为什么r值引用不能延长r值寿命？

Question

对于以下代码，我理解为什么在第一个示例中，临时对象的生存期得到延长，但为什么在第二个示例中没有发生该生存期延长的原因：

#include <iostream>

struct A
{
  A()                         { std::cout << "construct A:" << this << "\n"; }
  ~A()                        { std::cout << "destruct A:" << this << "\n"; }
  A & operator () (int i) &   { std::cout << "int&" << this << "\n"; return *this; }
  A && operator () (int i) && { std::cout << "int&&:" << this << "\n"; return std::move(*this); }
};

int main()
{
   {
       std::cout << "\nNOT DANGLING...\n";
       A && a = A()/*(32)(42)*/;
       std::cout << "using...\n";
       a(52)(62);                      // OK: No dangling reference
   }                                   // NORMAL: Temporary destroyed at end of scope
   {
       std::cout << "\nDANGLING...\n";
       A && a = A()(32)(42);           // PROBLEM: Temporary immediately destroyed
       std::cout << "using...\n";
       a(52)(62);                      // ERROR: Dangling reference
   }
   return 0;
}

Compiler Explorer的输出为：

NOT DANGLING...
construct A:0x7ffc7b1f08fe
using...
int&0x7ffc7b1f08fe
int&0x7ffc7b1f08fe
destruct A:0x7ffc7b1f08fe

DANGLING...
construct A:0x7ffc7b1f08ff
int&&0x7ffc7b1f08ff
int&&0x7ffc7b1f08ff
destruct A:0x7ffc7b1f08ff
using...
int&0x7ffc7b1f08ff
int&0x7ffc7b1f08ff

为什么在第二个示例中，临时临时对象被立即销毁了，而我却留下了一个悬空的指针？

（注意：这发生在使用C ++ 11或C ++ 17的GCC和Visual Studio上）