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Test time left: 02:46:08Understanding Unary operatorsWhat is the output of the code given below?#include <stdio.h>int main() { int i = 2; i = 3 + 3 * i++; printf("%d", i); return 0;}OptionsCompilation error12189

Question

Test time left: 02:46:08Understanding Unary operatorsWhat is the output of the code given below?#include <stdio.h>int main() { int i = 2; i = 3 + 3 * i++; printf("%d", i); return 0;}OptionsCompilation error12189

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Solution

The output of the given code will be 9.

Here's the step-by-step explanation:

  1. The initial value of i is 2.
  2. The expression 3 + 3 * i++ is evaluated. Here, the i++ is a post-increment operation, which means the current value of i is used in the expression and then i is incremented. So, the expression becomes 3 + 3 * 2 which equals 9.
  3. After the expression is evaluated, i is incremented to 3 due to the i++ operation.
  4. However, the new value of i (which is 3) is not used because it was a post-increment operation. The result of the expression (which is 9) is assigned to i.
  5. So, the final value of i is 9, which is printed out by the printf function.

This problem has been solved

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