Allow type assertion for "super" keyword in method.

Search Terms

super type assertion

Suggestion

Currently there is no way to provide a type assertion for "super" keyword, because the "super" keyword cannot be wrapped by parentheses, so either (<T>super).x or (super as T).x will result in an error: "'super' must be followed by an argument list or member access.ts(1034)".

The parentheses wrapping "super" which only for type assertion use should be allowed.

Use Cases

class A {
  // a decorator will add a instance method "foo" here.
}

class B extends A {
  bar() {
    (<A & {foo: () => void}>super).foo(); // doesn't work.
  }
}

class C extends A {
  bar() {
    (<A & {foo: () => void}>A.prototype).foo(); // this works.
  }
}

Checklist

My suggestion meets these guidelines:

• [●] This wouldn't be a breaking change in existing TypeScript/JavaScript code
• [●] This wouldn't change the runtime behavior of existing JavaScript code
• [●] This could be implemented without emitting different JS based on the types of the expressions
• [●] This isn't a runtime feature (e.g. library functionality, non-ECMAScript syntax with JavaScript output, etc.)
• [●] This feature would agree with the rest of TypeScript's Design Goals.

Reproduce the reported error with the TypeScript examples in the issue and compare the failing super assertion with the working A.prototype assertion. The work is done when the requested parenthesized type assertion around super is accepted without changing emitted JavaScript behavior, with coverage for the shown cases.