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: unassig...@gcc.gnu.org ReportedBy: daniel.krueg...@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>.