How to check if a class has an inherited function at compile-time?

Question

#include <vector>
#include <iostream>
#include <type_traits>

using namespace std;

template<typename Coll>
class has_push_back
{
    using coll_type = decay_t<Coll>;
    using True = char(&)[1];
    using False = char(&)[2];

    template<typename U, void(U::*)(const typename U::value_type&)>
    struct SFINAE {};

    template<typename T>
    static True Test(SFINAE<T, &T::push_back>*);

    template<typename T>
    static False Test(...);

public:
    enum { value = sizeof(Test<coll_type>(nullptr)) == sizeof(True) };
};

class MyColl : public vector<int> {};

int main()
{
    cout << has_push_back<vector<int>>::value << endl;
    cout << has_push_back<MyColl>::value << endl;
}

The program above will output:

1
0

It shows the template has_push_back doesn't work if the function push_back is inherited.

Is there a way to make it work even if it is inheried?

ildjarn · Accepted Answer · 2016-07-23 16:05:23Z

2

template<typename Coll>
struct has_push_back {
    template<
        typename T,
        typename = decltype(
            std::declval<T&>().push_back(std::declval<typename T::value_type>())
        )
    >
    static std::true_type Test(int);

    template<typename T>
    static std::false_type Test(long);

    using type = decltype(Test<Coll>(0));
    static constexpr bool value = type::value;
};

Online Demo

answered Jul 23, 2016 at 16:05

ildjarn

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Comments

Qaz · Accepted Answer · 2016-07-23 16:21:27Z

Here's a solution using void_t, which is standard in C++17 and also comes with additional utilities such as is_detected_exact in the Library Fundamentals v2 TS, taking most of the work out of has_push_back:

template<typename... Ts>
using void_t = void;

template<typename T>
using push_back_test = decltype(std::declval<T>().push_back(std::declval<typename T::const_reference>()));

template<typename T, typename = void>
struct has_push_back : std::false_type {};

template<typename T>
struct has_push_back<T, void_t<push_back_test<T>>> : std::is_same<push_back_test<T>, void> {};

Or with future utilities:

template<typename T>
using push_back_test = decltype(std::declval<T>().push_back(std::declval<typename T::const_reference>()));

template<typename T>
using has_push_back = std::experimental::is_detected_exact<void, push_back_test, T>;

If you'd like to learn about void_t in detail, I suggest checking out Walter Brown's CppCon 2015 talks.

skypjack · Accepted Answer · 2016-07-23 22:39:40Z

For the sake of completeness, I'd like to post another approach not previously mentioned.
This is based on functions' definitions and an alias declaration.
It follows a minimal, working example:

#include <vector>
#include <type_traits>
#include<utility>

using namespace std;

template<typename T, typename... U>
constexpr auto f(int)
-> std::conditional_t<false, decltype(std::declval<T>().push_back(std::declval<U>()...)), std::true_type>;

template<typename, typename...>
constexpr std::false_type f(char);

template<typename T, typename... U>
using has_push_back = decltype(f<T, U...>(0));

class MyColl : public vector<int> {};

int main() {
    static_assert(has_push_back<vector<int>, int>::value, "!");
    static_assert(has_push_back<MyColl, int>::value, "!");
}

Community · Accepted Answer · 2017-05-23 11:44:20Z

According to this answer, your code could look like this:

#include <type_traits>

// Primary template with a static assertion
// for a meaningful error message
// if it ever gets instantiated.
// We could leave it undefined if we didn't care.

template<typename, typename T>
struct has_push_back {
    static_assert(
        std::integral_constant<T, false>::value,
        "Second template parameter needs to be of function type.");
};

// specialization that does the checking

template<typename C, typename Ret, typename... Args>
struct has_push_back<C, Ret(Args...)> {
private:
    template<typename T>
    static constexpr auto check(T*)
    -> typename
        std::is_same<
            decltype( std::declval<T>().push_back( std::declval<Args>()... ) ),
            Ret    // ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
        >::type;  // attempt to call it and see if the return type is correct

    template<typename>
    static constexpr std::false_type check(...);

    typedef decltype(check<C>(0)) type;

public:
    static constexpr bool value = type::value;
};

All credits to @jork

You've probably used code from this answer but it isn't working with inherited functions.

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