Help getting to a derived class template given a pointer to a no-template base class

[chris.kemmerer]

I am having a problem with templates and I hope someone here can help.

I am writing a library that accepts data packets, parses them and
saves the information for later use. One member of the packet is an
enumeration that says what type of data the packet contains (int,
char, etc.). I have created classes similar to below. The problem I am
having is in trying to access the derived class given only a pointer
to the base class. I know that the pointer I am reading from the
vector points to the appropriate derived class but I have no way of
knowing it's underlying data type before hand so I can't explicitly
declare a variable of the correct derived class. Any help is
appreciated even if it is a definitive "Can't do it".

Thanks,

Chris

class PacketBase
{
virtual ~PacketBase() {}
...
};

template<typena me T>
class Packet : public PacketBase
{
std::vector<TVa lues() { return m_Values; }
std::vector<T m_Values;
...
};

class UsePackets
{
std::vector<Pac ketBase*m_Packe ts;
...
};

.... somewhere in the main code...
UsePackets foo;
foo.m_Packets.p ush_back(new Packet<int>);
foo.m_Packets.p ush_back(new Packet<short>);
PacketBase* packet = foo.m_Packets.a t(1);
packet->Values(); // can't access this function

[Victor Bazarov]

ch************@ att.net wrote:

I am having a problem with templates and I hope someone here can help.

It's not a problem with templates. It's a problem with understanding
(or not understanding) how inheritance works, I'm afraid.

I am writing a library that accepts data packets, parses them and
saves the information for later use. One member of the packet is an
enumeration that says what type of data the packet contains (int,
char, etc.). I have created classes similar to below.

"Similar"?

The problem I am
having is in trying to access the derived class given only a pointer
to the base class. I know that the pointer I am reading from the
vector points to the appropriate derived class but I have no way of
knowing it's underlying data type before hand so I can't explicitly
declare a variable of the correct derived class. Any help is
appreciated even if it is a definitive "Can't do it".

Thanks,

Chris

class PacketBase
{
virtual ~PacketBase() {}
...
};

template<typena me T>
class Packet : public PacketBase
{
std::vector<TVa lues() { return m_Values; }

Bad idea to return by value. BTW, is this function declared 'private'
intentionally?

std::vector<T m_Values;
...
};

class UsePackets
{
std::vector<Pac ketBase*m_Packe ts;
...
};

... somewhere in the main code...
UsePackets foo;
foo.m_Packets.p ush_back(new Packet<int>);
foo.m_Packets.p ush_back(new Packet<short>);
PacketBase* packet = foo.m_Packets.a t(1);
packet->Values(); // can't access this function

What are you trying to do? 'PacketBase' does not have 'Values'
member. Is that what your compiler is telling you? Well, it is
correct. Or is it telling you that the member is "unaccessib le"?

Read the FAQ 5.8 and follow its recommendations .

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

[chris.kemmerer]

On Sep 26, 4:50 pm, "Victor Bazarov" <v.Abaza...@com Acast.netwrote:

chris.kemme...@ att.net wrote:

I am having a problem with templates and I hope someone here can help.

It's not a problem with templates. It's a problem with understanding
(or not understanding) how inheritance works, I'm afraid.

I am writing a library that accepts data packets, parses them and
saves the information for later use. One member of the packet is an
enumeration that says what type of data the packet contains (int,
char, etc.). I have created classes similar to below.

"Similar"?

The problem I am
having is in trying to access the derived class given only a pointer
to the base class. I know that the pointer I am reading from the
vector points to the appropriate derived class but I have no way of
knowing it's underlying data type before hand so I can't explicitly
declare a variable of the correct derived class. Any help is
appreciated even if it is a definitive "Can't do it".

Thanks,

Chris

class PacketBase
{
virtual ~PacketBase() {}
...
};

template<typena me T>
class Packet : public PacketBase
{
std::vector<TVa lues() { return m_Values; }

Bad idea to return by value. BTW, is this function declared 'private'
intentionally?

std::vector<T m_Values;
...
};

class UsePackets
{
std::vector<Pac ketBase*m_Packe ts;
...
};

... somewhere in the main code...
UsePackets foo;
foo.m_Packets.p ush_back(new Packet<int>);
foo.m_Packets.p ush_back(new Packet<short>);
PacketBase* packet = foo.m_Packets.a t(1);
packet->Values(); // can't access this function

What are you trying to do? 'PacketBase' does not have 'Values'
member. Is that what your compiler is telling you? Well, it is
correct. Or is it telling you that the member is "unaccessib le"?

