operator==() using template template - c++

Question

EDIT: Prolog: I'm a victim of my own ignorance and also of late-night coding.
I'm writing a templated class using template template. It has an iterator, which means that I need to provide an appropriately templated operator==(). This is where I'm having trouble.
Representative code sample follows:
#include <iostream>
#include <typeinfo>
using namespace std;
namespace detail {
template <typename T> class foo {};
template <typename T> class bar {};
}
template <template<class> class A, template<class> class B>
struct basic_thing {
template <typename T> using target_type = A<B<T>>;
target_type<float> fmember;
target_type<int> imember;
struct iterator {
bool equal (const iterator& other) { return true; }
};
iterator begin () { return iterator{}; }
iterator end () { return iterator{}; }
};
template <template<class> class A, template<class> class B>
bool operator== (const typename basic_thing<A, B>::iterator& lhs, const typename basic_thing<A, B>::iterator& rhs) {
return lhs.equal(rhs);
}
int main ()
{
using Thing = basic_thing<detail::foo, detail::bar>;
Thing t;
cout << typeid(t.fmember).name() << endl;
cout << typeid(t.imember).name() << endl;
bool b = (t.begin() == t.end());
return 0;
}
My goal here is to provide a composable way to define basic_thing::target_type, and this pattern works for that purpose. But, I'm stuck at how to declare operator==() for basic_thing::iterator. Either this isn't very straightforward, or there's something obvious that I'm missing. (Likely the latter.)
g++-7.4.0 with -std=c++11 produces the following:
foo.cc: In function 'int main()':
foo.cc:39:23: error: no match for 'operator==' (operand types are 'basic_thing<detail::foo, detail::bar>::iterator' and 'basic_thing<detail::foo, detail::bar>::iterator')
bool b = (t.begin() == t.end());
~~~~~~~~~~^~~~~~~~~~
foo.cc:27:6: note: candidate: template<template<class> class A, template<class> class B> bool operator==(const typename basic_thing<A, B>::iterator&, const typename basic_thing<A, B>::iterator&)
bool operator== (const typename basic_thing<A, B>::iterator& lhs, const typename basic_thing<A, B>::iterator& rhs) {
^~~~~~~~
foo.cc:27:6: note: template argument deduction/substitution failed:
foo.cc:39:32: note: couldn't deduce template parameter 'template<class> class A'
bool b = (t.begin() == t.end());
^
What are some correct ways to do this? Is it even possible when template templates are involved?

Answer 1

The simpler way is to create it inside the struct directly (as member or friend function):
template <template<class> class A, template<class> class B>
struct basic_thing {
// ...
struct iterator {
bool equal (const iterator& other) { return true; }
bool operator ==(const iterator& rhs) const;
// friend bool operator ==(const iterator& lhs, const iterator& rhs);
};
};
With
template <template<class> class A, template<class> class B>
bool operator== (const typename basic_thing<A, B>::iterator& lhs,
const typename basic_thing<A, B>::iterator& rhs);
A and B are not deducible (on the left of ::).
so only callable the ugly way:
bool b = operator==<detail::foo, detail::bar>(t.begin(), t.begin());