#include <variant>
struct A
{
void foo(){}
};
struct B
{
void foo(){}
};
int main()
{
std::variant< A, B > v{ A{} };
v.foo(); // doesn't work
}
How do I use the std::variant value not knowing it's type but knowing it's properties? I believe this is called Generic Polymorphism equivalent to Duck Typing.
Totally valid use case. I imagine there's many ways to do it, but here's one:
std::visit([](auto&& val) { val.foo(); }, v);
The reason your initial code doesn't work is because A::foo is unrelated to B::foo, so to use them interchangeably you need a context where the type "containing" a foo member is deduced. In the visit example, we create such a context by making the callable a generic lambda.
