Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
1.5k views
in Technique[技术] by (71.8m points)

c++ - How to SFINAE out non-containers parameters

I have a template function that I want to enable only for standard containers (or containers compatible with standard containers, which at least provide a begin() member function). I'm SFINAE-ing out non-containers in the following way:

template<typename Container>
typename Container::value_type 
f(const Container& c,
    typename std::enable_if<
        std::is_same<
            decltype(*c.begin()),
            typename Container::value_type
        >::value
    >::type* = nullptr)
{
    // implementation here
}

The std::is_same and decltype don't look too elegant. Is there any better way of doing this?

PS: I need the SFINAE here because I have a different overload

template<typename Derived>
f(const Eigen::MatrixBase<Derived>& A)

and whenever I try f(some_Eigen_matrix), the Container overload ends up being picked up, then the compiler spits out an error because the type is lacking begin().

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

Using void_t, we can just make a type trait for having begin() and end() (and anything else you might want to check for, like typename T::iterator, you can just keep piling expressions on):

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

template <typename T>
struct is_std_container<T,
    void_t<decltype(std::declval<T&>().begin()),
           decltype(std::declval<T&>().end()),
           typename T::value_type
           >>
    : std::true_type { };

And then just SFINAE on that:

template <typename Container>
typename std::enable_if<
    is_std_container<Container>::value,
    typename Container::value_type
>::type 
f(const Container& c) { .. }

Also, if you really wanted to verify that begin() gives you back a T::iterator (or at least that they're equality comparable), you can do that too:

void_t<
    decltype(begin(std::declval<T&>()) == std::declval<typename T::iterator>())
>

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...