strange virtual function output

This is a discussion on strange virtual function output within the C++ Programming forums, part of the General Programming Boards category; Hello everyone, Why in the below code, pb->foo(); will output Final other than output Base? I have tested that the ...

  1. #1
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    strange virtual function output

    Hello everyone,


    Why in the below code,

    pb->foo();

    will output Final other than output Base? I have tested that the output is Final.

    My question is
    1. pb points Final instance, and the foo in Final is not virtual method;
    2. if function is virtual, we should invoke the foo based on the type of instance pointed to, if not virtual, we should invoke the foo based on the type which is the pionter type;
    3. since foo in Final is not final, so the output should be invoking the foo based on the type of pointer (which is Base), then the output should be Base?

    Code:
    #include <iostream>
    
    using namespace std;
     
    class Base {
    public:
        virtual int foo() {cout << "Base" << endl; return 0;}
    };
     
    class Derived: public Base
    {
    public:
        int foo() {cout << "Derived" << endl;return 0;}
    };
     
    class Final: public Derived{
    public:
      int foo() {cout << "Final" << endl;return 0;}
    };
     
    int main()
    {
        Final f;
        Base* pb = reinterpret_cast<Base*> (&f);
        pb->foo();
     
        return 0;
    }

    regards,
    George

  2. #2
    (?<!re)tired Mario F.'s Avatar
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    You probably wanted:

    Base* pb = &f;
    Last edited by Mario F.; 02-08-2008 at 07:30 AM. Reason: shameful lack of &
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  3. #3
    CSharpener vart's Avatar
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    http://msdn2.microsoft.com/en-us/lib...73(VS.60).aspx
    The virtual keyword is needed only in the base class's declaration of the function; any subsequent declarations in derived classes are virtual by default.
    The first 90% of a project takes 90% of the time,
    the last 10% takes the other 90% of the time.

  4. #4
    Kernel hacker
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    As vart says: Once you make a function virtual, it will be virtual in all derived classes whether you specify it virtual or not. It is (in my opinion) nice to specify virtual in the derived class, just so that you don't have to work back to the original base class to find out if the function is virtual or not, but it's not necessary from the compilers perspective.

    Note however that if you change the signature of the function (e.g. change argument count, argument type or return type), this function will:
    1. Hide the original function.
    2. Not be virtual unless you specify it.

    --
    Mats
    Compilers can produce warnings - make the compiler programmers happy: Use them!
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  5. #5
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    Thanks vart and Mats,


    My question is answered.

    Quote Originally Posted by matsp View Post
    As vart says: Once you make a function virtual, it will be virtual in all derived classes whether you specify it virtual or not. It is (in my opinion) nice to specify virtual in the derived class, just so that you don't have to work back to the original base class to find out if the function is virtual or not, but it's not necessary from the compilers perspective.

    Note however that if you change the signature of the function (e.g. change argument count, argument type or return type), this function will:
    1. Hide the original function.
    2. Not be virtual unless you specify it.

    --
    Mats

    regards,
    George

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