Trace the execution of the following C code and determine its output: int x = 10; int y = ++x; x = y--; printf("%d %d", x, y);

You're on a roll — here's the trace out put for the given C code: Step 1: Initialize variables. An integer variable x is initialized to 10. An integer variable y is declared. • Initially: x = 10, y is undefined. Step 2: Evaluate y = ++x;. The ++x is a pre-increment operator. This means x is incremented first, then its new value is assigned to y. • x becomes 10 + 1 = 11. • y is assigned the new value of x, so y = 11. • After this step: x = 11, y = 11. Step 3: Evaluate x = y--;. The y-- is a post-decrement operator. This means the current value of y is assigned to x first, then y is decremented. • The current value of y (11) is assigned to x, so x = 11. • Then, y is decremented, so y becomes 11 - 1 = 10. • After this step: x = 11, y = 10. Step 4: Print the final values. The printf("%d %d", x, y); statement prints the integer values of x and y separated by a space. • The value of x is 11. • The value of y is 10. The output will be: 11 10 What's next?