GATE | GATE-CS-2015 (Set 3) | Question 65

Consider the following C program:

# include <stdio.h>
int main( )
{
   int i, j, k = 0;
   j = 2 * 3 / 4 + 2.0 / 5 + 8 / 5;
   k  -= --j;
   for (i = 0; i < 5; i++)
   {
      switch(i + k)
      {
         case 1:
         case 2: printf("\n%d", i + k);
         case 3: printf("\n%d", i + k);
         default: printf("\n%d", i + k);
      }
   }
   return 0;
}

The number of times printf statement is executed is __________.

(A) 8
(B) 9
(C) 10
(D) 11


Answer: (C)

Explanation: The following statement makes j = 2

 j = 2 * 3 / 4 + 2.0 / 5 + 8 / 5; 

The following statement makes k = -1.

  
 k  -= --j; 

There is one important thing to note in switch is, there is no break. Let count of printf statements be ‘count’

For i = 0, the value of i+k becomes -1, default block 
           is executed, count = 1.
For i = 1, the value of i+k becomes 0, default block
           is executed, count = 2.
For i = 2, the value of i+k becomes 1, all blocks are 
           executed as there is no break, count = 5
For i = 3, the value of i+k becomes 2, three blocks 
           after case 1: are executed, count = 8
For i = 4, the value of i+k becomes 3, two blocks 
           are executed, count = 10 

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