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(" %d", i+k); case 3: printf(" %d", i+k); default: printf(" %d", i+k); } } return 0; }The number of times printf statement is executed is _________.

Let us trace the program step-by-step:

1. Evaluation of j:

   $$j = 2 \times 3 / 4 + 2.0 / 5 + 8 / 5$$

   • 2 * 3 / 4 involves integer arithmetic: $6 / 4 = 1$.

   • 2.0 / 5 involves floating-point arithmetic: $2.0 / 5.0 = 0.4$.

   • 8 / 5 involves integer arithmetic: $8 / 5 = 1$.

   • The expression evaluates to $1 + 0.4 + 1 = 2.4$.

   • Since j is of type int, it is truncated to 2.

2. Evaluation of k:

   • k -= --j; first decrements j from 2 to 1 (pre-decrement), so --j evaluates to 1.

   • k = k - 1 = 0 - 1 = -1.

3. Loop execution (i = 0 to 4): Note that the switch statement has no break statements, so execution falls through from matching cases.

   • i = 0: i + k = -1. Matches default:, executing printf1 time.

   • i = 1: i + k = 0. Matches default:, executing printf1 time.

   • i = 2: i + k = 1. Matches case 1:, falls through to case 2, case 3, and default, executing printf3 times.

   • i = 3: i + k = 2. Matches case 2:, falls through to case 3 and default, executing printf3 times.

   • i = 4: i + k = 3. Matches case 3:, falls through to default, executing printf2 times.

4. Total printf executions:

   $$1 + 1 + 3 + 3 + 2 = 10$$

Thus, the printf statement is executed 10 times.