Error after trying to get unique_ptr element from vector

Question

I read about how to insert a unique_ptr<T> into a vector<unique_ptr<T>>: Why can I not push_back a unique_ptr into a vector?

But how do I get a element back? I made an example:

#include <memory>
#include <vector>

class A {
  public:
    A(int x, int y){
      x_ = x;
      y_ = y;
    }
  private:
    int x_;
    int y_;
};


int main(){
  std::vector<std::unique_ptr<A>> vec_a;
  std::unique_ptr<A> tmp_a = std::unique_ptr<A>(new A(13,32));

  vec_a.push_back(std::move(tmp_a));

  vec_a.push_back(std::unique_ptr<A>(new A(3,2)));

  // error
  std::unique_ptr<A> a = vec_a.at(0);
}

I'm getting this error message:

main.cc: In function ‘int main()’:
main.cc:24:36: error: use of deleted function 
      ‘std::unique_ptr<_Tp, _Dp>::unique_ptr(const std::unique_ptr<_Tp, _Dp>&) 
      [with _Tp = A; _Dp = std::default_delete<A>]’
   std::unique_ptr<A> a = vec_a.at(0);
                                    ^
In file included from /usr/include/c++/4.8/memory:81:0,
                 from main.cc:1:
/usr/include/c++/4.8/bits/unique_ptr.h:273:7: error: declared here
       unique_ptr(const unique_ptr&) = delete;
       ^

What I'm doing wrong? How should I do it? And why does this not work?

Answer 1

unique_ptr cannot be copied or assigned. It needs to be move copied or move assigned. You can achieve this by use of std::move:

std::unique_ptr<A> a = std::move(vec_a.at(0));

The call to std::move essentially allows you to treat the RHS as an rvalue, which can then be used in unique_ptr'a move copy constructor.

Note that after this, you will have to be careful with the first element of vec_a.

If you don't want to take ownership of the pointer away from the vector, simply access the managed pointer directly:

A* a = vec_a.at(0).get();

You have to be careful not to use a after the object it points to has ceased to exist.

Answer 2

It's a unique_ptr, and you're attempting to make a copy. One way to fix is to get the raw pointer instead:

A* a = vec_a.at(0).get();

Keep in mind the vector will still own the pointer, and will still delete it when the vector goes out of scope.