3 ms·

Can someone give me an example of TypeScript declaration that is not "sound"? I thought that "It guarantees that (1) an expression that has a static type T ca

• lmm 12y ago
  Generics are bivariant in TypeScript and so unsound: class Base {} class Derived1 extends Base {} class Derived2 extends Base {} class Box<T> { constructor(public value: T) {} }; var a: Derived1 = new Derived1(); var b: Box<Base> = new Box<Base>(a); var c: Box<Derived2> = b; var d: Derived2 = c.value;
  • barrkel 12y ago
    TypeScript is structurally typed; there's nothing in your example that suggests that Derived1 and Derived2 have different types, instances of each are the same type.
    • lmm 12y ago
      Point, but add a differently-named field to each and it will still compile.
• xxgreg 12y ago
  My understanding is that gradually typed code, i.e. using "any", introduces unsoundness. var s : String = "bar"; var a : any = s; var i : int = a; Not sure, but I think Typescript's assignability rules, which use structural typing, also introduce unsoundness. class A { foo() { print('A'); } } class B { foo() { print('B'); } } var a : A = new A(); var b : B = a;