ternary type inference is wrong

[Spongman]

TypeScript Version: 2.4.0 / nightly (2.5.0-dev.201xxxxx)

Code

class Base { }
class Derived1 extends Base { }
class Derived2 extends Base { }

var t1 = (Math.random() < 0.5) ? "asdasd" : 34;
// t1: string | number

var t2 = (Math.random() < 0.5) ? new Derived1() : new Derived2();
// t2: Derived1

Expected behavior:

// t2: Derived1 | Derived2

I understand the shapes are the same, but this means the runtime types don't match up with the static type.

[kitsonk]

TypeScript uses a structural type system. To TypeScript, an instance of Derived1 and Derived2 look the same, so as it infers the type, it sees that there is no structural difference between Derived1 and Derived2 and so doesn't create a union.

Here, there is a structural difference between the two types, so you get what you expect:

class Base { }
class Derived1 extends Base {
	foo = 'bar'
}
class Derived2 extends Base {
	bar = 1
}

var t1 = (Math.random() < 0.5) ? "asdasd" : 34;
// t1: string | number

var t2 = (Math.random() < 0.5) ? new Derived1() : new Derived2();
// t2: Derived1 | Derived 2

JavaScript behaves in a structural way in everything except the instanceof operator. The discussion of some sort of nominal typing support for TypeScript is covered in issue #202 (as well as loads of other information on the subject).

[Spongman]

there is no structural difference between Derived1 and Derived2

really? (new Derived1()).constructor !== (new Derived2()).constructor

[RyanCavanaugh]

Please read the FAQ https://github.com/Microsoft/TypeScript/wiki/FAQ

[kitsonk]

really? (new Derived1()).constructor !== (new Derived2()).constructor

And for the record, those are pointers to functions that ermmm have the same structure... This is, I am afraid, basic TypeScript concepts.