TypeScript: force type error when anonymous object property doesn't meet type specification

Question

I've got an anonymous object with a TypeScript typed property. Something like this:

type Foo = {
  exists: true;
  bar: string;
};

doSomething({
  foo: {
    exists: true,
    bar: 'baz',
  } as Foo,
});

I'd like to type protect the foo property so if type Foo ever changes in the future (e.g. gets a new property), then the TypeScript compiler will warn me. While these cause compiler errors:

doSomething({
  foo: {
    exists: 'string', // Incorrect type
    bar: 'baz',
  } as Foo
});

doSomething({
  foo: {
    say: 'hello', // Not enough overlap with Foo
    bar: 'baz',
  } as Foo
});

This doesn't cause a type error, even though exists is missing:

doSomething({
  foo: {
    bar: 'baz',
  } as Foo
});

Likewise, this also doesn't cause a type error:

doSomething({
  foo: <Foo>{
    bar: 'baz',
  },
});

The only way I can force a type error when a property of Foo is missing is this:

const foo: {foo: Foo} = {
  foo: {
    bar: 'baz',
  },
};

doSomething(foo);

...but that requires me to unnecessarily create a local variable simply for the purpose of type safety. Is there a way to type protect the property of an anonymous object?

Note: doSomething()'s parameter accepts any and doesn't enforce any type requirements, otherwise I could rely on doSomething enforcing the type.

Tobias S. · Accepted Answer · 2022-10-11 17:53:36Z

2

TypeScript 4.9 will introduce the satisfies operator.

doSomething({
  foo: {
    bar: 'baz',
  } satisfies Foo,
});

Until then, you may use another function to validate foo at compile time.

const validFoo = (foo: Foo) => foo

doSomething({
  foo: validFoo({
    bar: 'baz',
  }),
});

Playground

edited Oct 11, 2022 at 17:53

answered Oct 11, 2022 at 17:43

Tobias S.

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