Re: Covariant return types doesn't work (with g++ 4.1.2)

From:

"Alf P. Steinbach" <alfps@start.no>

Newsgroups:

comp.lang.c++

Date:

Sun, 10 Feb 2008 19:18:19 +0100

Message-ID:

<13qufrks52hfucf@corp.supernews.com>

* mr.xiaofan.li@gmail.com:

I have been totally confused about the covariant return type feature
in C++. Below is an example code I wrote that doesn't compile with g++
4.1.2 (on Fedora 8)

class virt_base

Note that the term "virtual base (class)" has a special meaning in C++.

Your class "virt_base" is not used as a virtual base class.

{
public:
    virt_base()
    {
    }

    virtual ~virt_base()
    {

    }

    virtual virt_base* cut()
    {
        return new virt_base();
    }

    void say_hi()
    {
        cout <<"hi!!! " <<endl;
    }
};

class virt_derived
: public virt_base
{
public:
    virt_derived()
    {
    }

    ~virt_derived()
    {

    }

    virtual virt_derived* cut()
    {
        return new virt_derived();
    }

    void say_hi()
    {
        cout <<"HI!!!! " <<endl;
    }
};

int main()
{
    virt_base* my_virt_derived = new virt_derived();
    virt_derived* new_virt_derived = my_virt_derived->cut(); // g++
complains here: invalid
                                                                                        //
conversion from 'virt_base*' to
                                                                                        //
'virt_derived*'

Please be a little more conscientious with your quoting. This code
can't be copied and compiled. It must be manually reformatted.

The problem is simply that virt_base::cut() returns a virt_base*.

So you're initializing a virt_derived* pointer with a virt_base* pointer.

Declare new_virt_derived as

virt_base* new_virt_derived = ...

In addition, to get the result I think you expect, you'll have to
declare say_hi as virtual in virt_base.

new_virt_derived->say_hi();
}

Cheers, & hth.,

- Alf

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