new [], delete [] complexity

Question

I already know that the new[] operator first allocates memory and then calls the constructor for each element and that the delete[] operator first call

Answer 1

It depends on your exact syntax:

auto x = new unsigned[2];
auto y = new unsigned[2]();
::std::cout << x[0] << "\n" << x[1] << "\n" << y[0] << "\n" << y[1] << "\n";
delete[] x;
delete[] y;

gives the output (on my machine):

3452816845
3452816845
0
0

Because one will be default initialized and the other value initialized.

delete[] on the other hand is even simpler to understand: If your data type has a destructor, it will be called. The built in (and thus POD) types generally do not.