What are Implicitly Unwrapped Optionals in TypeScript?

When working in Swift, I usually handle class properties of another class as implicitly unwrapped optionals like this:

class MyClass {
  var prop: OtherClass!

  init(prop: OtherClass) {
    self.prop = prop
  }

  explodeProp() {
    // no need to unwrap it because it's implicitly unwrapped and always assigned in the initializer
    prop.explode() 
  }
}

However, when transitioning to TypeScript, it seems like I have to approach it differently:

export class MarkdownNode {
  tag = "";
  text = "";
  document: FoldingDocument | null = null; // most likely never really null
  parent: MarkdownNode | null = null; // might actually be null

  constructor(tag: string, text: string, document: FoldingDocument) {
    this.tag = tag;
    this.text = text;
    this.document = document;
  }

  addSibling(node: MarkdownNode) {
    if (!this.parent) {
      this.document!.nodes.push(node) // perhaps force-unwrapping is inevitable here?
    } else {
      this.parent.children.push(node);
    }
  }
}

I wonder if there exists a concept similar to implicitly unwrapped optionals in TypeScript or if it's something that might be introduced in the future?

Answer №1

Absolutely, Typescript indeed includes a feature like this, known as the Definite Assignment Assertion, and its syntax involves using an exclamation mark:

let x! string; // declaring a variable
class Foo {
  bar!: number; // declaring a property
}

It's worth noting that in TypeScript, definitely assigned variables/properties have type T rather than Optional<T> like in Swift. When left unassigned, they default to undefined. This means they offer a way to bypass type checking rules without affecting runtime behavior. You won't encounter errors like "got nil while implicitly unwrapping optional value," but instead just receive undefined. The error only occurs when attempting to use it, such as calling a method:

class Foo {
  var bar: String!
}
let x = Foo()
let y: String = x.bar // boom! Fatal error: Unexpectedly found nil while implicitly unwrapping an Optional value

In comparison, for Typescript:

class Foo {
  bar!: string;
}
let x = new Foo()
let y: string = x.bar // no issue, we simply assign 'undefined' to the string variable
someObject.name = y;
// Years later, somewhere far away
let nameLength = someObject.name.length; // boom! Uncaught TypeError: Cannot read property 'length' of undefined

Therefore, exercise caution with definite assignment as it can be a significant footgun.

Answer №2

In TypeScript (probably the same in Swift), it is possible to define a value within a constructor. Therefore, if you do not expect the value to be null, there is no need to include it in the type definition.

Check out this playground

class FoldingDocument {
    nodes: MarkdownNode[] = [];
}

class MarkdownNode {
  tag: string;
  text: string;
  document: FoldingDocument;
  parent: MarkdownNode | null = null; // could indeed be null
  children: MarkdownNode[] = [];

  constructor(tag: string, text: string, document: FoldingDocument) {
    this.tag = tag;
    this.text = text;
    this.document = document;
  }

  addSibling(node: MarkdownNode) {
    if (!this.parent) {
      this.document.nodes.push(node) // no choice but to force-unwrap?
    } else {
      this.parent.children.push(node)
    }
  }
}

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