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Vendor: Kinaxis

Exam Code: KX3-003 Dumps

Questions and Answers: 104

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NEW QUESTION: 1
Given the following incorrect program:
class MyTask extends RecursiveTask<Integer> {
final int low;
final int high;
static final int THRESHOLD = /* . . . */
MyTask (int low, int high) { this.low = low; this.high = high; }
Integer computeDirectly()/* . . . */
protected void compute() {
if (high - low <= THRESHOLD)
return computeDirectly();
int mid = (low + high) / 2;
invokeAll(new MyTask(low, mid), new MyTask(mid, high));
Which two changes make the program work correctly?
A. The compute () method must be changed to return an Integer result.
B. The MyTask class must be modified to extend RecursiveAction instead of RecursiveTask.
C. The midpoint computation must be altered so that it splits the workload in an optimal manner.
D. Results must be retrieved from the newly created MyTask Instances and combined.
E. The THRESHOLD value must be increased so that the overhead of task creation does not dominate the cost of computation.
F. The computeDirectly () method must be enhanced to fork () newly created tasks.
Answer: A,D
Explanation:
D: the compute() method must return a result.
A: These results must be combined (in the lineinvokeAll(new MyTask(low, mid), new MyTask(mid, high));)
Note 1:A RecursiveTask is a recursive result-bearing ForkJoinTask.
Note 2: The invokeAll(ForkJoinTask<?>... tasks) forks the given tasks, returning when isDone holds for each task or an (unchecked) exception is encountered, in which case the exception is rethrown.
Note 3: Using the fork/join framework is simple. The first step is to write some code that performs a segment of the work. Your code should look similar to this:
if (my portion of the work is small enough) do the work directly else split my work into two pieces invoke the two pieces and wait for the results Wrap this code as a ForkJoinTask subclass, typically as one of its more specialized types RecursiveTask(which can return a result) or RecursiveAction.

NEW QUESTION: 2
次のスキャンのパケットキャプチャを考えます。

次のうちどれがスキャンの出力から推測される可能性が最も高いですか?
A. 192.168.1.55はファイルサーバーです。
B. 192.168.1.55はWebサーバーをホストしています。
C. 192.168.1.55はLinuxサーバーです。
D. 192.168.1.115はWebサーバをホストしています。
Answer: A

NEW QUESTION: 3
Transparent page sharing is employed in a compute system. What occurs if a write to a shared page is attempted?
A. Data is overwritten on the page.
B. A private copy of the page is created.
C. A shared copy of the page is created.
D. The shared page size is extended.
Answer: B


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