The problem involves tracing the execution of a C-like code snippet and determining the final values of the variables $a$, $b$, and $c$. The code uses pre-decrement, post-increment operators, and a ternary conditional operator. We are given the initial values of $x$ and $y$, and we need to evaluate the expressions for $a$, $b$, and $c$ in sequence, considering the side effects of the increment/decrement operators.

1. Problem Description

The problem involves tracing the execution of a C-like code snippet and determining the final values of the variables $a$, $b$, and $c$. The code uses pre-decrement, post-increment operators, and a ternary conditional operator. We are given the initial values of $x$ and $y$, and we need to evaluate the expressions for $a$, $b$, and $c$ in sequence, considering the side effects of the increment/decrement operators.

2. Solution Steps

First, we are given the initial values:

$x = 10$

$y = 9$

Next, we evaluate the expression for $a$:

$a = (--x == y++) ? --x : ++y;$

The conditional expression is $--x == y++$.

$--x$ means $x$ is decremented before its value is used. So, $x$ becomes $10 - 1 = 9$.

Then, the expression becomes $9 == y++$. The current value of $y$ is $9$, so $9 == 9$ is true.

$y++$ means the value of $y$ is incremented after its value is used in the comparison. Thus, after the comparison, $y$ becomes $9 + 1 = 10$.

Since the condition is true, the value of $a$ will be $--x$. $x$ is currently $9$.

$--x$ means $x$ is decremented before its value is assigned to $a$. So, $x$ becomes $9 - 1 = 8$, and $a = 8$.

Now we have:

$x = 8$

$y = 10$

$a = 8$

Next, we evaluate the expression for $b$:

$b = x++;$

$x++$ means the value of $x$ is incremented after its value is assigned to $b$. So, $b = 8$ and $x$ becomes $8 + 1 = 9$.

Now we have:

$x = 9$

$y = 10$

$a = 8$

$b = 8$

Finally, we evaluate the expression for $c$:

$c = y;$

The value of $y$ is $10$, so $c = 10$.

Now we have:

$x = 9$

$y = 10$

$a = 8$

$b = 8$

$c = 10$

3. Final Answer

$a = 8$

$b = 8$

$c = 10$