How to deduce the return type of a function object from parameters list?

Question

I\'m trying to write a projection function that could transform a vector into a vector. Here is an example:

auto v =          


        

Answer 1

Option #1:

Basic decltype() usage:

template 
auto select(const std::vector& c, F f)
    -> std::vector
{
    using R = decltype(f(c[0]));
    std::vector v;
    std::transform(std::begin(c), std::end(c), std::back_inserter(v), f);
    return v;
}

Option #2:

Basic std::result_of usage:

template ::type>
std::vector select(const std::vector& c, F f)
{
    std::vector v;
    std::transform(std::begin(c), std::end(c), std::back_inserter(v), f);
    return v;
}

Option #3:

Advanced decltype() usage and perfect-forwarding (see notes*):

template 
auto select(const std::vector& c, F&& f)
    -> std::vector::type&>()(*c.begin()))>::type>
{
    using R = typename std::decay::type&>()(*c.begin()))>::type;
    std::vector v;
    std::transform(std::begin(c), std::end(c)
                 , std::back_inserter(v)
                 , std::forward(f));
    return v;
}

Option #4:

Advanced std::result_of usage and perfect-forwarding (see notes*):

template ::type&(typename std::vector::const_reference)>::type>::type>
std::vector select(const std::vector& c, F&& f)
{
    std::vector v;
    std::transform(std::begin(c), std::end(c)
                 , std::back_inserter(v)
                 , std::forward(f));
    return v;
}

---

* Note: Options #3 and #4 assume that the std::transform algorithm takes a function object by-value, and then uses it as a non-const lvalue. This is why one can see this strange typename std::decay::type& syntax. If the function object is supposed to be called within the select function itself, and the result type is not going to be used as a container's template argument (for the purpose of what the outer-most std::decay is used), then the correct and portable syntax for obtaining the return type is:

/*#3*/ using R = decltype(std::forward(f)(*c.begin()));

/*#4*/ typename R = typename std::result_of::const_reference)>::type