C++0x "move from" container

Question

In C++0x, we get an efficiency boost concerning containers with std::move:

SomeExpensiveType x = /* ... */;
vec.push_back(std::move(x));

But I can't find anything going the other way. What I mean is something like this:

SomeExpensiveType x = vec.back(); // copy!
vec.pop_back(); // argh

This is more frequent (the copy-pop) on adapter's like stack. Could something like this exist:

SomeExpensiveType x = vec.move_back(); // move and pop

To avoid a copy? And does this already exist? I couldn't find anything like that in n3000.

I have a feeling I'm missing something painfully obvious (like the needlessness of it), so I am prepared for "ru dum". :3

Answer 1

I might be total wrong here, but isn't what you want just

SomeExpensiveType x = std::move( vec.back() ); vec.pop_back();

Assuming SomeExpensiveType has a move constructor. (and obviously true for your case)

Answer 2

Generally for expensive types I think you'd want to push either a wrapper class or smart pointer into the container instead. This way you are avoiding expensive copies and instead only doing the cheap copies of the smart pointer or the wrapper class. You can also use raw pointers too if you want haha.

class ExpensiveWrapper
{
public:
   ExpensiveWrapper(ExpensiveClass* in) { mPtr = in; }

   // copy constructors here....

private:
   ExpensiveWrapper* mPtr;

};

Answer 3

For completeness (and anyone stumbling on this question without a C++1x compiler), an alternative that already exists today:

SomeExpensiveType x;
std::swap(x, vec.back()); 
vec.pop_back();

It just requires a specialization of std::swap to exist for the element type.

Answer 4

template<class C>
auto pop_back(C& c) -> typename std::decay<decltype(c.back())>::type
{
  auto value (std::move(c.back()));
  c.pop_back();
  return value;  // also uses move semantics, implicitly
  // RVO still applies to reduce the two moves to one
}