Spread operator breaks bounded polymorphism

[Mr-Wallet]

/* @flow */
type Un = A | B
type A = { x: string };
type B = { y: string };

function myFunc<T: Un>(foo: T): T {
  const bar = { ...foo };
  return bar;
}

I would expect flow to say, "Well, T is a plain object, and you spread T into an object and returned it, so it must be a T. All good!"

Instead I get object literal. Could not decide which case to select.

[erikdesjardins]

To be pedantic, T is an arbitrary subtype of Un, which includes classes that have an x: string or y: string property (try flow), so the provided code shouldn't typecheck.

However, there is still a problem here, because it also happens without generics (try flow).

[Mr-Wallet]

This is indeed just a special case of "the spread operator doesn't work so well right now", but I felt it bore mention just to make sure it didn't get missed.

[villesau]

This will fix the issue: #7298

[nmote]

This no longer errors in master