#include <stdlib.h>
#include <stdio.h>

#define square1( a ) a * a

#define square2( a ) ((a) * (a))

/*
 * Another potential problem with macros is that argument 
 * expressions that are not carefully parenthesized can produce 
 * erroneous results due to operator precedence and binding.
 *
 * For example, 2*square1( 3 + 4 ) expands to
 * 
 * result = 2 * 3 + 4 * 3 + 4, which is interprete by the preprocessor as
 * 
 * result = (2 * 3) + (4 * 3) + 4  
 *
 * but 2*square2( 3 + 4 ) expands to
 * 
 * result = 2 *((3 + 4) * (3 + 4))
 */

int square(int a);

int main(void)
{
  int result;

  printf("Sample 1: using macro: square1( a ) a * a on the input expression '3+4'\n");
  result = 2*square1(3+4);
  printf("Value of result: %d\n", result);
  
  printf("Sample 2: using macro: square2( a ) ((a) * (a)) on the input expression '3+4'\n");
  result = 2*square2(3+4);
  printf("Value of result: %d\n", result);
  
  result = 2*square(3+4);
  printf("Value of result from the function call square(): %d\n", result);

  return EXIT_SUCCESS;
}

int square(int a)
{
  return a*a;
}