debug message in a template class

[thomas]

template <class T>
class BufferQueue{
public:
int BufferQueue<T>: :pushToQueue(T pkt, int timeout);
}

template<class T>
int BufferQueue<T>: :pushToQueue(T pkt, ACE_Time_Value *timeout){
cout<<"in pushToQueue"<<e ndl; --------Line p
return 0;
}

Suppose I have two instances BufferQueue<A_a , BufferQueue<B_b ;
I want to know which instance I'm in by printing some debug messages
out in Line p
like
"in pushToQueue of A", etc.

How can I tell the compiler or change Line p to make it?

[Barry]

On Sep 30, 2:46*pm, thomas <FreshTho...@gm ail.comwrote:

template <class T>
class BufferQueue{
public:
* * * * * int BufferQueue<T>: :pushToQueue(T pkt, int timeout);

}

template<class T>
int BufferQueue<T>: :pushToQueue(T pkt, ACE_Time_Value *timeout){
* * * * * cout<<"in pushToQueue"<<e ndl; * * * * * ** --------Line p
* * * * * return 0;

}

Suppose I have two instances BufferQueue<A_a , BufferQueue<B_b ;
I want to know which instance I'm in by printing some debug messages
out in Line p
like
"in pushToQueue of A", etc.

How can I tell the compiler or change Line p to make it?

You can define a static memeber function like T::GetName() which
returns
the name of the class, so

cout << "in pushToQueue " << T::GetName() << endl;

Or you may have typeid(T).name( ) output as message.

[thomas]

>
You can define a static memeber function like T::GetName() which
returns
the name of the class, so

cout << "in pushToQueue " << T::GetName() << endl;

How can you write the "GetName" method?

>
Or you may have typeid(T).name( ) output as message.- Hide quoted text -

- Show quoted text -

Yeah.. The typeid function should work. Thanks.

[James Kanze]

On Sep 30, 11:18 am, thomas <FreshTho...@gm ail.comwrote:

[...]

Or you may have typeid(T).name( ) output as message.- Hide
quoted text -

Yeah.. The typeid function should work. Thanks.

It should, but whether it does or not is a quality of
implementation issue. As far as the standard is concerned, an
implementation which always returns "" is fully conforming. In
practice, most implementations (with the notable exception of
g++) do return something useful.

--
James Kanze (GABI Software) email:ja******* **@gmail.com
Conseils en informatique orientée objet/
Beratung in objektorientier ter Datenverarbeitu ng
9 place Sémard, 78210 St.-Cyr-l'École, France, [phone]

[Barry]

On Sep 30, 5:18*pm, thomas <FreshTho...@gm ail.comwrote:

You can define a static memeber function like T::GetName() which
returns
the name of the class, so

cout << "in pushToQueue " << T::GetName() << endl;

How can you write the "GetName" method?

template <class T>
struct MyClass
{
void MyFunc() const
{
std::cout << T::GetName() << std::endl;
}
};

Here every instatiation of T is requred
to have a static function named "GetName";

struct T_type1
{
static char const* GetName() { return "T_type1"; }
};

int main()
{
MyClass<T_type1 c1;
c1.MyFunc();
}

>

Or you may have typeid(T).name( ) output as message.- Hide quoted text -

- Show quoted text -

Yeah.. The typeid function should work. Thanks.