Can you declare a object literal type that allows unknown properties in typescript?

Question

Essentially I want to ensure that an object argument contains all of the required properties, but can contain any other properties it wants. For example:

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Answer 1

Yes, you can. Try this:

interface IBaz {
    baz: number;
    [key: string]: any;
}

function foo(bar: IBaz) : number {
    return bar.baz;
}

foo({ baz: 1, other: 2 });