operator< redefinition on a std::pair<T1,T2> in class template - c++

Question

I've got the following situation in a class template case:
template<class T1,class T2>
class targetClass{
public:
typedef typename std::pair<T1, T2> ToSortType;
typedef typename std::set<ToSortType> ContainerSort;
void bar(ToSortType a, ToSortType b);
private:
ContainerSort container;
bool operator<(const ToSortType& rhs) const;
}
template<class T1,class T2>
void targetClass<T1,T2>::bar(ToSortType a, ToSortType b){
container.insert(a);
container.insert(b);
}
template <class T1,class T2>
bool targetClass<T1,T2>::operator<(const ToSortType& rhs) const
{
return this->first < rhs.first;
}
In main function something like this:
targetClass<int,T2> anObjectTarget;
T2 a;
T2 b;
anObjectTarget.bar(std::make_pair(0,a),std::make_pair(1,b));
Where T2 is a user-defined type which generally does not have a defined operator<
In this particular case, std::set has to compare std::pair<int,T2> by firstly check an operator< (and others, maybe) for int type and then for T2. In this case, the compiler can not find a suitable operator for T2. In the previous snippet then I make a redefinition of the operator concerned but the compiler complains in this way:
/usr/include/c++/7/bits/stl_pair.h:456: error: no match for ‘operator<’ (operand types are ‘const T2’ and ‘const T2’)
|| (!(__y.first < __x.first) && __x.second < __y.second); }
~~~~~~~~~~~^~~~~~~~~~~~
I've never redefined an operator before but looking to the documentation it looks correct to me (but not to the compiler).

Answer 1

The operator you overloaded is a member of targetClass<T1,T2> and takes two ToSortType as parameter. Thats not how an overload of the < operator works. Consider that for instances of class type the follwing two are equivalent:
a < b
a.operator<(b)
ie operators are just syntactic sugar for calling special member functions. The operator you wrote could only be called like
targetClass<T1,T2> t;
T1 a;
T2 b;
t.operator<(a,b);
but what the set tries to call is a < b, ie a.operator(b) and that apparently does not exist (std::pair<T1,T2> can only be comared via < when both T1 and T2 can).
Long story short: You cannot use your operator to compare two instances of ToSortType.
I would not recommend to try to overload the operator< for std::pair<T1,T2>, but rather use a custom type:
template<class T1,class T2>
class targetClass{
public:
struct value_type {
T1 first;
T2 second;
bool operator<(const value_type& other) {
return first < rhs.first;
}
}
using container_type = std::set<value_type>;
void bar(const value_type& a,const value_type& b);
private:
container_type container;
};
If you want to stay with std::pair then you can use the fact that std::set allows you to chose the type of the comparator. However, first I have to explain a bit not to confuse you because the following may appear to be contradicting the above (it does not). The default comparator that set uses is std::less<Key>, that is a type with an operator() that compares two element of type Key, it is something similar (but not exactly) like this:
template <typename Key>
struct less {
bool operator() (const Key& a,const Key& b) const {
return a < b;
}
};
And this is the place where the compiler cannot find a < for your Key type (which is std::pair<T1,T2>). You can use your own comparator:
template <typename T1,typename T2>
struct my_comparator {
bool operator() (const std::pair<T1,T2>& a, const std::pair<T1,T2>& b) const {
return a.first < b.first;
}
};
And then your set is
using container_type = std::set<typename std::pair<T1,T2>,typename my_comparator<T1,T2>>;