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 C code snippet, we need to analyze the expression in the printf statement carefully, taking into consideration the order of evaluation and the behavior of post-increment (i++) and pre-increment (++i).

Step 1: Break Down the Problem

1. We have an integer variable i initialized to 5.
2. In the printf statement, we are printing three values: i, i++, and ++i.
3. We need to evaluate how each of these expressions will contribute to the output, considering the effect of increments on i.

Step 2: Relevant Concepts

• Post-Increment Operator (i++): Returns the current value and then increments i.
• Pre-Increment Operator (++i): Increments i first and then returns the new value.

Step 3: Analysis and Detail

1. Initially, i = 5.
2. The evaluation of i in printf uses the current value of i, so the first value output will be 5.
3. The expression i++ will use the current value of i (which is still 5 for this evaluation), and then increment i to 6. Thus, the second value will be 5.
4. Before evaluating ++i, i is now 6. The expression ++i increments i to 7 and then returns the value 7. Hence, the third value will be 7.

Step 4: Verify and Summarize

Putting these evaluations together, the output of the printf statement will be:

• First value: 5
• Second value: 5
• Third value: 7

Thus, the final output of the code will be:

Final Answer

5 5 7