UiPath UiPath-ADAv1 Q&A - in .pdf

  • UiPath-ADAv1 pdf
  • Exam Code: UiPath-ADAv1
  • Exam Name: UiPath Automation Developer Associate v1 Exam
  • PDF Version: V13.25
  • Q & A: 85 Questions and Answers
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  • Exam Code: UiPath-ADAv1
  • Exam Name: UiPath Automation Developer Associate v1 Exam
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  • Exam Code: UiPath-ADAv1
  • Exam Name: UiPath Automation Developer Associate v1 Exam
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NEW QUESTION: 1
An application is waiting for notification of changes to a tmp directory using the following code statements:
Path dir = Paths.get("tmp")
WatchKey key = dir.register (watcher, ENTRY_CREATE, ENTRY_DELETE, ENTRY_MODIFY) ;
In the tmp directory, the user renames the file testA to testB,
Which statement is true?
A. The events received and the order of events are consistent across all UNIX platforms.
B. The events received and the order of events are platform dependent.
C. The events received and the order of events are consistent across all Microsoft Windows versions.
D. The events received and the order of events are consistent across all platforms.
Answer: D
Explanation:
Most file system implementations have native support for file change notification. The Watch Service API takes advantage of this support where available. However, when a file system does not support this mechanism, the Watch Service will poll the file system, waiting for events.
Note:WatchKey : When a Watchable entity is registered with a WatchService a key which is a WatchKey is generated. Initially the key is in ready state waiting to be notified of any events on the Watchable entity. Once an event occurs the key goes into signaled state and allows to access the events using its pollEvents method. After processing the poll events the key has to be reset by invoking its reset method.
Reference: The Java Tutorials,Watching a Directory for Changes

NEW QUESTION: 2
Which statement about Link Aggregation when implemented on a Cisco Wireless LAN Controller is true?
A. The EtherChannel must be configured in "mode active".
B. One functional physical port is needed to pass client traffic.
C. To pass client traffic, two or more ports must be configured.
D. When enabled, the WLC bandwidth drops to 500 Mbps.
Answer: B

NEW QUESTION: 3
분할 및 비교를 사용하여 숫자 배열을 올바른 순서로 정렬하는 정렬 알고리즘을 개발 중입니다.
피벗보다 작은 항목은 왼쪽으로 가고, 피벗보다 큰 항목은 오른쪽으로 이동하도록 배열을 분할하는 메소드를 작성합니다.
파티셔닝 방법에는 다음과 같은 서명이 있습니다.
- static int 파티션 (int [] numbers, int left,
- int right, int pivotIndex)
다음 알고리즘 중 Partition 메서드를 사용하여 배열을 정렬하는데 사용해야 하는 알고리즘은 무엇입니까?
A. static int [] QuickSort (int [] numbers,
int 왼쪽, int 오른쪽)
{
if (오른쪽> 왼쪽)
{
int pivotIndex = 왼쪽 + (오른쪽 - 왼쪽) / 2;
pivotIndex = 파티션 (
숫자, 왼쪽, 오른쪽, pivotIndex);
QuickSort (
숫자, 왼쪽, pivotIndex);
QuickSort (
숫자, pivotIndex + 1, 오른쪽);
}
숫자를 반환;
}
B. static int [] QuickSort (int [] numbers,
int 왼쪽, int 오른쪽)
{
if (오른쪽> 왼쪽)
{
int pivotIndex = 왼쪽 + (오른쪽 - 왼쪽) / 2;
pivotIndex = 파티션 (
숫자, 왼쪽, 오른쪽, pivotIndex);
QuickSort (
숫자, 왼쪽, pivotIndex - 1);
QuickSort (
숫자, pivotIndex + 1, 오른쪽);
}
숫자를 반환;
}
C. static int [] QuickSort (int [] numbers,
int 왼쪽, int 오른쪽)
{
if (오른쪽> 왼쪽)
{
int pivotIndex = 왼쪽 + (오른쪽 - 왼쪽) / 2;
pivotIndex = 파티션 (
숫자, 왼쪽, 오른쪽, pivotIndex);
QuickSort (
숫자, 왼쪽, pivotIndex - 1);
QuickSort (
숫자, pivotIndex, 오른쪽);
}
숫자를 반환;
}
D. static int [] QuickSort (int [] numbers,
int 왼쪽, int 오른쪽)
{
if (오른쪽> 왼쪽)
{
int pivotIndex = 왼쪽 + (오른쪽 - 왼쪽) / 2;
pivotIndex = 파티션 (
숫자, 왼쪽, 오른쪽, pivotIndex);
QuickSort (
숫자, 왼쪽, pivotIndex + 1);
QuickSort (
숫자, pivotIndex + 1, 오른쪽);
}
숫자를 반환;
}
Answer: B

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