在另一个模板基类中实现抽象基类函数 [英] implementing an abstract base class function in another template base class

问题描述

我有一个Visual Studio 2008 C ++项目，其中我有一个类继承自一个抽象基类和另一个模板类，实现抽象基类中的一个函数。例如：

I have a Visual Studio 2008 C++ project where I have a class that inherits from an abstract base class and another template class that implements a function in the abstract base class. For example:

class Foo;

struct Buzz
{
    virtual ~Buzz() {};
    virtual void Buzz_Do( Foo* ) = 0;
};

class Base
{
public:
    virtual ~Base() {};
    virtual void Base_Do( Buzz* ) = 0;
};

template< class T >
class Bar
{
public:
    virtual void Base_Do( Buzz* v )
    {
        v->Buzz_Do( static_cast< T* >( this ) );
    };
};

class Foo : public Base, public Bar< Foo >
{
};

int _tmain(int argc, _TCHAR* argv[])
{
    Foo c;
    return 0;
}

$ b <

Unfortunately, this yields the compiler error:

1>MyApp.cpp(39) : error C2259: 'Foo' : cannot instantiate abstract class
1>        due to following members:
1>        'void Base::Base_Do(Buzz *)' : is abstract
1>        MyApp.cpp(17) : see declaration of 'Base::Base_Do'

a public using Bar < Foo> :: Base_Do; 到 class Foo ，但是没有帮助。

I've tried adding a public using Bar< Foo >::Base_Do; to class Foo, but that did not help.

是否有办法使这个工作，或者我需要在 Foo 中实现一个具体的 Base_Do（）并不使用 Bar<> ？

Is there a way to make this work or will I need to put a concrete Base_Do() implementation in Foo and not use Bar<>?

谢谢，
PaulH

Thanks, PaulH

推荐答案

如果可能，让 Bar<> 派生自Base，
和Foo只有 Bar< T> 。

If possible let the Bar<> derive from Base, and Foo derive only from Bar<T>.

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