string, from: T, to: T) {
    super(field);
    this.from = from;
    this.to = to;
  }

  getFilter(): IRange<T> {
    return {
      field: super.field,
      from: this.from,
      to: this.to
    };
  }
}

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