forwarding reference with containers

Question

It's possible that give std::vector<T> e.g to function by forwarding reference? I know that

template<typename T>
void f(const std::vector<T>&&){} // is rvalue
template<typename T>
void f(const std::vector<T>&){} // is lvalue

but how I can make function for lvalue and rvalue (with std::vector as parametr)?

I am a little rusty on this but I believe that a in a function of the form: `void f(T&&)` will only be a "universal reference" if T is an "unknown" type (like a template parameter) — DarthRubik, Jul 03 '17 at 22:59
Are you asking if the parameters of the template functions, as you've shown them, are universal references? (as per Scott Meyers' definition, who I believe invented the term) — Benjamin Lindley, Jul 03 '17 at 23:05
Please use full sentences so that we can tell what you're asking. — Lightness Races in Orbit, Jul 03 '17 at 23:07
Meyers coined the term in his effective modern c++ book. The c++ powers that be now call them forwarding references. — Paul Rooney, Jul 03 '17 at 23:09

score 1 · Accepted Answer · answered Jul 03 '17 at 23:11

1

I think you want

// traits for detecting std::vector
template <typename> struct is_std_vector : std::false_type {};
template <typename T, typename A>
struct is_std_vector<std::vector<T, A>> : std::true_type {};

template<typename T>
std::enable_if_t<is_std_vector<std::decay_t<T>>::value>
f(T&&); // Take any std::vector

answered Jul 03 '17 at 23:11

Jarod42

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I though about this but I had make sure. Thanks. – 21koizyd Jul 03 '17 at 23:13
Okey, I was implemented that but i have problem with return value from function if I want, i tried `typename std::enable_if_t>::value>::type`, but doesn't work in this case. Type of course (`T` - input) – 21koizyd Jul 04 '17 at 02:03

forwarding reference with containers

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