Typescript: Understanding the Optional Operator and Dealing with Null Ambiguity

Seeking guidance on writing clean code, particularly regarding a property with a backing field:

get value(): string { return this._value; }
set value(v: string) { this._value = v; }
private _value?: string;

It has come to my attention that this code is considered invalid because _value is undefined by default. Must I explicitly specify the type as string | undefined each time? This seems cumbersome. Additionally, if I need to allow null values, it would require further modification:

get value(): string | undefined | null { return this._value; }
set value(v: string | null) { this._value = v; }
private _value?: string | null;

Is there a more elegant solution to handle this scenario? Or is this the standard approach?

Answer №1

One option is to use the typeofthis._value approach to automatically deduce the type without explicitly defining it each time.

class T {
  get value(): typeof this._value {
    return this._value;
  }
  set value(v: typeof this._value) {
    this._value = v;
  }
  private _value?: string | null;
}

Check out TypeScript Playground

Explicit types may not be essential in this scenario, as TypeScript can derive them on its own. The following code snippet maintains type safety without specifying types for every instance.

class T {
  get value() {
    return this._value;
  }
  set value(v) {
    this._value = v;
  }
  private _value?: string | null;
}

Answer №2

If a situation like this arises, I would lean towards using a type declaration.

type TValue = string | undefined | null;
get value(): TValue { return this._value; }
set value(v: TValue) { this._value = v; }
private _value: TValue;

If all is well, I might even forego specifying the return type.

get value() { return this._value; }
set value(v) { this._value = v; }
private _value: string | undefined | null;

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