typescript - Is it possible to define an interface/type to which excess property checks are always applied?

Question

In TypeScript, excess property checks are applied to object literals, but not beyond that. (This is intentional documented behavior.) So given:

interface Example {
    id: number;
    name: string;
}
// ...
function doSomething(ex: Example) {
    // ...
}

This fails (playground link):

doSomething({
    id: 42,
    name: "Life, the Universe, and Everything",
    extra: "not allowed" // Error 2345: Object literal may only specify known properties, and 'extra' does not exist in type 'Example'.
});

but this doesn't (playground link):

const ex = {
    id: 42,
    name: "Life, the Universe, and Everything",
    extra: "not allowed"
};
doSomething(ex);

Is it possible to define the interface in some way that excess properties are forbidden even when not using an object literal? E.g., a "final" interface of sorts?

If not, it's easily done at runtime, but if there's a way to do it in the type definition, that would be useful.

Answer 1

What you are looking for is exact types. This feature has been in discussion since forever on GH but it has not come any closer to actually being implemented. The concern is (at least from what I heard at tsConf from Anders) is that exact types would create a bifurcated type system, where you could not assign from exact to inexact types and backwards, and this would make the dev experience much worse.

There are work arounds, but all are flawed in some way, you can read the GH issue to find all sorts of versions that might work well enough depending on what you are trying to do.