Incorporate bang operator hack

[weswigham]

Fixes half of #22934

With this change, the ! operator only applies the NonNullable<T> type if T is an instantiable type (or union thereof) constrained to null or undefined (or unconstrained).

This does not, however, fix the underlying conceptual problem, which is that when you write

class X {
    fn(): void {
        // Type 'this' is not assignable to type 'NonNullable<this>'.
        //  Type 'X' is not assignable to type 'NonNullable<this>'.
        const x: NonNullable<this> = this;
        //    ~
    }
}

One would expect the assignment to work, as the this type can never include null or undefined (barring shenanigans like X & undefined).

[Jessidhia]

undefined is a value that this can legitimately have at runtime in strict mode code, though. But the --noImplicitThis mode probably catches most of the code that doesn't account for this.

[weswigham]

@Kovensky AFAIK a class method we consider to have a strongly typed this in strict mode and don't allow calling it without one (and one compatable with the this type from the original declaration at that). The type of this is a method is an implicit type variable extending the containing class - such a definition does not allow for null or undefined values. It's the fact that it's a type parameter that makes it not work like you'd expect. If you tried to assign an X to a NonNullable<X>, it'd work fine because NonNullable<X> immediately simlifies to just X.

[typescript-bot]

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