boost::variant; std::unique_ptr and copy

Question

This Question Determined That a Non-Copyable Type Can\'t Be Used With Boost Variant

Tree class

template 

        

Answer 1

using Mixed = boost::variant< 
  std::unique_ptr<char>,
  std::unique_ptr<short>,
  std::unique_ptr<int>,
  std::unique_ptr<unsigned long>
>;

int main() {    
  auto md = std::unique_ptr<int>(new int(123));
  Mixed mixed = std::move(md);
  std::cout << *boost::get< std::unique_ptr<int> >(mixed) << std::endl;
  return 0;
}

unique_ptr is move-only and can be used in variant. The above example can compile and work (C++11).

Answer 2

Concerning the call to boost::bind(), you should use boost::ref() when passing an object by reference to a function template that accepts the corresponding argument by value, otherwise a copy will be attempted (which results in a compiler error in this case, since the copy constructor is inaccessible):

boost::bind(TreeVisitor(), boost::ref(tree), val, keyIndex);
//                         ^^^^^^^^^^^^^^^^

However, there is a bigger problem here: boost::variant can only hold types which are copy-constructible. From the Boost.Variant online documentation:

The requirements on a bounded type are as follows:

• CopyConstructible [20.1.3].

• Destructor upholds the no-throw exception-safety guarantee.

• Complete at the point of variant template instantiation. (See boost::recursive_wrapper<T> for a type wrapper that accepts incomplete types to enable recursive variant types.)

Every type specified as a template argument to variant must at minimum fulfill the above requirements. [...]