JavaScript this, Prototypes and Classes, what does console.log(Named.label()) print after const Named = class MyClass?

Answer, C. MyClass

A. Named

Assumes the inner name is just an alias for whatever variable the class expression was assigned to on the outside, so it would reflect the outer binding name instead. The name given in a class expression is a separate binding visible only inside the class body, it always refers to the class itself, not to the outer variable it happens to be stored in.

B. ReferenceError

Assumes referencing the internal name from a static method fails because that name is only meant for recursive references and not for general use. The internal binding created by a named class expression works like any normal reference to the class, it can be used for recursion or for anything else, including simply reading its name property.

C. MyClass

Correct. A named class expression makes its own name available inside its body only, so the static method reads MyClass, which is the class own name regardless of the outer variable it was assigned to.

D. undefined

Assumes the internal class name is not actually usable from inside the class own static methods, resulting in a missing value. A named class expression name is fully valid and readable from any method defined in that class body, static or not, exactly like an ordinary function reference.

Writing class MyClass { ... } as an expression assigned to Named creates an internal binding named MyClass that is visible only from within the class body itself, separate from the outer variable Named. Inside label, MyClass refers to the class itself, and its name property reflects that internal name, not the variable it was ultimately stored in. So Named.label() reads MyClass.name, giving the string MyClass.