CS256 - Fundamentals of Structured Programming

Displaying exercises/e5/solution/h2.c

#include <stdio.h>

/**
 * Read three integers (say, a, b and c) using scanf.  Do not print a prompt.
 * The inputs will be either zero or one.  Produce the output "pass" or "fail"
 * depending on the values of a, b and c using the following
 * table:
 *
 *   a | b | c | output
 *  -------------------
 *   0 | 0 | 0 | "pass"
 *   0 | 0 | 1 | "fail"
 *   0 | 1 | 0 | "pass"
 *   0 | 1 | 1 | "fail"
 *   1 | 0 | 0 | "pass"
 *   1 | 0 | 1 | "fail"
 *   1 | 1 | 0 | "fail"
 *   1 | 1 | 1 | "pass"
 * 
 * Example input/output:
 * > ./h2
 * 0 1 0
 * pass
 */
int main(void)
{
  int a, b, c;
  
  scanf("%d %d %d", &a, &b, &c);

/*
  // Simple if-else chain method:
  if (a == 0 && b == 0 && c == 0) printf("pass\n");
  else if (a == 0 && b == 1 && c == 0) printf("pass\n");
  else if (a == 1 && b == 0 && c == 0) printf("pass\n");
  else if (a == 1 && b == 1 && c == 1) printf("pass\n");
  else printf("fail\n");
  
  // More concise version that assumes that a, b and c can only be 0 or 1:
  // Recall that 0 is "false" and non-zero is "true".
  if      (!a && !b && !c) printf("pass\n");
  else if (!a &&  b && !c) printf("pass\n");
  else if ( a && !b && !c) printf("pass\n");
  else if ( a &&  b &&  c) printf("pass\n");
  else printf("fail\n");
  
  // Even more concise, combines all the if expressions into one using ||:
  if ((!a && !b && !c) || (!a &&  b && !c) || ( a && !b && !c) || ( a &&  b &&  c))
    printf("pass\n");
  else
    printf("fail\n");
*/

  // Assumes that a, b and c can only ever be 0 or 1:
  if (a == 0) {
    // b has no effect on pass/fail when a == 0:
    if (c == 0) printf("pass\n");
    else printf("fail\n");
  } else {
    // When a == 1, then pass when b and c are the same.
    if (b == c) printf("pass\n");
    else printf("fail\n");
  }

  return 0;
}