Which two elements occur once in array

This problem was asked by Facebook.

Given an array of integers in which two elements appear exactly once and all other elements appear exactly twice, find the two elements that appear only once.

For example, given the array [2, 4, 6, 8, 10, 2, 6, 10], return 4 and 8. The order does not matter.

Follow-up: Can you do this in linear time and constant space?

My Solution(C++):

#include <iostream>
#include <unordered_map>

// int *occurOnce(int *arr){
//   // linear time linear space
//   int *A = new int[2];
//   std::unordered_map<int, int> H;
//   for (int i=0; i<sizeof(arr); i++){
//     H[arr[i]]++;
//   }
//   int k = 0;
//   for (int i=0; i<sizeof(arr); i++){
//     if (H[arr[i]]==1){
//       A[k] = arr[i];
//       k++;
//     }
//   }
//   return A;


int *occurOnce(int *arr){
  // linear time constant space using bit ops
  int xored = 0;
  for (int i=0; i<sizeof(arr); i++){
    xored = xored^arr[i];
  }
  // now the question boils down to this - xored is the xor of which 2 items
  // in the array
  int right_most_set_bit = xored&(~(xored-1));
  // this is the first bit from right with different values (one has 0 other has 1)
  // then we can form two groups based on whether or not their right_most_set_bit is 1 or 0. each group's xor are the answer
  int x = 0;
  int y = 0;
  for (int i=0; i<sizeof(arr); i++){
    if ((right_most_set_bit&arr[i])==0){
      x^=arr[i];
    }else{
      y^=arr[i];
    }
  }
  int *A = new int[2];
  A[0] = x;
  A[1] = y;
  return A;
}

void test(){
  int arr[] = {2, 4, 6, 8, 10, 2, 6, 10};
  int *ans = occurOnce(arr);
  std::cout << ans[0] << " " << ans[1] << std::endl;
  int arr2[] =  {2, 7, 8, 3, 5, 3, 7, 8};
  int *ans2 = occurOnce(arr2);
  std::cout << ans2[0] << " " << ans2[1] << std::endl;
}

int main(){
  test();
  return 0;
}