In a B+- tree where each node can hold at most four key values, a root to leaf path consists of the following nodes:
\( A = (49, 77, 83, -) \)
\( B = (7, 19, 33, 44) \)
\( C = (20^*, 22^*, 25^*, 26^*) \)
The *-marked keys signify that these are data entries in a leaf. Assume that a pointer between keys \( k_1 \) and \( k_2 \) points to a subtree containing keys in \([ k_1, k_2 )\), and that when a leaf is created, the smallest key in it is copied up into its parent. A record with key value 23 is inserted into the B+- tree. The smallest key value in the parent of the leaf that contains 25* is __________ . (Answer in integer)
Step 1: Identify the Correct Leaf Node - Key 23 is inserted into leaf \( C \), which currently contains \( (20^*, 22^*, 25^*, 26^*) \). - After insertion, \( C \) will contain \( (20^*, 22^*, 23^*, 25^*, 26^*) \).
Step 2: Leaf Node Splitting - Since each node can hold at most 4 keys, the leaf splits into two: - First leaf: \( (20^*, 22^*, 23^*) \) - Second leaf: \( (25^*, 26^*) \) - The smallest key of the second leaf (\( 25 \)) is pushed up into its parent (\( B \)).
Step 3: Identify Parent Update - The updated keys in \( B \) are now \( (7, 19, 25, 33, 44) \). - Since \( B \) also exceeds the allowed 4 keys, it splits into two nodes: - First node: \( (7, 19) \) - Second node: \( (33, 44) \) - The smallest key of the second node (\( 33 \)) is pushed up into \( A \). Thus, the answer is \( 33 \).
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:
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 hierarchical cache system with the following access times:
\[ \begin{array}{|c|c|c|} \hline \textbf{Cache Level} & \textbf{Hit Rate} & \textbf{Access Time} \\ \hline L1 & 90\% & 1 \text{ ns} \\ L2 & 80\% & 10 \text{ ns} \\ L3 & 100\% & 100 \text{ ns} \\ \hline \end{array} \]Find \( T_{avg} \) for hierarchical or simultaneous access.
What is the output of the following C code?
void foo(int *p, int x) {
*p = x;
}
void main() {
int *z;
int a = 20, b = 25;
z = a; // Incorrect: Should be z = a;
foo(z, b);
printf("%d", a);
}
Issue: The statement z = a; is invalid because a is an integer, and z is a pointer.