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Consider:

std::tuple<int , const A&> func (const A& a) 
{
  return std::make_tuple( 0 , std::ref(a) );
}

Is the std::ref required for writing correct and portable code? (It compiles fine without it)

Background:

If I remove std::ref my code builds fine without any warnings (g++-4.6 -Wall), but doesn't run correctly.

In case of interest the definition of A:

struct A {
  std::array<int,2> vec;
  typedef int type_t;

  template<typename... OPs,typename... VALs>
  A& operator=(const std::pair< std::tuple<VALs...> , std::tuple<OPs...> >& e) {
    for( int i = 0 ; i < vec.size() ; ++i ) {
      vec[i] = eval( extract(i,e.first) , e.second );
    }
  }
};
See Question&Answers more detail:os

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1 Answer

std::ref does not make a reference, so in your code sample it doesn't do what you expect. std::ref creates an object that behaves similarly to a reference. It may be useful, for example, when you want to instantiate a functor, and pass a reference-like version of it to a standard library algorithm. Since algorithms take functors by value, you can use std::ref to wrap the functor.


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