Nindi Singh wrote:
class Base {
public:
virtual void func()=0;
};
class Derived:public Base {
public:
void func(){}
};
[snip]
I am curious the method 'func' in class Derived is described as virtual. Its
not. If I am missing something, then how can I tell whether a function 'may' be
overloaded in a derived class or not.
Yes, it is declared virtual. It does not matter whether the virtual
keyword is present. If there is a function with the same signature in
the parent class that is declared virtual it implicitly makes your
function virtual. This is a C++ rule.
If instead you write
class Base {
public:
virtual void func()=0;
};
class Derived: public Base {
public:
void func(int) {}
};
then Derived::func is not virtual because it does not match the
signature from Base::func. Instead you will be left with an abstract
class because Base::func is pure-virtual so it must be defined in
Derived to make it a concrete type.
-Brad
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