How do I enable a function template if a class has a specific member function?

Question

I wrote the following template function, which checks whether an arbitary container contains a specific element:

template<template<class, class...> class container_t, class item_t, class... rest_t>
bool contains(const container_t<item_t, rest_t...> &_container, const item_t &_item) {
    for(const item_t &otherItem : _container) {
        if(otherItem == _item) { return true; }
    }
    return false;
}

This works well for most containers. However for all kinds of sets (and maps) it is sub optimal since there we could use:

template<template<class, class...> class set_t, class item_t, class... rest_t>
bool contains(const set_t<item_t, rest_t...> &_set, const item_t &_item) {
    return _set.count(_item) > 0;
}

Obviously we can't use both templates simultaneously because of ambiguity. Now I am looking for a way to use std::enable_if to enable the to first template if container_t does not provide a count member function and the second template if it does. However I can't figure out how to check for a specif member function (using C++11).

Answer 1

C++20

// helper concept
template <typename Container, typename Item>
concept containing = requires (const Container& container, const Item& item) {
    { container.contains(item) } -> std::convertible_to<bool>;
};

// first overload, using std::ranges::contains from <algorithm>
template<std::ranges::input_range Container, typename Item>
bool contains(const Container &container, const Item &item) {
    return std::ranges::contains(container, item);
}

// if possible, use the .contains() member function directly
template<std::ranges::input_range Container, typename Item>
requires containing<Container, Item>
bool contains(const Container &container, const Item &item) {
    // note: in C++20, the old .count() technique has been obsoleted
    //       by a proper .contains() member function
    return container.contains(item);
}

The second overload will win in overload resolution because it is more constrained.

C++11

In C++11, you can write the following, using tag dispatch:

// Create tags for tag dispatch
struct contains_find_t {};
struct contains_count_t : contains_find_t {};

// fallback with no SFINAE
template<typename Container, typename Item>
bool do_contains(const Container &container, const Item &item, contains_find_t) {
    return std::find(container.begin(), container.end(), item) != container.end();
}

// preferred funtion template, with SFINAE in trailing return type
template<typename Container, typename Item>
auto do_contains(const Container &container, const Item &item, contains_count_t)
  -> decltype(container.count(item) != 0) {
    return container.count(item) != 0;
}

template<typename Container, typename Item>
bool contains(const Container &container, const Item &item) {
    return do_contains(container, item, contains_count_t{});
}

The inheritance hierarchy of contains_find_t and contains_count_t ensures that the second overload is a better match. The decltype-based SFINAE eliminates the second overload if the container does not support count().

Answer 2

C++14 feature, reimplemented:

template<class...>struct voider{using type=void;};
template<class...Ts>using void_t=typename voider<Ts...>::type;

A mini metaprogramming library:

template<class...>struct types{using type=types;};
namespace details {
  template<template<class...>class Z, class types, class=void>
  struct can_apply : std::false_type {};
  template<template<class...>class Z, class...Ts>
  struct can_apply< Z, types<Ts...>, void_t< Z<Ts...> > >:
    std::true_type
  {};
}
template<template<class...>class Z, class...Ts>
using can_apply = details::can_apply<Z,types<Ts...>>;

can_apply< some_template, args... > inherits from true_type iff some_template<args...> is a valid expression (in the immediate context).

Now for your problem:

template<class T, class I>
using dot_count_type = decltype( std::declval<T>().count(std::declval<I>()) );

template<class T, class I>
using has_dot_count = can_apply<dot_count_type, T, I>;

and has_dot_count is a traits class that inherits from true_type iff T.count(I) is a valid expression.

namespace details {
  template<class C, class I>
  bool contains(std::false_type, C const& c, I const& i) {
    for(auto&& x:c) {
      if(x == i) { return true; }
    }
    return false;
  }
  template<class C, class I>
  bool contains(std::true_type, C const& c, I const& i) {
    return c.count(i) != 0;
  }
}
template<class C, class I>
bool contains( C const& c, I const& i ) {
  return details::contains( has_dot_count<C const&,I const&>{}, c, i );
}

which uses tag dispatching instead of SFINAE.

Using find seems like a better idea than .count as an aside. In fact, in one case you should use .find the other find. In both cases you should use using std::end; auto e = end(c);.

As an aside, MSVC 2013 (I don't know about 2015) doesn't handle this kind of SFINAE used above. They call it "expression SFINAE". They have custom extensions to detect the existence of a member function. But that is because they are far from C++11 standard compliant.