Why can I not push_back a unique_ptr into a vector?

Question

You need to move the unique_ptr:

vec.push_back(std::move(ptr2x));

unique_ptr guarantees that a single unique_ptr container has ownership of the held pointer. This means that you can't make copies of a unique_ptr (because then two unique_ptrs would have ownership), so you can only move it.

Note, however, that your current use of unique_ptr is incorrect. You cannot use it to manage a pointer to a local variable. The lifetime of a local variable is managed automatically: local variables are destroyed when the block ends (e.g., when the function returns, in this case). You need to dynamically allocate the object:

std::unique_ptr<int> ptr(new int(1));

In C++14 we have an even better way to do so:

make_unique<int>(5);

std::unique_ptr has no copy constructor. You create an instance and then ask the std::vector to copy that instance during initialisation.

error: deleted function 'std::unique_ptr<_Tp, _Tp_Deleter>::uniqu
e_ptr(const std::unique_ptr<_Tp, _Tp_Deleter>&) [with _Tp = int, _Tp_D
eleter = std::default_delete<int>, std::unique_ptr<_Tp, _Tp_Deleter> =
 std::unique_ptr<int>]'

The class satisfies the requirements of MoveConstructible and MoveAssignable, but not the requirements of either CopyConstructible or CopyAssignable.

The following works with the new emplace calls.

std::vector< std::unique_ptr< int > > vec;
vec.emplace_back( new int( 1984 ) );

See using unique_ptr with standard library containers for further reading.