Question

I am compiling using g++. This is valid but I want to convert to the C++ operators new() and delete() because this is considered best practice.

  int *scratch = (int *)malloc(size * sizeof(int));
  free(scratch);

What is the equivalent using new and delete? This is my guess.

 int **scratch = new int*[size];
 delete[] scratch;

This is for an array of pointers.

The double star is a reference. It is a pointer to a pointer.

Answer 1

You probably want to do:

std::vector<int> scratch(size);

But for the record, what you tried should be:

int* scratch = new int[size];
delete[] scratch;

Note that if size is known at compile-time, you also simply do:

int scratch[size];

Edit

For an array of pointers to int, the desired syntax is:

int** scratch = new int*[size];
delete[] scratch;

Note that the the pointers inside the array are not refering to allocated memory and that you must allocate every pointed int before being able to use them.

That is, you may do:

int a = 3;
int b = 5;

scratch[0] = &a;
scratch[1] = &b;
scratch[2] = &a;
scratch[3] = new int(5); // Well, I don't see why anyone would like to do that but if you do, don't forget to also call delete on the pointer when you are done with it.

Anyway, unless you have some particular constraints, the most C++ way of doing things is probably still a std::vector<int*> as dealing with pointers of pointers becomes a bit unmaintainable after some levels of indirection.