Is there a difference in C++ between copy initialization and direct initialization? (C++11 version)

Question

I just saw a similar question, but it's from 5 years ago.

Now we have C++11, introducing new moving semantics which may change how lines in the question above are interpreted. So, how would these lines behave in the new standard?

struct POD {
  double doge; // so plain, much data, wow
}

class CopyCtible {
public:
  CopyCtible();
  CopyCtible( int );
  CopyCtible( CopyCtible const& );
  CopyCtible( CopyCtible&& ) = delete;
  ~CopyCtible();
}

class CopyCtibleEx {
public:
  CopyCtibleEx();
  CopyCtibleEx( int );
  explicit CopyCtibleEx( CopyCtibleEx const& );
  CopyCtibleEx( CopyCtibleEx&& ) = delete;
  ~CopyCtibleEx();
}

class MoveCtible {
public:
  MoveCtible();
  MoveCtible( int );
  MoveCtible( MoveCtible const& ) = delete;
  MoveCtible( MoveCtible&& );
  ~MoveCtible();
}

class BothCtible {
public:
  BothCtible();
  BothCtible( int );
  BothCtible( BothCtible const& );
  BothCtible( BothCtible&& );
  ~BothCtible();
}

POD pod1 = POD();
POD pod2 = pod2;

CopyCtible cc1 = CopyCtible();
CopyCtible cc2 = 10;

CopyCtibleEx cce2 = 20;

MoveCtible mc1 = MoveCtible();
MoveCtible mc2 = 30;

BothCtible bc1 = BothCtible();
BothCtible bc2 = 42;

I intentionally removed less interesting lines (like MoveCtible mc3 = mc1; - oh come on, it's boringly obvious a good compiler will just tell you to lay your keyboard and walk away), but, if you less hardcore folks wish, I may add them too.