Question

I just wanted to confirm the difference here, take this as an example:

class Gate
{
   public:
           Gate(); //Constructor
           void some_fun();
   private:
           int one, two;
           ptr p1;
           Gate* next;
};
typedef Gate* ptr;

Gate::Gate()
{
  one = 0;
  two = 0;
}

void Gate::some_fun()
{
  p1 = new Gate;
  p1 = p1->next;
  p1 = new Gate();
}

In my example, I have created 2 new nodes of “Gate” and the only difference between them is that the first node does not have the variables “one and two” initialized, while the second one does.

Answer 1

C++ has two classes of types: PODs and non-PODs (“POD” stands for “plain old data” … a somewhat misleading hint).

For non-PODs, there is no difference between new T and new T(). The difference only affects PODs, for which new T doesn’t initialise the memory, whereas new T() will default-initialise it.

So what are PODs? All built-in C++ types (int, bool …) are.

Furthermore, certain user-defined types are as well. Their exact definition is somewhat complicated but for most purposes it’s enough to say that a POD cannot have a custom constructor (as well as certain other functions) and all its data members must themselves be PODs. For more details, refer to the linked FAQ entry.

Since your class isn’t a POD, both operations are identical.