Say I have the following:
struct Base
{
virtual void callback() { /* */ }
};
struct Derived : public Base
{
void callback() { /* */ }
};
Base* obj = new Derived;
static_cast<Derived*>(obj)->callback();
Will the call to callback invoke via the vtable or a straight function call as it's not marked as virtual in the function signature?
Derived::callback
is implicitlyvirtual
because theBase::callback
function with a the same return type and parameters is declared asvirtual
.That said, whether or not the vtable is used depends on the compiler and what optimizations it performs. Since the compiler can know here that
obj
points to aDerived
object, it knows what the final overrider for thecallback
virtual function is, so it could potentially make the call without looking up the function in the vtable. Whether or not it does so depends on the compiler and optimization settings, though.