c program to reverse an array

c program to reverse an array :- This program reverses the array elements. For example if a is an array of integers with three elements
such that
a[0] = 1
a[1] = 2
a[2] = 3
then on reversing the array will be
a[0] = 3
a[1] = 2
a[0] = 1
Given below is the c code to reverse an array .

C programming code

#include <stdio.h>
 
int main()
{
   int n, c, d, a[100], b[100];
 
   printf("Enter the number of elements in array\n");
   scanf("%d", &n);
 
   printf("Enter the array elements\n");
 
   for (c = 0; c < n ; c++)
      scanf("%d", &a[c]);
 
   /*
    * Copying elements into array b starting from end of array a
    */
 
   for (c = n - 1, d = 0; c >= 0; c--, d++)
      b[d] = a[c];
 
   /*
    * Copying reversed array into original.
    * Here we are modifying original array, this is optional.
    */
 
   for (c = 0; c < n; c++)
      a[c] = b[c];
 
   printf("Reverse array is\n");
 
   for (c = 0; c < n; c++)
      printf("%d\n", a[c]);
 
   return 0;
}

Output of program:

Reverse array by swapping(without using additional memory)

#include <stdio.h>
 
int main() {
  int array[100], n, c, t, end;
 
  scanf("%d", &n);
  end = n - 1;
 
  for (c = 0; c < n; c++) {
    scanf("%d", &array[c]);
  }
 
  for (c = 0; c < n/2; c++) {
    t          = array[c];
    array[c]   = array[end];
    array[end] = t;
 
    end--;
  }
 
  printf("Reversed array elements are:\n");
 
  for (c = 0; c < n; c++) {
    printf("%d\n", array[c]);
  }
 
  return 0;
}

c program to reverse an array using pointers

#include<stdio.h>
#include<stdlib.h>
 
void reverse_array(int*, int);
 
main()
{
   int n, c, *pointer;
 
   scanf("%d",&n);
 
   pointer = (int*)malloc(sizeof(int)*n);
 
   if( pointer == NULL )
      exit(EXIT_FAILURE);
 
   for ( c = 0 ; c < n ; c++ )
      scanf("%d",(pointer+c));
 
   reverse_array(pointer, n);
 
   printf("Original array on reversal is\n");
 
   for ( c = 0 ; c < n ; c++ )
      printf("%d\n",*(pointer+c));
 
   free(pointer);
 
   return 0;
}
 
void reverse_array(int *pointer, int n)
{
   int *s, c, d;
 
   s = (int*)malloc(sizeof(int)*n);
 
   if( s == NULL )
      exit(EXIT_FAILURE);
 
   for ( c = n - 1, d = 0 ; c >= 0 ; c--, d++ )
      *(s+d) = *(pointer+c);
 
   for ( c = 0 ; c < n ; c++ )
      *(pointer+c) = *(s+c);
 
   free(s);
}