How the destructor is called for heap object without delete operator?

Question

Can somebody explain here how destructor is called for object created in heap via new operator when the corresponding delete is not called. In addition since in below code we are catching object via const reference and in destructor we are changing object value (i.e. setting n=0), so how is that possible.

class A
{
    private:
        int n;
    public:
        A()
        {
            n=100;
            std::cout<<"In constructor..."<<std::endl;
        }
        ~A()
        {
            n=0;
            std::cout<<"In destructor..."<<std::endl;
        }
};
int main()
{
  try
  {
      throw *(new A());
  }
  catch(const A& obj)
  {
      std::cout<<"Caught...."<<std::endl;
  }

 return 0;
}

Output from program (running on http://cpp.sh/3jm4x):

In constructor...
Caught....
In destructor...

And you know that the destructor is called because...? Have you tried running in a debugger, setting a breakpoint in the destructor to see where it's called from? — Some programmer dude
Even though it might be elided, I suggest you add a copy-constructor as well. It might help you understand a little more. Or use some static counter that you increase in the constructor and check in the destructor. — Some programmer dude

yuri kilochek yuri kilochek · Accepted Answer · 2018-12-07T06:54:08

throw makes a copy of the object, which is then destroyed automatically after catch. It is this destruction that you observe. The original heap-allocated object is indeed never destroyed.

How the destructor is called for heap object without delete operator?

2 Answers