Solution

The correct answer is 27

Concept:

In this problem, we are dealing with a recursive function in C that prints a '*' character based on certain conditions. Understanding the flow of recursive calls and how the parameters change is crucial to solving this problem.

Explanation:

Consider the function in the provided C code:

void Cal(int a, int b) {
    if (b != 1) {
        if (a != 1) {
            printf("*");
            Cal(a / 2, b);
        } else {
            b = b - 1;
            Cal(10, b);
        }
    }
}

Let's trace the function call Cal(10, 10);:

1. Cal(10, 10); - Since b != 1 and a != 1, it prints '*' and calls Cal(5, 10);

2. Cal(5, 10); - Again, b != 1 and a != 1, it prints '*' and calls Cal(2, 10);

3. Cal(2, 10); - Again, b != 1 and a != 1, it prints '*' and calls Cal(1, 10);

4. Cal(1, 10); - Here, a == 1, so it decrements b to 9 and calls Cal(10, 9);

The process repeats with Cal(10, 9); similarly, and continues reducing the value of b by 1 each time a becomes 1, until eventually b becomes 1. For each b value from 10 down to 2, the function makes three recursive calls (dividing a by 2 each time until a == 1) and prints '*' three times.

So, the number of '*' printed can be calculated as: 3 (for b = 10) + 3 (for b = 9) + ... + 3 (for b = 2) = 9 * 3 = 27.

Hence, the correct answer is 27.