Iniitializing an array of unknown size

Every one is telling you to use a vector, but nobody is showing you how to do that. Here's how:

#include <vector>

std::vector<int> target_function(int size)
{
    std::vector<int> v(size);
    v[0] = size;
    return v;
}

int main()
{
    std::vector<int> moves = target_function( my_favorite_int );
}

You can return a pointer instead of an array:

int* moves = target_function();

But don't return a pointer to something you created on the stack as it will go out of scope when the function returns. You can dynamically allocate the array on the heap.

I would suggest not using such hacks. There is std::vector ready for you to use. If you really want to go this way, here's the code that does what you want:

int *allocate(int size)
{
  int *res = new int[size];
  res[0] = size;
  return res;
}


// Prints "Yes, it's 42":
int *myArray = allocate(42);
if (myArray[0] == 42)
  std::cout << "Yes, it's 42!" << std::endl;

Usually you would use a pointer to an dynamically allocated array:

int* target_function() {
  int result* = new int[123];
  result[0] = 123;
  return result;
}

int *moves = target_function();
std::cout << moves[0] << " moves" << std::endl;

That being said, generally it is more practical and less error prone to use a standard library container like std::vector<int> instead. In C++ this is basically always the better choice than a raw array.

Such array cannot be an automatic variable. It needs to be a pointer to a dynamically created array as Mark said.

The short answer is that you can't return an array. You can return a pointer to dynamically allocated memory though:

int* moves = target_function();
// do stuff with moves
delete[] moves;

The target_function() will have to use new to allocate the memory.

Note that this is not ideal from a memory management point of view, because it's easy to forget to call delete[] on the returned array. Instead, consider returning a std::vector<int>.