Deduce return type on base class virtual method

Assuming the following couple of classes Base and Derived, is there a way to either a) declare a virtual method value() on the base class without a specific return type (which I doubt it's possible) or 2) have different templated types on the same vector/container (which is why I am using pointers to a non-template abstract base class)? I am using C++11 which doesn't allow me to use auto without a return type.

#include <vector>
#include <iostream>

class Base {
 public:
    Base() { /* */ }

    // virtual auto value() const = 0;  // <- need return type!!
};

template <typename T>
class Derived : public Base{
 public:
    explicit Derived(const T& _val) : member(_val) { /* */ }

    T value() const { return member; }
 protected:
    T member;
};


void doStuff(const std::vector<Base*>& _data) {
    // ...
    for (auto& obj : _data) {
        std::cout << obj->value() << std::endl;  // error - ‘class Base’ has no member named ‘value’
    }
}

int main(int argc, char const *argv[]) {
    std::vector<Base*> data;
    Derived<float> foo(0.5);
    Derived<int>   bar(47);
    data.push_back(&foo);
    data.push_back(&bar);

    doStuff(data);

    return 0;
}