ADL does not find templated free function - c++

Question

I tried to understand how ADL works, at least its basics, and created following piece of code:
#include <iostream>
#include <string>
#include <vector>
using std::pair;
using std::string;
using std::vector;
namespace My {
using std::to_string;
/** With these forward declarations it works ...
string to_string(char arg);
const string& to_string(const string& str);
template <typename T>
string to_string(const vector<T>& rhs);
template <typename T1, typename T2>
string to_string(const pair<T1, T2>& rhs);
struct A;
string to_string(const A& rhs);
*/
string to_string(char arg)
{
return string(1, arg);
}
const string& to_string(const string& str)
{
return str;
}
template <typename T>
string to_string(const vector<T>& rhs)
{
string str("");
for (const auto& e : rhs) {
str += to_string(e) + " "; //< this fails with `pair<..>`
/// `(to_string)(e)` would fail even with `A`
}
return str;
}
template <typename T1, typename T2>
string to_string(const pair<T1, T2>& rhs)
{
return to_string(rhs.first) + " " + to_string(rhs.second);
}
struct A {
static int counter;
string to_string() const
{
using My::to_string; //< avoid using `A::to_string`
return to_string("<A>[") + to_string(counter++) + to_string("]");
}
};
int A::counter = 0;
string to_string(const A& rhs)
{
return rhs.to_string();
}
}
int main(int /*argc*/, const char* /*argv*/[])
{
using My::to_string;
using My::A;
using std::cout;
using std::endl;
cout << to_string(3.1415) << endl;
cout << to_string(pair<int, char>{5, 'a'}) << endl;
cout << to_string(pair<double, pair<int, int>>{3.14, {1, 2}}) << endl;
cout << to_string(vector<int>{1, 2, 3}) << endl;
cout << to_string(pair<string, vector<int>>{"key", {1, 2, 3}}) << endl;
cout << to_string(pair<string, A>{"key", {}}) << endl;
cout << to_string(vector<A>{{}, {}, {}}) << endl;
/// this will fail to compile
cout << to_string(vector<pair<string, int>>{{"a", 1}, {"b", 2}}) << endl;
return 0;
}
I figured out that, inside of My, using std::to_string will use std:: free function if it exists and will use My:: otherwise. Then, outside of namespace My, it is sufficient to do using My::to_string to cover both cases. Fine so far.
Then I used the same technique inside member function A::to_string to avoid preferring the function itself instead of free functions (the same will apply for any other member function).
Finally, I was a little surprised that to_string(vector<A>) compiles, although A is not forward-declared. That's where ADL kicks in, as I've understood. Disabling it (enclosing to_string into brackets) will cause compilation failure.
Now after the long story, here goes my question: why ADL does not work in this case for templated function, i.e. to_string(pair<T1, T2>)? And more importantly, how to fix it? I would be glad if it was not necessary to perform forward declarations, as in my use case the definition of to_string(vector<T>) is located in some base header file and the latter definitions are in different headers and are not supposed to be known at this time.
EDIT:
I tried to somehow "fake" needed forward declarations by some templates or even with some SFINAE, but it either led to ambiguities or to the same result. Finally, I came up with solution which uses member function to_string (but it could be any other name) of desired class. This requires to always implement this function if compatibility with to_string is desired, which in case of STL containers requires to inherit them and add the member function. But I believe that in this case ADL will never fail.
Here are modified parts of the code:
template <typename T,
typename Fun = decltype(&T::to_string),
typename = std::enable_if_t<
std::is_member_function_pointer<Fun>::value>>
string to_string(const T& rhs)
{
return rhs.to_string();
}
template <typename T>
struct Vector : public vector<T> {
using vector<T>::vector;
string to_string() const
{
using My::to_string; //< avoid using `Vector::to_string`
string str("");
for (const auto& e : *this) {
str += to_string(e) + " ";
}
return str;
}
};
template <typename T1, typename T2>
struct Pair : public pair<T1, T2> {
using pair<T1, T2>::pair;
string to_string() const
{
using My::to_string; //< avoid using `Pair::to_string`
return to_string(this->first) + " " + to_string(this->second);
}
};
However, one has to replace vectors and pairs with Vectors and Pairs. (Free function to_string(A) is not needed any more).
Other solutions, comments?

Answer 1

why ADL does not work in this case for templated function, i.e. to_string(pair<T1, T2>)?
ADL works by inspecting the namespaces associated with the types involved in a given call. It then will consider the appropriate overloads found in those namespaces and select the best one.
to_string(vector<pair<string, int>>{{"a", 1}, {"b", 2}})
This call will select the overload My::to_string(const vector<T>&) and this in turns calls to_string(std::pair<std::string, int>).
ADL inspect then the namespaces associated with std::pair, std::string and int to look for the overload to_string(std::pair<...>). Since no such overload is defined in the namespace std it needs to find the definition previous to the call, but the overload My::to_string(const pair<T1, T2>&) is defined after. That's why you need to forward declare it.
Notice that you need to forward declare it because you also have this:
to_string(pair<string, vector<int>>)
If, on the other hand, you had something like:
to_string(vector<pair<string, My::A>>{{"a", {}}, {"b", {}}})
Then, one of the associated namespaces for the call will be My itself and the overload to_string(std::pair<...>) will be found without the need to forward declare it.