Sign Up

Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.

Have an account? Sign In

Have an account? Sign In Now

Sign In

Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.

Sign Up Here

Forgot Password?

Don't have account, Sign Up Here

Forgot Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Have an account? Sign In Now

You must login to ask a question.

Forgot Password?

Need An Account, Sign Up Here

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Sign InSign Up

The Archive Base

The Archive Base Logo The Archive Base Logo

The Archive Base Navigation

  • Home
  • SEARCH
  • About Us
  • Blog
  • Contact Us
Search
Ask A Question

Mobile menu

Close
Ask a Question
  • Home
  • Add group
  • Groups page
  • Feed
  • User Profile
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Buy Points
  • Users
  • Help
  • Buy Theme
  • SEARCH
Home/ Questions/Q 94441
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 10, 20262026-05-10T23:30:07+00:00 2026-05-10T23:30:07+00:00

Does anyone know why typedefs of class names don’t work like class names for

  • 0

Does anyone know why typedefs of class names don’t work like class names for the friend declaration?

class A { public: };  class B : public A { public:    typedef A SUPERCLASS; };  typedef A X;  class C { public:    friend class A;             // OK    friend class X;             // fails    friend class B::SUPERCLASS; // fails }; 
  • 1 1 Answer
  • 0 Views
  • 0 Followers
  • 0
Share
  • Facebook
  • Report

Leave an answer
Cancel reply

You must login to add an answer.

Forgot Password?

Need An Account, Sign Up Here

1 Answer

  • Voted
  • Oldest
  • Recent
  • Random
  1. 2026-05-10T23:30:08+00:00Added an answer on May 10, 2026 at 11:30 pm

    It can’t, currently. I don’t know the reason yet (just looking it up, because i find it interesting). Update: you can find the reason in the first proposal to support typedef-names as friends: http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2003/n1520.pdf . The reason is that the Standard only supported elaborated-type-specifiers. It’s easy to allow only those, and say if the entity declared as friend is not declared yet, it will be made a member of the surrounding namespace. But this means that if you want to use a template parameter, you would have to do (a class is required then for example)

    friend class T; 

    But that brought additional problems, and it was figured not worth the gain. Now, the paper proposes to allow additional type specifiers to be given (so that this then allows use of template parameters and typedef-names).

    The next C++ version (due to 2010) will be able to do it.

    See this updated proposal to the standard: http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2005/n1791.pdf . It will not allow only typedef names, but also template parameters to be used as the type declared as friend.

    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Ask A Question

Stats

  • Questions 262k
  • Answers 262k
  • Best Answers 0
  • User 1
  • Popular
  • Answers
  • Editorial Team

    How to approach applying for a job at a company ...

    • 7 Answers
  • Editorial Team

    How to handle personal stress caused by utterly incompetent and ...

    • 5 Answers
  • Editorial Team

    What is a programmer’s life like?

    • 5 Answers
  • Editorial Team
    Editorial Team added an answer Why not just find . -name '*.mp3' -exec afplay '{}'… May 13, 2026 at 11:54 am
  • Editorial Team
    Editorial Team added an answer Its an anonymous function. Its considered a callback. In PHP… May 13, 2026 at 11:54 am
  • Editorial Team
    Editorial Team added an answer Even if you're using a CultureInfo object, DateTime.Parse will try… May 13, 2026 at 11:54 am

Related Questions

Does anyone know why typedefs of class names don't work like class names for
Does anyone know why using-declarations don't seem to work for importing type names from
Does anyone know why this will not compile? I've tried both VS 2008 and
I'm trying to use a typedef from a subclass in my project, I've isolated
I am trying to call the internal Windows NT API function NtOpenProcess. I know

Trending Tags

analytics british company computer developers django employee employer english facebook french google interview javascript language life php programmer programs salary

Top Members

Explore

  • Home
  • Add group
  • Groups page
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Users
  • Help
  • SEARCH

Footer

© 2021 The Archive Base. All Rights Reserved
With Love by The Archive Base

Insert/edit link

Enter the destination URL

Or link to existing content

    No search term specified. Showing recent items. Search or use up and down arrow keys to select an item.