Extending a generic parameter in Typescript

Question

I want to create a utility function which creates a checklist by adding an isChecked knockout observable property to each item in an array. This function should look like this:

createCheckList<T>(allEntities: T[], selected: T[]) : CheckListItem<T> {
    ...
}

I am returning a CheckListItem<T> because this interface should extend T to add the isChecked property. However, typescript will not allow me to do this:

interface CheckListItem<T> extends T {
    isChecked: KnockoutObservable<boolean>;
}

Gives me the error:

An interface may only extend another class or interface.

Is there any way to do this?

ominug · Accepted Answer · 2017-12-16 21:16:49Z

7

Since TypeScript 1.6 you can use intersection types:

type CheckListItem<T> = T & {
  isChecked: KnockoutObservable<boolean>;
};

answered Dec 16, 2017 at 21:16

ominug

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Ryan Cavanaugh · Accepted Answer · 2013-09-24 20:47:17Z

3

There isn't a way to express this in TypeScript.

You can obviously do this instead:

interface CheckListItem<T> {
    isChecked: KnockoutObservable<boolean>;
    item: T;
}

This has the advantage of not breaking the object when it happens to have its own isChecked property.

answered Sep 24, 2013 at 20:47

Ryan Cavanaugh

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1 Comment

Grokys Over a year ago

Thanks. Yes, I had considered that but it's rather clumsy IMHO. I've gone with returning an 'any[]' instead as I'm only using the results for knockout binding where type-safety doesn't buy me anything anyway.

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