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

To determine the output of the given code snippet, we should analyze the printf function's arguments and the behavior of the post-increment (i++) and pre-increment (++i) operators in C.

Step 1: Understand Variable Initialization

• We initialize int i = 5;.

Step 2: Analyze the printf Function

• The statement to evaluate is:

  printf("%d %d %d", i, i++, ++i);
  

• The printf function is using three arguments:
  • The value of i
  • The value of i after the post-increment
  • The value of i after the pre-increment

Step 3: Determine the Order of Evaluation

• In C, the order of evaluation for function arguments is unspecified. So, we need to understand what happens step by step:
  1. The current value of i is 5 when evaluated first.
  2. Post-increment (i++) will use the current value of i (which is still 5 in this step) and then increment it. Therefore, this part evaluates to 5, but i becomes 6.
  3. Pre-increment (++i) will then increment i (now 6) to 7, but it uses the incremented value. So, this part evaluates to 7.

Step 4: Collect Results

So we have:

• First argument: i = 5
• Second argument: i++ = 5 (but now i becomes 6)
• Third argument: ++i = 7 (after incrementing 6)

Step 5: Prepare the Output

Putting it all together, the final output will be:

5 5 7

Final Answer

The output of the code will be 5 5 7.