Re: Template multiple inheritance Organization: Roundhouse Consulting, Ltd.

From:
=?ISO-8859-1?Q?Daniel_Kr=FCgler?= <daniel.kruegler@googlemail.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Thu, 19 May 2011 07:53:28 CST
Message-ID:
<ir2aji$ptv$1@dont-email.me>
On 2011-05-18 23:20, dec4106 wrote:

Sure. Do the same thing you'd do if you had two non-template bases.

--
Pete


That's what I thought but either I don't understand it (obviously...)
or the compiler doesn't support it. Here's a more complete example,
compiled with VS 2008:

class MyData
{};

struct BaseType
{
     BaseType() {}
     virtual ~BaseType() {}
     virtual void execute() = 0;
     virtual void connectTo(BaseType * dest, int outputPort, int
inputPort) = 0;
};

template<typename T>
struct Provider : virtual public BaseType
{
     virtual ~Provider() {}
     virtual T getValue() const = 0;
};

template<typename T, int port>
struct PortProvider : public Provider<T>
{
     virtual ~PortProvider() {}
     virtual T getValue() const
     {
         return getPortValue();
     }
     virtual T getPortValue() const = 0;
};

class BS :
     public PortProvider<MyData, 1>,
     public PortProvider<MyData, 2>
{
public:
     BS() : PortProvider<MyData, 1>(), PortProvider<MyData, 2>()
     {}

     virtual ~BS()
     {}

     virtual MyData PortProvider<MyData, 1>::getPortValue() const
     {}

     virtual MyData PortProvider<MyData, 2>::getPortValue() const
     {}

     virtual void execute()
     {}

     virtual void connectTo(BaseType * dest, int outputPort, int
inputPort)
     {}
};

int _tmain(int argc, _TCHAR* argv[])
{
     BS * bs1(new BS);
     return 0;
}

Error messages include:

error C3241: 'MyData BS::getPortValue(void)' : this method was not
introduced by 'PortProvider<T,port>'
error C2259: 'BS' : cannot instantiate abstract class due to following
members:
1> 'MyData PortProvider<T,port>::getPortValue(void) const' : is
abstract


This is not a compiler defect, you are misusing the language. In the
function definitions of getPortValue() above you cannot use the base
class qualifier. In this case you have to provide a single override:

class BS :
     public PortProvider<MyData, 1>,
     public PortProvider<MyData, 2>
{
public:
     virtual MyData getPortValue() const;
     virtual void execute();
     virtual void connectTo(BaseType * dest, int outputPort, int
inputPort);
};

And yes, you have now to decide, what this single override of
getPortValue() returns (In this example the answer is probably easy,
because none of the base classes has yet an implementation for
getPortValue()).

If you additionally want to provide each base class functionality
separately, you can add a non-virtual function template or separate
functions to BS, e.g. like this:

class BS :
     public PortProvider<MyData, 1>,
     public PortProvider<MyData, 2>
{
public:
     virtual MyData getPortValue() const;
     template<int port>
     MyData getPortValue() const;
     virtual void execute();
     virtual void connectTo(BaseType * dest, int outputPort, int
inputPort);
};

If both base classes *would* provide an implementation for
getPortValue(), one possible return statement could be:

return PortProvider<MyData, port>::getPortValue();

This allows you to write this kind of code:

int main()
{
     BS bs;
     bs.getPortValue(); // OK
     bs.getPortValue<1>(); // OK
     bs.getPortValue<2>(); // OK
     bs.getPortValue<3>(); // Error
}

HTH & Greetings from Bremen,

Daniel Kr?gler

--
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