typescript - Is there an option to make spreading an object strict?

Question

I was wondering if there is a compiler option or something similar to make spreading objects strict.

Please see following example to understand what I mean:

interface Foo {
  a: string;
}

interface Bar {
  a: string;
  b: number;
}

const barObject: Bar = { a: "a string", b: 1 };

// should give a warning because Bar has more properties (here b) than Foo
const fooObject: Foo = { ...barObject };

// actually prints 1
console.log((fooObject as any).b); 

Is something like this possible?

Answer 1

Interesting question. According to this issue, the result of the spread operator is intended to not trigger excess property checks.

// barObject has more props than Foo, but spread does not trigger excess property checks
const fooObject: Foo = { ...barObject };

// b is explicit property, so checks kicks in here
const foo: Foo = { ...bar, b: 2 }

There aren't exact types for TypeScript currently, but you can create a simple type check to enforce strict object spread:

// returns T, if T and U match, else never
type ExactType<T, U> = T extends U ? U extends T ? T : never : never

const bar: Bar = { a: "a string", b: 1 };
const foo: Foo = { a: "foofoo" }

const fooMergeError: Foo = { ...bar as ExactType<typeof bar, Foo> }; // error
const fooMergeOK: Foo = { ...foo as ExactType<typeof foo, Foo> }; // OK

With a helper function, we can reduce redundancy a bit:

const enforceExactType = <E>() => <T>(t: ExactType<T, E>) => t

const fooMergeError2: Foo = { ...enforceExactType<Foo>()(bar) }; // error

Code sample