Creating a Variant class and std::map<Variant, Variant> - c++

Question

I craeted a simple Variant class to store string, integer, double, etc. I'm trying to use a map of type std::map<Variant, Variant> but I'm getting this strange error:
In file included from /usr/include/c++/7/string:48:0,
from /home/dev/proj/cpp/common/Variant.h:3,
from /home/dev/proj/cpp/common/Event.h:3,
from /home/dev/proj/cpp/common/Event.cpp:1:
/usr/include/c++/7/bits/stl_function.h: In instantiation of 'constexpr bool std::less<_Tp>::operator()(const _Tp&, const _Tp&) const [with _Tp = Variant]':
/usr/include/c++/7/bits/stl_map.h:511:32: required from 'std::map<_Key, _Tp, _Compare, _Alloc>::mapped_type& std::map<_Key, _Tp, _Compare, _Alloc>::operator[](std::map<_Key, _Tp, _Compare, _Alloc>::key_type&&) [with _Key = Variant; _Tp = Variant; _Compare = std::less<Variant>; _Alloc = std::allocator<std::pair<const Variant, Variant> >; std::map<_Key, _Tp, _Compare, _Alloc>::mapped_type = Variant; std::map<_Key, _Tp, _Compare, _Alloc>::key_type = Variant]'
/home/dev/orwell/cpp/common/Event.cpp:33:18: required from here
/usr/include/c++/7/bits/stl_function.h:386:20: error: no match for 'operator<' (operand types are 'const Variant' and 'const Variant')
{ return __x < __y; }
~~~~^~~~~
This is my Variant class:
class Variant
{
public:
enum class Type
{
Integer,
Double,
String
};
Variant()
{
}
Variant(int integer)
{
this->type = Type::Integer;
setInteger(integer);
}
Variant(std::string string)
{
this->type = Type::String;
setString(string);
}
Variant(double _double)
{
this->type = Type::Double;
setDouble(_double);
}
Type type;
This is where the error is happening:
void Event::add(std::string key, std::string value) {
this->map[key] = Variant(value); //problem here
}

Answer 1

std::map is a sorted array. To do that it uses the < operator.
Thus, if you want to use Variant in a map (I believe this only applies to keys), you will need to supply an operator<() for it. You can find some examples here.
Or, you'll need a comparison function. That can work too.