How to provide std::map as an argument to a template template parameter

Question

I have a class that needs to use some sort of map. By default, I want to use std::map, but I also want to give the user the ability to use something different if they want (e.g. std::unordered_map or maybe even a user created one).

So I have code that looks like

#include <map>

template<class Key, template<class, class> class Map = std::map>
class MyClass {
};

int main() {
  MyClass<int> mc;
}

But then, g++ complains

test.cpp:3:61: error: template template argument has different template parameters than its corresponding template template parameter
template<class Key, template<class, class> class Map = std::map>
                                                            ^
test.cpp:8:14: note: while checking a default template argument used here
  MyClass<int> mc;
  ~~~~~~~~~~~^
/Library/Developer/CommandLineTools/usr/bin/../include/c++/v1/map:781:1: note: too many template parameters in template template argument
template <class _Key, class _Tp, class _Compare = less<_Key>,
^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
test.cpp:3:21: note: previous template template parameter is here
template<class Key, template<class, class> class Map = std::map>
                    ^~~~~~~~~~~~~~~~~~~~~~
1 error generated.

So it looks like g++ is unhappy that std::map has default arguments.

Is there a way I can allow Map to be any sort of template that can accept at least two template arguments?

I would prefer to stick with C++98 if I can, but I'm open to C++11.

Answer 1

The problem is that your template template parameter has only two template parameters, as opposed to map, which has four.

template<class Key, template<class, class, class, class> class Map = std::map>
class MyClass {
};

Or

template<class Key, template<class...> class Map = std::map>
class MyClass {
};

Should compile.
However, to avoid such problems, try to take the map type instead, and extract the key type via the corresponding member typedef. E.g.

template <class Map>
class MyClass {
    using key_type = typename Map::key_type;
};

Answer 2

Your code will compile in C++17. A long-standing defect report of the C++ Core Working Group (CWG 150) was resolved (by P0522R0) in time for C++17.

cppreference.com also discuss this here, and include a helpful example:

template<class T> class A { /* ... */ };
template<class T, class U = T> class B { /* ... */ };
template <class ...Types> class C { /* ... */ };

template<template<class> class P> class X { /* ... */ };
X<A> xa; // OK
X<B> xb; // OK in C++17 after CWG 150
         // Error earlier: not an exact match
X<C> xc; // OK in C++17 after CWG 150
         // Error earlier: not an exact match

Testing with my version of GCC (8.3.0), I find that using the -std=c++17 flag will successfully compile your program; while using earlier versions of C++ (e.g. -std=c++14 or -std=c++11) will fail.