Question posted in 
It sounds like a strange question, but when I use the following example code in Visual Studio 2022, IntelliSense is warning me that the member iMember in the parent class is inaccessible, even though the code itself seems ok to me and is running without any problems.
Example Code:
template<class S, class T> class Child;
template<class S, class T>
class Base {
int iMember = 0;
public:
Base(const Child<S, T>& child);
Base(int iMember) : iMember(iMember) {}
void Print() { std::cout << iMember << std::endl; }
};
template<class S, class T>
class Child : public Base<S, T>{
public:
Child(int i) : Base<S,T>(i) {}
};
//the following constructor causes the IntelliSense warning that Base<S,T>::iMember is inaccessible
template<class S, class T>
Base<S, T>::Base(const Child<S, T>& child) : iMember(child.iMember) {}
int main() {
Child<int, int> cc(234);
Base<int, int> bb(cc);
cc.Print(); //prints 234
}
I have no idea why IntelliSense is making this warning, but maybe my code has an undefined behavior.
2
Answers
The program is well-defined i.e there is no UB in the given code.
This is a false positive. In the constructor initializer list of the base class, we can safely/correctly write
child.iMemberbecause we’re in the scope of the base class.It is highly likely that the IntelliSense warning you are encountering appears to be a false positive rather than an actual issue with your code. The code you provided correctly initializes the
Base<S, T>::iMemberin the constructorBase<S, T>::Base(const Child<S, T>& child)through the member initializer list, which is a valid and appropriate practice.It’s not uncommon for static analysis tools like IntelliSense to occasionally generate false positives or negatives. In this case, since your code compiles and runs without any problems and there are no runtime issues, it is reasonable to dismiss the IntelliSense warning as a tooling quirk.