{"action":"closed","issue":{"url":"https://api.github.com/repos/nzakas/understandinges6/issues/151","labels_url":"https://api.github.com/repos/nzakas/understandinges6/issues/151/labels{/name}","comments_url":"https://api.github.com/repos/nzakas/understandinges6/issues/151/comments","events_url":"https://api.github.com/repos/nzakas/understandinges6/issues/151/events","html_url":"https://github.com/nzakas/understandinges6/issues/151","id":64702770,"number":151,"title":"`super()` not allowed in derived methods","user":{"login":"getify","id":150330,"avatar_url":"https://avatars.githubusercontent.com/u/150330?v=3","gravatar_id":"","url":"https://api.github.com/users/getify","html_url":"https://github.com/getify","followers_url":"https://api.github.com/users/getify/followers","following_url":"https://api.github.com/users/getify/following{/other_user}","gists_url":"https://api.github.com/users/getify/gists{/gist_id}","starred_url":"https://api.github.com/users/getify/starred{/owner}{/repo}","subscriptions_url":"https://api.github.com/users/getify/subscriptions","organizations_url":"https://api.github.com/users/getify/orgs","repos_url":"https://api.github.com/users/getify/repos","events_url":"https://api.github.com/users/getify/events{/privacy}","received_events_url":"https://api.github.com/users/getify/received_events","type":"User","site_admin":false},"labels":[],"state":"closed","locked":false,"assignee":null,"milestone":null,"comments":3,"created_at":"2015-03-27T06:03:23Z","updated_at":"2015-04-04T17:56:41Z","closed_at":"2015-04-04T17:56:41Z","body":"> If you're calling a prototype method with the exact same name, then you can also call super as a function\r\n\r\nThis is no longer true. It was removed from ES6 a few months ago. You have to do `super.foo()` in a derived class method, not just `super()`."}}