[Bug libstdc++/48635] New: [C++0x] unique_ptr<T, D&> moves the deleter instead of copying it

[daniel.kruegler at googlemail dot com]

http://gcc.gnu.org/bugzilla/show_bug.cgi?id=48635

           Summary: [C++0x] unique_ptr<T, D&> moves the deleter instead of
                    copying it
           Product: gcc
           Version: 4.7.0
            Status: UNCONFIRMED
          Severity: normal
          Priority: P3
         Component: libstdc++
        AssignedTo: unassigned@gcc.gnu.org
        ReportedBy: daniel.kruegler@googlemail.com


The following program using 4.7.0 20110409 (experimental) should print "copy",
but it prints "move" instead:

//-------------
#include <memory>
#include <utility>
#include <iostream>

struct Deleter {
  Deleter() = default;
  Deleter(const Deleter&) = default;
  Deleter(Deleter&&) = default;
  Deleter& operator=(const Deleter&) {
    std::cout << "copy" << std::endl;
    return *this;
  }
  Deleter& operator=(Deleter&&) {
    std::cout << "move" << std::endl;
    return *this;
  }
  template<class T>
  void operator()(T* p) const
  {
    delete p;
  }
};

int main() {
  Deleter d1, d2;
  std::unique_ptr<int, Deleter&> p1(new int, d1), p2(nullptr, d2);
  p2 = std::move(p1);
}
//-------------

The reason for the failure is, that the library implementation of the
move-assignment operator of std::unique_ptr (and of it's templated variant)
uses std::move to transfer the deleter from the source to the target, but as of
[unique.ptr.single.asgn] p. 2 and p. 6 it shall use std::forward instead. Doing
it correctly will ensure that the deleter is copied and not moved in this case,
which is an intended design aim of std::unique_ptr with deleters that are
lvalue-references.

The necessary fix is to replace the usage of std::move at the two places by
std::forward<deleter_type>.