CODEWITHSUNDEEP
c++templatessfinae
Victor Martin
Victor Martin

Reputation: 89

SFINAE not telling correctly whether a method exists

I am trying to replicate the (I guess) typical SFINAE example to tell whether a type has a certain method. My code is basically the one found in the accepted answer of a previous popular question on the topic:

#include <iostream>
#include <vector>
#include <set>
#include <map>
#include <list>

template <typename T>
class has_push_back_sfinae {
    typedef int8_t yes;
    typedef int16_t no;

    template <typename C> static constexpr yes test(decltype(&C::push_back));
    template <typename C> static constexpr no test(...);    

 public:
    static constexpr bool value = (sizeof(test<T>(0)) == sizeof(yes));
};

template <typename T>
constexpr bool has_push_back = has_push_back_sfinae<T>::value;

int main() {
    std::cout << std::boolalpha;
    std::cout << has_push_back<int> << std::endl;
    std::cout << has_push_back<std::set<int>> << std::endl;
    std::cout << has_push_back<std::map<int, char>> << std::endl;
    std::cout << has_push_back<std::vector<char>> << std::endl;
    std::cout << has_push_back<std::list<int>> << std::endl;
    std::cout << has_push_back<std::string> << std::endl;
    return 0;
}

Basically, has_push_back<T> is meant to be true if and only if push_back is a method of T. I expected that my output had 3 false lines, followed by 3 true ones. However, this is the actual output:

false
false
false
false
false
true

Am I missing something?

(PS: Is there any more standard or nicer way to write such a class?)

Upvotes: 2

Views: 538

Answers (2)

Patrick Roberts
Patrick Roberts

Reputation: 51946

HolyBlackCat has already written a great answer explaining why the attempted approach didn't work and what more standard C++17 SFINAE should look like. I just wanted to add my two cents by providing an additional approach available in C++20.

Is there any more standard or nicer way to write such a class?

By defining a concept, it's very straight-forward to constrain your types to those where the member method push_back() is well-formed:

template <typename T>
concept has_push_back = requires (T& t, const T::value_type& v) {
    { t.push_back(v) };
};

static_assert(not has_push_back<int>);
static_assert(not has_push_back<std::set<int>>);
static_assert(not has_push_back<std::map<int, char>>);
static_assert(has_push_back<std::vector<char>>);
static_assert(has_push_back<std::list<int>>);
static_assert(has_push_back<std::string>);

Try it on godbolt.org: Demo.

Upvotes: 3

HolyBlackCat
HolyBlackCat

Reputation: 96791

This method only works for non-overloaded functions. If it's overloaded, you can't make a pointer-to-member to it, unless you cast it to a specific pointer type right away.

Such casting is error-prone, so I instead recommend to try to call the function:

/*...*/ constexpr yes test(decltype(void(
    std::declval<C &>().push_back(std::declval<const typename C::value_type &>())), int{}));

IMO, the approach to SFINAE suggested in the linked answer is unnecessarily verbose. Here's what I would do:

template <typename T, typename = void> struct has_push_back_sfinae : std::false_type {};
template <typename T> struct has_push_back_sfinae<T,
    decltype(void(std::declval<T &>().push_back(std::declval<const typename T::value_type &>())))
> : std::true_type {};
template <typename T> inline constexpr bool has_push_back = has_push_back_sfinae<T>::value;

Upvotes: 6

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