C++11 - is initialization of a reference not (part of) an expression?

Consider

struct A {
   ~A() {std::cout << "A's dtor called\n";}
};

A f1(const A& a) 
{
   return a;
}

int main()
{
   {
      const A& r = f1(A());
      std::cout << "XXX\n";
   }
   std::cout << "YYY\n";
}

Output:

A's dtor called
XXX
A's dtor called
YYY

Now let

const A& f2(const A& a) 
{
   return a;
}

and consider the output of the same main-function except that f1 is replaced with f2:

Output this time:

A's dtor called
XXX
YYY

Why is that? Shouldn't - in the second case - the temporary A() that is passed through f2 "live" until the entire expression const A& r = f2(A()) has been executed? In that case I'd expect that the lifetime of the temporary should be extended until r dies (as in the first case), but it dies already before cout << "XXX" gets executed. Or is the initialization of r not part of the expression into which the evaluation of f2 is involved, so that A() is already dead before it can be bound to r?