pure virtual function calls / optimization

[alexander.stippler]

Hi

I wonder if their will be a performance penalty in the following
situation due to virtual function calls:

class Interface
{
public:
virtual void methodA() = 0;
virtual void methodB() = 0;
virtual void methodC() = 0;
virtual void methodD() = 0;
};

class Implementation
{
public:
void methodA() { /* do someting */ }
void methodB() { /* do someting */ }
void methodC() { /* do someting */ }
void methodD() { /* do someting */ }

};

IMHO there is no need for class Implementation to have a virtual method
table, since there is just one possibility for every method. It's a
special case, I know. Other derived classes may have virtual functions.
What do current compilers do? Are there optimizations for such
situations or did I just miss something?

regards,
Alex

 

[alexander.stippler]

Hi

I wonder if their will be a performance penalty in the following
situation due to virtual function calls:

class Interface
{
public:
virtual void methodA() = 0;
virtual void methodB() = 0;
virtual void methodC() = 0;
virtual void methodD() = 0;
};

class Implementation
{
public:
void methodA() { /* do someting */ }
void methodB() { /* do someting */ }
void methodC() { /* do someting */ }
void methodD() { /* do someting */ }

};

IMHO there is no need for class Implementation to have a virtual method
table, since there is just one possibility for every method. It's a
special case, I know. Other derived classes may have virtual functions.
What do current compilers do? Are there optimizations for such
situations or did I just miss something?

regards,
Alex

Sorry, of course the Implementation class is publicly derived from the
Interface class.

class Implementation
: public Interface
{
//...
};

 

[Ian Collins]

Hi

I wonder if their will be a performance penalty in the following
situation due to virtual function calls:

class Interface
{
public:
virtual void methodA() = 0;
virtual void methodB() = 0;
virtual void methodC() = 0;
virtual void methodD() = 0;
};

class Implementation
{
public:
void methodA() { /* do someting */ }
void methodB() { /* do someting */ }
void methodC() { /* do someting */ }
void methodD() { /* do someting */ }

};

IMHO there is no need for class Implementation to have a virtual method
table, since there is just one possibility for every method. It's a
special case, I know. Other derived classes may have virtual functions.
What do current compilers do? Are there optimizations for such
situations or did I just miss something?

The may well be an over head, but it will be small. Whether or not the
compiler can perform any optimisations such as inlining depends on the
compiler and the use of the functions.

 

[Rolf Magnus]

Hi

I wonder if their will be a performance penalty in the following
situation due to virtual function calls:

class Interface
{
public:
virtual void methodA() = 0;
virtual void methodB() = 0;
virtual void methodC() = 0;
virtual void methodD() = 0;
};

class Implementation
{
public:
void methodA() { /* do someting */ }
void methodB() { /* do someting */ }
void methodC() { /* do someting */ }
void methodD() { /* do someting */ }

};

IMHO there is no need for class Implementation to have a virtual method
table, since there is just one possibility for every method.

If that is the whole program. There might be classes derived from
Implementation. If you access an object of such a derived class through a
pointer or reference to Implementation, how would the compiler know which
function to call?

 

[alexander.stippler]

If that is the whole program. There might be classes derived from
Implementation. If you access an object of such a derived class through a
pointer or reference to Implementation, how would the compiler know which
function to call?

The members of Implementation are no longer virtual. So this should be
clear.

Alex

 

[Rolf Magnus]

The members of Implementation are no longer virtual. So this should be
clear.

You're mistaken. If a member function of a base class is virtual, that
member function is automatically virtual in every derived class, even if
not explicitly specified.

 

[alexander.stippler]

You're mistaken. If a member function of a base class is virtual, that
member function is automatically virtual in every derived class, even if
not explicitly specified.

ah ok. I did not know this. So just forget the whole question. It does
no longer make sense.
Thanks

Alex

 

[Mike Smith]

The members of Implementation are no longer virtual. So this should be
clear.

They were declared virtual in Interface, and would therefore still be
virtual in Implementation. You can't "un-virtualize" a function.