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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

Question

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
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Solution

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.

This problem has been solved

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