Read the FAQ 5.8 and follow its recommendations .

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

I'm sorry, I was making up these classes to show the issue with as
little superfluous stuff as possible. The Values() functions is
declared public.

I am currently passing by value but that is not a requirement and I
can easily change it to pass by reference.

I know that PacketBase does not have a Values() function because it
can't, PacketBase doesn't know anything about the template type. I
could do away with all this indirection if UsePackets could hold a
vector of Packet class pointers but I didn't think that was possible
since Packet is a class template and each Packet instance can have a
different type.

Where do I find FAQ 5.8?

Thanks,

Chris

[Victor Bazarov]

ch************@ att.net wrote:

[..]
Where do I find FAQ 5.8?

See http://www.parashift.com/c++-faq-lite/ And "Where to I find
the FAQ for this newsgroup?" should be the first question you
ask when you walk in. Of course, you wouldn't have to do that
if you bothered to read the newsgroup for at least a day before
posting...

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

[shazled]

On Sep 26, 9:32 pm, chris.kemme...@ att.net wrote:

>
class PacketBase
{
virtual ~PacketBase() {}
...

};

template<typena me T>
class Packet : public PacketBase
{
std::vector<TVa lues() { return m_Values; }
std::vector<T m_Values;
...

};

class UsePackets
{
std::vector<Pac ketBase*m_Packe ts;
...

};

... somewhere in the main code...
UsePackets foo;
foo.m_Packets.p ush_back(new Packet<int>);
foo.m_Packets.p ush_back(new Packet<short>);
PacketBase* packet = foo.m_Packets.a t(1);
packet->Values(); // can't access this function

Would something like this work?

class PacketBase
{
virtual ~PacketBase() {}
virtual std::vector<boo st::any>& Values() const =0;
...

};

template<typena me T>
class Packet : public PacketBase
{
std::vector<boo st::any>& Values() const { return m_Values; }
std::vector<boo st::any m_Values;
...

};

class UsePackets
{
std::vector<Pac ketBase*m_Packe ts;
...

};

... somewhere in the main code...
UsePackets foo;
foo.m_Packets.p ush_back(new Packet<int>);
foo.m_Packets.p ush_back(new Packet<short>);
PacketBase* packet = foo.m_Packets.a t(1);
vector<boost::a nywoot = packet->Values();

You may also be able to wrap the vector itself in boost::any but I've
never done anything like that.

Saul

[shazled]

On Sep 27, 2:26 pm, chris.kemme...@ att.net wrote:

>
A related question that may actually solve my first problem. The real
issue is being able to create a type from the enumeration I get from
the data packet. If I could somehow morph that into a usable typename
I could dynamically cast the base pointer to the proper derived
template. Something of the form:
typedef typeid(???).nam e T;
PacketBase* base = foo.m_Packets.a t(1);
Packet<T>* derived = dynamic_cast<Pa cket<T>*>(base) ;
The problem is ??? isn't an object, just an enumeration.

Sorry, I didn't previous realise that the original problem is actually
that you need to store different types in the same container. Its best
not to throw away the type information at all. One possible way is
using vector<boost::v ariant<int,shor t,etc to hold the packets along
with boost::static_v isitor when you want to process them.

Saul

[Adam Nielsen]

I know that the pointer I am reading from the

vector points to the appropriate derived class but I have no way of
knowing it's underlying data type before hand so I can't explicitly
declare a variable of the correct derived class.

Couldn't you use dynamic_cast?

UsePackets foo;
foo.m_Packets.p ush_back(new Packet<int>);
foo.m_Packets.p ush_back(new Packet<short>);
PacketBase* packet = foo.m_Packets.a t(1);
packet->Values(); // can't access this function

Packet<int*pi = dynamic_cast< Packet<int>* >(packet);
if (pi) pi->Values();
else {
Packet<short*ps = dynamic_cast< Packet<short>* >(packet);
if (ps) ps->Values();
}

Not the nicest code and there's probably a far better way, but it is a
quick solution.

Cheers,
Adam.