int x = 126, y = 105

do {

if(x > y) x = x - y;

else y = y - x;

} while(x!=y);

printf("%d", x);

The output of the given C code segment is ______.

The correct answer is 21.

Let's trace the execution of this C code segment step-by-step to understand why the output is 21.

The given code implements the subtraction-based version of Euclid's algorithm for finding the Greatest Common Divisor (GCD) of two integers, $x$ and $y$.

Initially, the variables are initialized as follows:
$x=126$
$y=105$

The code uses a do-while loop, which will execute the body first and then check the condition while (x != y).

Iteration 1:
We enter the loop with $x=126$ and $y=105$.
We check the condition: if (x > y). Since 126 is greater than 105, this condition is true.
We update $x$:
$x=x-y=126-105=21$.
At the end of the iteration, we check the loop condition: while (x != y). Since $x=21$ and $y=105$, they are not equal, so the loop continues.

Iteration 2:
We enter the loop with $x=21$ and $y=105$.
We check the condition: if (x > y). Since 21 is not greater than 105, the else branch is executed.
We update $y$:
$y=y-x=105-21=84$.
We check the loop condition: while (x != y). Since $x=21$ and $y=84$ are not equal, the loop continues.

Iteration 3:
We enter the loop with $x=21$ and $y=84$.
Since $x$ is not greater than $y$, we update $y$:
$y=y-x=84-21=63$.
Since $x=21$ and $y=63$ are not equal, the loop continues.

Iteration 4:
We enter the loop with $x=21$ and $y=63$.
Since $x$ is not greater than $y$, we update $y$:
$y=y-x=63-21=42$.
Since $x=21$ and $y=42$ are not equal, the loop continues.

Iteration 5:
We enter the loop with $x=21$ and $y=42$.
Since $x$ is not greater than $y$, we update $y$:
$y=y-x=42-21=21$.
We check the loop condition: while (x != y). Since $x=21$ and $y=21$ are now equal, the loop terminates.

After exiting the loop, the program prints the value of $x$ using printf("%d", x);.
Since the final value of $x$ is 21, the output printed to the console is 21.