Error when pass std::map as template template argu

Question

I defined a function like this, in which there is a template template class

template<typename Key, typename Value, template <typename, typename> class Map>
    struct ForEachOf {
        void operator()(const Map<Key, Value>& map, std::function<void (Key, Value)> func) {
            for(const auto& pair : map) {
                func(pair.first, pair.second);
            }
        }
    };

std::map<int, string> m { {1, "foo"}, {3, "bar"}};
ForEachOf<int, string, std::map> forEachOf;
    forEachOf(m, [](int key, string value) {
        cout << key << value;
    });

However, above code is not able to compile. the error is like:

error: template template argument has different template parameters
      than its corresponding template template parameter
    ForEachOf<int, string, std::map> forEachOf;
/Applications/Xcode.app/Contents/Developer/Toolchains/XcodeDefault.xctoolchain/usr/bin/../include/c++/v1/__tree:1119:5: note: too many template
      parameters in template template argument
    template <class, class, class, class> friend class _LIBCPP_TYPE_VIS_ONLY map;
note: previous template template parameter is here
    template<typename Key, typename Value, template <typename, typename> class Map>

Then How to pass std::map as the template template parameter here?

Answer 1

You don't really need template template parameter in the first place. You could make it

template<typename Map>
struct ForEachOf {
  void operator()(const Map& map,
      std::function<void (typename Map::key_type, typename Map::mapped_type)> func) const {
            for(const auto& pair : map) {
                func(pair.first, pair.second);
            }
        }
    };

or even

template <typename F>
void operator() (const Map& map, F func) const { ... }

Answer 2

Your problem is that ::std::map does not take only two template parameters.

The solution is simply to add a template that takes only two parameters instead of four:

template<typename key, typename value>
using mymap = std::map<key, value>;

(See it)

Or, alternatively add the missing arguments with their defaults:

template<typename Key, typename Value, template <typename, typename, typename, typename> class Map>
struct ForEachOf {
    void operator()(const Map<Key, Value, ::std::less<Key>, ::std::allocator<std::pair<const Key, T> >>& map, std::function<void (Key, Value)> func) {
        for(const auto& pair : map) {
            func(pair.first, pair.second);
        }
    }
};

(See it)

Which can similarly be written by using a variadic type template:

template<typename Key, typename Value, template <typename...> class Map>
struct ForEachOf {
     void operator()(const Map<Key, Value>& map, std::function<void (Key, Value)> func) {
        for(const auto& pair : map) {
            func(pair.first, pair.second);
        }
    }
};

(See it)

Of course, you could also just create a template function that takes the map and deduces everything:

#include <map>
#include <string>
#include <iostream>

template<typename T, typename F>
void ForEachOf(T&& map, F&& func) {
    for(auto& pair : map) {
        func(pair.first, pair.second);
    }
}

int main(void) {
    std::map<int, std::string> m { {1, "foo"}, {3, "bar"}};
    ForEachOf(m, [](auto key, auto value) {
        ::std::cout << key << value;
    });
}

(See it)