Broken instanceof type guarding

[nin-jin]

TypeScript Version: 2.4.0

Code

class A { foo: string }
class B { bar: string }

function test(x : A | B) {
  if (x instanceof A) {
    console.log(x.foo);
  } else {
    console.log(x.bar); // Expected typeof x == A|B
  }
}

test({ foo: "Hello, world" }); // undefined

Expected behavior:

Error at this line:

    console.log(x.bar);

Because x can implement A (and not implement B) interface but do not be instance of A class.

Actual behavior:

No errors. Wrong result.

[ikatyang]

Duplicate of #17344

[mhegazy]

Automatically closing this issue for housekeeping purposes. The issue labels indicate that it is unactionable at the moment or has already been addressed.

[nin-jin]

My issue is not about nominal typing. My issue is about wrong logic. If x is instance of class A, then x implements interface A. But if x is not instance of A, then x can implement and can implement not interface A. Be attentive!

https://en.wikipedia.org/wiki/Set_theory

[mhegazy]

@nin-jin, you are correct, except that most uses of this pattern assume that A and B are disjoint, and a check for inclusion in A implicitly checks for exclusion from B . you can see more about the rational for the current design in #1719 (comment)

[nin-jin]

In most uses else block assume that B is any. Why you broke logic for incorrect uses? You will take more and more issues about it until fix it.