c++ - Perfectly capturing a perfect forwarder (universal reference) in a lambda -

so have perfect forwarder, , want appropriately capture in lambda, such r-values copied in, , l-values captured reference. using std::forward doesn't job, evidenced code:

#include<iostream>  class testclass { public:    testclass() = default;    testclass( const testclass & other ) { std::cout << "copy c" << std::endl; }    testclass & operator=(const testclass & other ) { std::cout << "copy a" << std::endl; } };  template< class t> void testfunc(t && t)    { [test = std::forward<t>(t)](){}(); }  int main() {    testclass x;    std::cout << "please no copy" << std::endl;    testfunc(x);    std::cout << "done" << std::endl;     std::cout << "copy here" << std::endl;    testfunc(testclass());    std::cout << "done" << std::endl; } 

compiling

g++ -std=c++14 main.cpp 

produces output

please no copy copy c done copy here copy c done 

in perfect world, have "copy c" appear in rvalue case, not lvalue case.

my work around use helper function overloaded l- , r- values, wondering if there better way.

cheers!

you may use following:

[test = std::conditional_t<              std::is_lvalue_reference<t>::value,              std::reference_wrapper<std::remove_reference_t<t>>,              t>{std::forward<t>(t)}] 

live demo

but providing helper function seems more readable