What will be the output of the following C code?

Steps to Solve

1. Initialize the variables

   The code starts by initializing the variables:

   • int x = 3;
   • int y; (uninitialized at this point)
   • float z; (also uninitialized)

2. Assign the value to y

   The next line executes y = x + 3;, which assigns the value of y:

   • Since x is 3, $$ y = 3 + 3 = 6 $$

3. Print the value of y

   Now the first printf executes:

   printf("%d\n", y);
   

   • This will print 6.

4. Update the value of y

   Next, y is updated: $$ y = 3xx + 2*x + 1 $$

   • Substituting x (3): $$ y = 333 + 2*3 + 1 = 27 + 6 + 1 = 34 $$

5. Print the updated value of y

   The second printf executes:

   printf("%d\n", y);
   

   • This will print 34.

6. Calculate z with a multiplication

   Next, we compute z: $$ z = x * 0.3 $$

   • Substituting x (3): $$ z = 3 * 0.3 = 0.9 $$

7. Print the value of z

   The third printf executes:

   printf("%f\n", z);
   

   • This will print 0.900000.

8. Update z based on y and x

   The code updates z: $$ z = y / x $$

   • With y=34 and x=3: $$ z = 34 / 3 \approx 11.3333 $$

9. Print the new value of z

   The fourth printf is executed:

   printf("%f\n", z);
   

   • This will print approximately 11.333333.

10. Cast y to float and update z

    Finally, z is updated to: $$ z = (float) y / x $$

    • Calculating again using y=34 and x=3: $$ z = \frac{34}{3} \approx 11.3333 $$

11. Print the casted new value of z

    The last printf executes:

    printf("%f\n", z);
    

    • This will again print approximately 11.333333.

6
34
0.900000
11.333333
11.333333