Reputation: 2027
Assume the snippet below. How can I make this compiling/working? I do not want to move print
to the String/Float
class, because in my real world situation this function is combining a lot of data.
So basically I want a pointer/member to "any type (in this case string/float)" then use it, and call dynamically something else (in this case print
)?
I assume that this does not work (among others) because it cannot determine at compile time which type T
of ptr
will have at compile time.
What is the general pattern to solve such kind of problems?
#include <iostream>
template<typename T>
class AbstractClass {
virtual T getValue()=0;
};
class StringClass : public AbstractClass<std::string> {
std::string getValue() override {
return "A";
}
};
class FloatClass : public AbstractClass<float> {
float getValue() override {
return 1;
}
};
class Processor {
public:
AbstractClass<T>* ptr;
void doIt() {
ptr=new StringClass();
print(ptr->getValue());
delete ptr;
ptr=new FloatClass();
print(ptr->getValue());
delete ptr;
}
void print(std::string i) {
std::cout << "String "<<i<<std::endl;
}
void print(float i) {
std::cout << "Float "<<i<<std::endl;
}
}
int main() {
Processor a;
a.doIt();
}
Upvotes: 0
Views: 70
Reputation: 10880
If you want an object that's 'one of' a given set of types, you can use std::variant<Ts...>
. Mathematically, it represents discriminated/tagged union. This way, you don't need a pointer, neither a base class. Example:
#include <iostream>
#include <variant>
class StringClass {
std::string getValue() override {
return "A";
}
};
class FloatClass {
float getValue() override {
return 1;
}
};
using ClassWithGetValue = std::variant<StringClass, FloatClass>;
class Processor {
public:
ClassWithGetValue v;
void doIt() {
v = StringClass();
std::visit([&](auto&& v1) {
print(v1.getValue());
});
v = FloatClass();
std::visit([&](auto&& v1) {
print(v1.getValue());
});
}
void print(std::string i) {
std::cout << "String "<<i<<std::endl;
}
void print(float i) {
std::cout << "Float "<<i<<std::endl;
}
}
int main() {
Processor a;
a.doIt();
}
Upvotes: 1