CODEWITHSUNDEEP
typescripttypestypeguards
Nikola Stojaković
Nikola Stojaković

Reputation: 2305

User defined type guards without passing a parameter

Let's say I have these classes:

export abstract class AbstractSubjectChild<T extends Subject>
{
    protected parent: T | undefined;

    public hasParent()
    {
        return this.parent != null;
    }

    public setParent(parent: T): void
    {
        this.parent = parent;
    }

    public getParent(): T | undefined
    {
        return this.parent;
    }
}

class Child extends AbstractSubjectChild<Post>
{   
}

And I want to do something like this:

const child = new Child();

if (child.hasParent()) {
    const post: Post = child.getParent();
}

Is there a way to tell TS compiler to infer the type based on hasParent() result without having to explicitly use as Post everywhere?

Upvotes: 0

Views: 47

Answers (1)

bryan60
bryan60

Reputation: 29335

export abstract class AbstractSubjectChild<T extends Subject>
{
    protected _parent?: T;

    public set parent(parent: T): void
    {
        this._parent = parent;
    }

    public get parent(): T | undefined
    {
        return this._parent;
    }
}

write your class that way and you can do:

if (child.parent) {
  const post = child.parent;
  // everywhere in this block scope is now aware of the type of post
}

Upvotes: 2

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