Optimal Efficiency: dynamic_cast pointer or reference?

[Tomás]

dynamic_cast can be used to obtain a pointer or to obtain a reference.

If the pointer form fails, then you're left with a null pointer.

If the reference form fails, then an exception is thrown.

Would "Feed1" or "Feed2" be preferable in the following:

#include <iostream>

#include <typeinfo>

using std::bad_cast;

using std::cout;
using std::endl;

class Mammal
{
public:

Mammal()
{
cout << "Constructo r: Mammal" << endl;
}

virtual ~Mammal()
{
cout << "Destructor : Mammal" << endl;
}

/*************** *************** ******
Virtual destructor because Mammal
has to be polymorphic.
*************** *************** ******/
};
class Panda : public Mammal {
public:

Panda()
{
cout << "Constructo r: Panda" << endl;
}

~Panda()
{
cout << "Destructor : Panda" << endl;
}
};
void TellZooKeeperTh atWeFedAPanda()
{
cout << "Hey Zoo Keeper, we just fed a Panda." << endl;
}

void Feed1(Mammal &mammal)
{
if ( dynamic_cast< Panda* > ( &mammal ) )
{
TellZooKeeperTh atWeFedAPanda() ;
}
}

void Feed2(Mammal &mammal)
{
try
{
dynamic_cast< Panda& > (mammal);

TellZooKeeperTh atWeFedAPanda() ;
}
catch ( bad_cast ) { }
}

int main()
{
Mammal panda;

Feed1( panda );

Feed2( panda );
}

-Tomás

[Victor Bazarov]

Tomás wrote:

[...efficiency difference question...]

Let me ask a leading question in response: what efficiency difference
_do_ you _notice_ or _encounter_ in your application/system that prompts
you to ask this efficiency question? Hint: if you can't see any, there
is none.

V
--
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