C Programming: Understanding Logical Operators and Short-Circuit Evaluation

The output of the code will be:
x=1 y=0
x=2 y=0
\nExplanation:\

• In the first line, the expression !x||y++ is evaluated. x is 1, so !x is false (0). Since the OR operator (||) short-circuits, the y++ part is not executed. So, y remains 0. Therefore, the first print statement outputs x=1 y=0.\
• In the second line, the expression !y&&++x is evaluated. y is 0, so !y is true (1). But, since the AND operator (&&) short-circuits, the ++x part is not executed. So, x remains 1. Therefore, the second print statement outputs x=1 y=0.