CSE 130  Fall 2009

                           Solutions to Ex-1


   1. (a) N (b) Y (c) Y (d) Y (c) N 

   2. (a) 1 (b) 4, % is done first (c) False (d) False
      (a, b) are worth 2 points, and (c, d) are 3 each.

   3. (a) Y (b) When C1 is true and C2 is false.

      (c)  if (C1 && C2) { S1; }
             else if (C1 && !C2) { S2; }
                     // No else for the second if.

   4. (a) Code-I: 5   Code-II: Nothing  (b) None of these
  
      (c) For Code-I, i increases and n decreases after every
      iteration. Initially i <= j, so loop starts. But eventually
      n becomes negative, making j < i. This terminates the loop.
      For Code-II, n < i is always false (i is n-1) for any value
      of n supplied by the user. So this loop will not be done even
      once. So it is terminated by default.

   5. (a) R = a;       // What is left in a is remainder.

      (b) int a, b Q, R;

          for (Q = 0; a >= b; Q++)
             { a = a - b; }
 
          R = a;
      
   6. (a) Either a comma missing between array and a[3], or
      one of two variables should not be there. Compiler will treat
      array as a variable name.

      (b) !== is not correct. This is an operator error. Correct
      operator should be != or just ==.

   7. int main(void)
      { float x; int i;   // Can be on the same line.

        scanf("%f", &x);     // (2+4+2) points
        i = (int)(x + 0.5);	// There are other answers.
        printf("%d", i);

        return 0;
      }

   8. (a) The example was given for an array of size 4. 
      Here size is 10.

      // Initialize array A
      for (i = 0; i < 10; i++ )        // (1+3 points)
          A[i] = (float)(i * i * i)/m;

      // Copy A to B in reverse order.
      for (i = 0; i < 10; i++ )    // (1+3 points)
          B[9 - i] = A[i];         // There are several answers.
   


    
   9.     int flip (int n)      // Header, 1.5 points

           { int i, j, m;       // Local variables. 1 point

             i = n/10;          // Extract left digit. 1 point
             j = n%10;          // Extract right digit 1 point
             m = j*10 + i;      // Flip digits  1.5 points
             return m;          // 1 point
           }

   10. The part of the original program is included.

   int main (void)

     { int m, n, count = 0, i, j;     // Use m, n in your program.

       // Complete nested while loops. Assume m and n have
       // been initialized.

        i = 1;
        while ( i <= n )
             {
               j = 1;                        // 0.5 point
               while (j <= n)                // 1 point
                 { if (i+j == m) count++;    // 2 points
                   j = j+1; }                // 0.5 point
               i = i+1;                      // 0.5 point
              } // Close outer loop          // 0.5 point

        // No need to write a statement for printing count.
         return 0;
      }