In the following, I expected class Child
's protected field member _AorB
to be of type B
, and not A
, but reality shows otherwise.
What am I mis-understanding, and how can I adjust the code for the desired behavior?
class A{
public:
void doit(){
std::cout<<" this is A!"<<std::endl;
}
};
class B{
public:
void doit(){
std::cout<<" this is B!"<<std::endl;
}
};
class Parent{
public:
void doit(){
_AorB.doit();
}
protected:
A _AorB;
};
class Child: public virtual Parent{
protected:
B _AorB;
};
int main()
{
cout<<"Hello World";
auto c = Child();
c.doit(); // I expected this to print "This is B" because c is Child(), and Child class's _AorB is of type B.
return 0;
}
CodePudding user response:
You can make such changes:
template <typename AorB>
class Parent{
public:
void doit(){
_AorB.doit();
}
protected:
AorB _AorB;
};
class Child: public virtual Parent<B> {
}
Also take a look at What are the rules about using an underscore in a C identifier?
- Reserved in any scope, including for use as implementation macros:
- identifiers beginning with an underscore followed immediately by an uppercase letter
CodePudding user response:
273K's answer is excellent.
Depending on what kind of problem you are trying to solve and how the data is held in the hierarchy, you could use a std::variant<A, B>
to allow "flippy" behavior based on the type, and access that member variable through a virtual getter member function.
#include <iostream>
#include <variant>
class A {
public:
void doit() {
std::cout << " this is A!\n";
}
};
class B {
public:
void doit() {
std::cout << " this is B!\n";
}
};
class Parent {
public:
virtual ~Parent() = default;
void doit() {
auto ab = get_AorB();
std::visit([](auto arg) { arg.doit(); }, ab);
}
virtual auto get_AorB() -> std::variant<A, B> {
return _a;
}
protected:
A _a;
};
class Child : public virtual Parent {
protected:
B _b;
auto get_AorB() -> std::variant<A, B> override {
return _b;
}
};
int main() {
std::cout << "Hello World";
auto c = Child();
c.doit(); // "this is B!"
}