C++ Iterate Over All Generics At Once - c++

Question

The following code:
#include <cstdio>
#include <iostream>
#include <vector>
template<class type>
struct A {
type i;
A(type j){
i = j;
}
};
template<class type>
std::vector<A<type>*> ve;
int main(int argc, char** args){
ve<int>.push_back(new A<int>(1));
ve<int>.push_back(new A<int>(2));
ve<char>.push_back(new A<char>('a'));
ve<char>.push_back(new A<char>('b'));
for(unsigned int i = 0; i < ve<int>.size(); i++)
std::cout << ve<int>[i]->i << std::endl;
for(unsigned int i = 0; i < ve<char>.size(); i++)
std::cout << ve<char>[i]->i << std::endl;
return 0;
}
outputs:
1
2
a
b
How can I reproduce the same output, but with only one for loop?
Yes I understand the preprocessor made two different vectors here.
Show me how to combine them or show me a generic for loop.

Answer 1

Solution:
#include <cstdio>
#include <iostream>
#include <vector>
struct B {
virtual void memberOut(){}
};
template<class type>
struct A : B{
type i;
A(type j):B(){
i = j;
}
void memberOut(){
std::cout << i << std::endl;
}
};
std::vector<B*> ve;
int main(int argc, char** args){
ve.push_back(new A<int>(1));
ve.push_back(new A<int>(2));
ve.push_back(new A<char>('a'));
ve.push_back(new A<char>('b'));
for(unsigned int i = 0; i < ve.size(); i++)
ve[i]->memberOut();
return 0;
}
Notes:
Either keep it as two containers, or make virtual functions. Both have their uses.
Using a template on a variable just makes two variables.
You can also save another vector of type_index and use typeid to remember what pointer type a void pointer is later on, but that's just a sloppier version of virtual functions. Example:
std::vector<void*> values
std::vector<std::type_index> types
if(types[i]==typeid(A<int>*))
Side Notes:
Try not to get off-base when writing comments.
The objective of comments is to get the question answered, not debate.