Covariant assignability of type guard functions is unsound

TypeScript Version: master (af8e44a)

Search Terms: user-defined type guard predicate assignable assignability unsound covariant invariant

Code

class A {
    a: string;
}
class B extends A { 
    b: string;
}
function isB(x: {}): x is B { 
    return x instanceof B;
}
const isA: (x: {}) => x is A = isB;  // allowed, should be compile error

const x = <A | string>new A();
if (!isA(x)) {
    console.log(x.slice());  // runtime error
}

Expected behavior: Assignability error where marked.

Actual behavior: Successful compilation, runtime error.

Playground Link: link

Related Issues: #24865

If we want a kind of type guard function that is covariant, we need to not narrow when it returns false, i.e., a "true" result would be sufficient but not necessary for the value to be of the tested type. And we'd need a different syntax for the type of a type guard function that is necessary and sufficient compared to a type guard function that is sufficient but not necessary. One idea for the latter is (x: {}) => x is A | false. (A type guard function that is necessary but not sufficient would then be (x: {}) => x is A | true.) Getting all these variants supported would help us support conjunctions and disjunctions of type guard calls with other boolean expressions in #24865. If you like, this issue can be for the removal of the unsound assignability rule and I can file a separate suggestion for the new kinds of type guard functions.

Beginne mit der verlinkten Playground-Reproduktion und der markierten Zuweisung von isB zum Typ isA. Verfolge, wie der Compiler die Zuweisungsverträglichkeit von Typprädikatsfunktionen prüft, und verwende den verwandten Issue als Kontext. Als abgeschlossen gilt die Aufgabe, wenn die Zuweisung einen Fehler erzeugt und die demonstrierte unsichere Eingrenzung verhindert wird.