problem to understand " new virtual "
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Thread: problem to understand " new virtual "

  1. #1
    Join Date
    Sep 2013
    Posts
    16

    problem to understand " new virtual "

    to explain i will write a little code:
    Code:
    class A {
    public virtual void f() { print "1"; }
    }
    
    class B : A {
    public override void f() { print "2"; } 
    }
    
    
    main() {
    A bb = new B();
    bb.f() ;
    in first case it will print 2
    but if i replace the word " override" with " new" then it will print 1
    why is it? why when i set method as new virtual it exicute the method that implemented in the base class ( generic)
    how it work exactly?

  2. #2
    Arjay's Avatar
    Arjay is offline Moderator / MS MVP Power Poster
    Join Date
    Aug 2004
    Posts
    11,359

    Re: problem to understand " new virtual "

    Search bing or google for "c# new virtual tutorial".

    In general, when learning C#, you are going to need to do or find several examples on the internet or in a book in order to understand the concepts. Please try to find the answers on your own before posting to the forum. I'm definitely not saying we don't want to help, but in many cases with regard to the initial learning of C#, an online tutorial is going to be able to explain the subject in more depth than we can. If one tutorial doesn't make things clear, then do other tutorials (because each one will explain things slightly different).

    Btw, one of the best things that you can do as a beginning programmer is to learn how to find the information yourself. That way, when you are stuck with a difficult problem and under a time crunch, you won't have to wait for someone else to answer your question. It's never good to have to tell a boss, "I don't know what the problem is and I'm waiting for help on the internet."

  3. #3
    Join Date
    Oct 2004
    Location
    Rocket City
    Posts
    219

    Re: problem to understand " new virtual "

    A bb = new B();
    Did you intend to use:
    Code:
    B bb = new B();

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