/*
 * Copyright 2011 Steven Gribble
 *
 *  This file is the solution to an exercise problem posed during
 *  one of the UW CSE 333 lectures (333exercises).
 *
 *  333exercises is free software: you can redistribute it and/or modify
 *  it under the terms of the GNU General Public License as published by
 *  the Free Software Foundation, either version 3 of the License, or
 *  (at your option) any later version.
 *
 *  333exercises is distributed in the hope that it will be useful,
 *  but WITHOUT ANY WARRANTY; without even the implied warranty of
 *  MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
 *  GNU General Public License for more details.
 *
 *  You should have received a copy of the GNU General Public License
 *  along with 333exercises.  If not, see <http://www.gnu.org/licenses/>.
 */

// Lecture 2 Extra Exercise #1:
//
// Write a function that:
//  - accepts an array of 32-bit unsigned integers, and a length
//  - reverses the elements of the array in place
//  - returns void  (nothing)

#include <stdio.h>     // needed for the definition of NULL, printf
#include <stdint.h>    // needed for the definition of uint32_t
#include <assert.h>    // needed for the definition of assert()

// To be painfully explicit, I'm using the C99 stdint.h header
// and the uint32_t type.  I could have used "unsigned int"
// instead of uint32_t as well, but if you have some assumption
// about the # of bytes in an integer type, it's better to be
// explicit.
void ReverseArray(uint32_t *arr, unsigned int len) {
  int i;
  // check the obvious corner case of no reversal needed
  if (len <= 1)
    return;

  // make sure the caller passed us a valid array
  assert(arr != NULL);

  // do the reversal.  think through the termination
  // condition carefully.  If there are an even number
  // of elements (2, 4, 6, ...), we want to loop through
  // the first half, swapping with the second half, so
  // we want (len/2) iterations.  If there are an odd
  // number of elements, we want to loop through
  // (len-1)/2.  Since we're doing integer division,
  // for odd lens, (len-1/2) == (len/2).  So, it's the
  // same termination condition for both odd and even cases!
  for (i = 0; i < (len/2); i++) {
    uint32_t tmp;

    tmp = arr[i];
    arr[i] = arr[len-i-1];
    arr[len-i-1] = tmp;
  }
}

// here's some code to test our function

void PrintArray(uint32_t *arr, int len) {
  int i;

  for (i = 0; i < len; i++) {
    if (i == 0)
      printf("%u", arr[0]);
    else
      printf(" %u", arr[i]);
  }
  printf("\n");
}

int main(int argc, char **argv) {
  uint32_t arr[3] = {1, 2, 3};
  uint32_t arr2[4] = {1, 2, 3, 4};

  PrintArray(arr, 3);
  ReverseArray(arr, 3);
  PrintArray(arr, 3);

  PrintArray(arr2, 4);
  ReverseArray(arr2, 4);
  PrintArray(arr2, 4);

  return 0;
}