Consider the following ANSI C function:
int SimpleFunction(int Y[], int n, int x)
{
int total = Y[0], loopIndex;
for (loopIndex = 1; loopIndex <= n - 1; loopIndex++)
total = x * total + Y[loopIndex];
return total;
}Let \( Z \) be an array of 10 elements with \( Z[i] = 1 \), for all \( i \) such that \( 0 \le i \le 9 \). The value returned by SimpleFunction(\( Z, 10, 2 \)) is \(\underline{\hspace{2cm}}\).
Step 1: Understand initialization.
Initially,
\[
\text{total} = Y[0] = 1
\]
Step 2: Understand loop behavior.
The loop runs from \( \text{loopIndex} = 1 \) to \( 9 \).
In each iteration, the update rule is:
\[
\text{total} = 2 \times \text{total} + 1
\]
Step 3: Observe the recurrence relation.
Starting from \( \text{total}_0 = 1 \), the recurrence generates the sequence:
\[
1,\; 3,\; 7,\; 15,\; 31,\; 63,\; 127,\; 255,\; 511,\; 1023
\]
Step 4: Final value after 9 iterations.
After the final iteration, the value of \text{total} becomes:
\[
1023
\]
% Final Answer
Final Answer: \[ \boxed{1023} \]
Consider the following code:
main() {
int x = 126, y = 105;
{
if (x > y)
x = x - y;
else
y = y - x;
}
while (x != y)
printf("%d", x);
}
Consider the following code:
int a;
int arr[] = {30, 50, 10};
int *ptr = arr[10] + 1;
a = *ptr;
(*ptr)++;
ptr = ptr + 1;
printf("%d", a + arr[1] + *ptr);
Consider the following code:
main() {
int x = 126, y = 105;
{
if (x > y)
x = x - y;
else
y = y - x;
}
while (x != y)
printf("%d", x);
}
Given the following cache parameters:
\[ \begin{array}{|c|c|} \hline \textbf{Tag} & 4 \, \text{bits} \\ \textbf{Index} & 12 \, \text{bits} \\ \textbf{Block Size} & 1 \, \text{byte} \\ \hline \end{array} \]Find the size of the main memory and the size of the cache memory.
Consider the following process information for Shortest Remaining Time First (SRTF) scheduling:
\[ \begin{array}{|c|c|c|} \hline \textbf{Process} & \textbf{Arrival Time (AT)} & \textbf{Burst Time (BT)} \\ \hline P1 & 0 & 10 \\ P2 & 1 & 13 \\ P3 & 2 & 6 \\ P4 & 8 & 9 \\ \hline \end{array} \]Find the turnaround time for each process.