警告：返回对本地临时对象的引用

Question

Compiling the following:

namespace platform {
  struct event {};
  struct keyboard_event : public event {};

  const platform::event& wait_event()
  {
    return platform::keyboard_event(); 
  }
}

int main(int argc, const char* argv[])
{
  const platform::event& event = platform::wait_event();
  return 0;
}

Yields the following warning with clang

main.cc:7:12: warning: returning reference to local temporary object [-Wreturn-stack-address]
    return platform::keyboard_event(); 
           ^~~~~~~~~~~~~~~~~~~~~~~~~~

Which makes sense but the new wording in C++11 seems to imply that returning a const reference to a value extends it's lifetime until the reference goes out of scope?

The current draft isn't loading for me so I'll cite cppreference.com instead:

The lifetime of a temporary object may be extended by binding to a const lvalue reference or to an rvalue reference (since C++11), see reference initialization for details.

Answer 1

No, "returning a reference" does not magically extend any lifetime.

The only time that lifetime is extended is when a prvalue (or an xvalue referring to a member of a prvalue) is bound to a reference variable , and the lifetime of the prvalue is extended to that of the variable:

struct Foo{};

{
    const auto & r = Foo{};   // Foo object not destroyed at semicolon...
    // ...
}
// ... but is destroyed only here.

Your prvalue is not bound to any variable, and hence no lifetime is extended.

(Also note that non-static class data members don't count as "variables" for this consideration, so you also can't extend lifetimes via constructor initializer lists if your class happens to have reference members.)