The output of the code will be ____. int i=5; printf(“%d %d %d”, i, i++, ++i); ans. 5 5 7 7 6 6 7 6 5 7 6 7

To determine the output of the provided code, we need to analyze how the variables are evaluated in the printf statement. The key here is to understand how the post-increment (i++) and pre-increment (++i) operators work.

1. Variable Initialization:

   • int i = 5;

2. Function Call: We look at the printf function: printf("%d %d %d", i, i++, ++i);

3. Evaluation Order:

   • The arguments in printf are evaluated from right to left, but the order of evaluation of the expressions is unspecified in C. However, both post-increment and pre-increment will alter the value of i.

4. Detailed Breakdown:

   • i: Initially, i = 5.
   • i++: This returns the current value of i (which is 5) and then increments i by 1, so after this operation, i becomes 6.
   • ++i: Since i is now 6, this increments it to 7 and returns 7.

5. Final Outcome: The values passed to printf, based on the described order:

   • The value for i: 6 (after evaluating i++),
   • The value for i++: 5 (before it increments),
   • The value for ++i: 7 (after it increments).

Based on this interpretation, we find that the printed output will be 7 6 5.

Final Answer

Thus, the output of the code will be 7 6 5.