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

Consider the following C program:

 `# include ` `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

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 ```