From: "marcandre (Marc-Andre Lafortune)" <ruby-core@...>
Date: 2012-11-13T06:31:14+09:00
Subject: [ruby-core:49266] [ruby-trunk - Feature #7339][Rejected] Version of super that doesn't raise when super undefined


Issue #7339 has been updated by marcandre (Marc-Andre Lafortune).

Status changed from Open to Rejected

Indeed, `defined?(super)` is probably what you were looking for.

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Feature #7339: Version of super that doesn't raise when super undefined
https://bugs.ruby-lang.org/issues/7339#change-32823

Author: prijutme4ty (Ilya Vorontsov)
Status: Rejected
Priority: Normal
Assignee: 
Category: 
Target version: 


=begin
I propose new method try_super (it's possibly bad name, any suggestions) which would work like a super except not raising in case that super method undefined. It can be useful in such a situation - when module makes smth useful even in absence of super class.
I use an example with Module#prepend, but similar example with Module#include can be created with a bit more effort.

  module MyLogging
    def info(*args)
      $stderr.puts "Hello! You've an info message"
      super # with super it raises when super undefined
    end
  end

  require 'logger'
  class Logger
    prepend MyLogging
  end


  class Foo
    prepend MyLogging
  end

  Logger.new.info 'message'
  Foo.new.info 'message'


In an example Foo.new.info raises an exception while really it shouldn't in my opinion. So try_super can be used

If it doesn't broke many libraries (I believe not many of them uses super in cases when it raises), super can be renamed to 'super!' and more mild version can be 'super'.

=end



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