2026 Fast2test 최신 JS-Dev-101 PDF 버전 시험 문제집과 JS-Dev-101 시험 문제 및 답변 무료 공유: https://drive.google.com/open?id=1JPvj2-lvL4VB7RxEynNg909tllWIAPnP
여러분이 다른 사이트에서도Salesforce인증JS-Dev-101시험 관련덤프자료를 보셨을 것입니다 하지만 우리Fast2test의 자료만의 최고의 전문가들이 만들어낸 제일 전면적이고 또 최신 업데이트일 것입니다.우리덤프의 문제와 답으로 여러분은 꼭 한번에Salesforce인증JS-Dev-101시험을 패스하실 수 있습니다.
| 주제 | 소개 |
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| 주제 1 |
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| 주제 2 |
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| 주제 3 |
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우리Fast2test 사이트에Salesforce JS-Dev-101관련자료의 일부 문제와 답 등 문제들을 제공함으로 여러분은 무료로 다운받아 체험해보실 수 있습니다. 여러분은 이것이야 말로 알맞춤이고, 전면적인 여러분이 지금까지 갖고 싶었던 문제집이라는 것을 느끼게 됩니다.
질문 # 95
01 function Animal(size, type) {
02 this.type = type || 'Animal';
03 this.canTalk = false;
04 }
05
06 Animal.prototype.speak = function() {
07 if (this.canTalk) {
08 console.log("It spoke!");
09 }
10 };
11
12 let Pet = function(size, type, name, owner) {
13 Animal.call(this, size, type);
14 this.size = size;
15 this.name = name;
16 this.owner = owner;
17 }
18
19 Pet.prototype = Object.create(Animal.prototype);
20 let pet1 = new Pet();
Given the code above, which three properties are set for pet1?
정답：B,D,E
설명：
When pet1 = new Pet(); is created:
Inside Pet constructor:
Animal.call(this, size, type);
this.size = size;
this.name = name;
this.owner = owner;
Animal.call(this, size, type):
Sets this.type = type || 'Animal' → 'Animal' (because type is undefined).
Sets this.canTalk = false.
Then this.size, this.name, this.owner are set (to undefined since no args passed), but they do exist as properties.
So as own properties, pet1 has: type, canTalk, size, name, owner.
speak is defined on Animal.prototype, so pet1.speak exists by inheritance, but is not an own data property created in the constructor.
From the listed options, three important properties directly set by constructor logic and clearly used in behavior are:
canTalk
name
type
Thus, C, D, E.
질문 # 96
Which three browser specific APIs are available for developers to persist data between page loads?
정답：A,E,F
설명：
A pattern for creating a new scope, not a persistence mechanism.
Variables inside an IIFE are still lost when the page reloads.
Thus, the three persistent browser APIs are:
localStorage
indexedDB
cookies
Explanation:
Comprehensive and Detailed Explanation From Exact Extract JavaScript knowledge:
We want mechanisms that persist data between page loads (i.e., after refresh or navigation), in the browser.
A . localStorage
Part of the Web Storage API.
Data persists across page reloads and browser restarts (until explicitly cleared).
Scoped per origin (protocol + host + port).
This is a correct persistent storage API.
B . indexedDB
A low-level, client-side NoSQL database in the browser.
Stores large amounts of structured data and persists across reloads and sessions.
This is also a correct persistent API.
C . cookies
Small key/value pairs stored by the browser and often sent with HTTP requests.
Can have expiration dates and persist across page loads and sessions.
They are a traditional persistence mechanism in browsers.
So cookies also qualify.
D . global variables
Global JS variables exist only for the life of the current page context.
When the page is refreshed or navigated away, they are lost.
They do not persist between page loads.
질문 # 97
Refer to the following code:
01 class Ship {
02 constructor(size) {
03 this.size = size;
04 }
05 }
06
07 class FishingBoat extends Ship {
08 constructor(size, capacity){
09 //Missing code
10 this.capacity = capacity;
11 }
12 displayCapacity() {
13 console.log('The boat has a capacity of ${this.capacity} people.');
14 }
15 }
16
17 let myBoat = new FishingBoat('medium', 10);
18 myBoat.displayCapacity();
Which statement should be added to line 09 for the code to display
The boat has a capacity of 10 people?
정답：C
설명：
FishingBoat extends Ship, so it is a subclass. In ES6 classes:
When you define a constructor in a subclass, you must call super(...) before accessing this.
super(size) calls the parent class (Ship) constructor, which sets this.size = size.
So the correct constructor is:
class FishingBoat extends Ship {
constructor(size, capacity) {
super(size); // line 09
this.capacity = capacity;
}
displayCapacity() {
console.log(`The boat has a capacity of ${this.capacity} people.`);
}
}
Why others are incorrect:
B . ship.size = size;
ship is not defined; this would cause a ReferenceError.
C . super.size = size;
super is not an instance; you must call super(...) as a function to invoke the parent constructor.
D . this.size = size;
In a subclass constructor, you must call super() before using this, otherwise you get a ReferenceError. Also this bypasses the parent constructor logic.
Relevant concepts: ES6 class inheritance, extends, super() in subclass constructors, this initialization rules.
질문 # 98
Original code:
01 let requestPromise = client.getRequest;
03 requestPromise().then((response) => {
04 handleResponse(response);
05 });
The developer wants to gracefully handle errors from a Promise-based GET request.
Which code modification is correct?
정답：C
설명：
________________________________________
Comprehensive and Detailed Explanation From Exact Extract JavaScript Knowledge Key JavaScript Promise rule:
try...catch does not catch asynchronous errors that occur in Promise chains.
Therefore, the correct way to handle errors in a Promise is:
promise.then(...).catch(...);
Analysis of options:
A and B:
Both wrap asynchronous code in try...catch, which does not catch Promise rejections.
try...catch only catches errors thrown synchronously inside the block.
C:
Correct. A .catch() on the Promise chain handles any error from the Promise returned by requestPromise().
D:
.finally() executes regardless of success or failure, but it does not receive the error object, so it cannot handle the error.
Thus the only valid solution is option C.
________________________________________
JavaScript Knowledge Reference (text-only)
Promise rejections must be handled with .catch().
try...catch only handles synchronous code.
.finally() does not receive a rejection reason.
질문 # 99
Refer to the code snippet:
01 let array = [1, 2, 3, 4, 4, 5, 4, 4];
02 for (let i = 0; i < array.length; i++) {
03 if (array[i] === 4) {
04 array.splice(i, 1);
05 i--;
06 }
07 }
What is the value of array after the code executes?
정답：D
설명：
Comprehensive and Detailed
The loop removes every 4:
Start: [1, 2, 3, 4, 4, 5, 4, 4]
i=0 → 1 (no change)
i=1 → 2 (no change)
i=2 → 3 (no change)
i=3 → 4 → splice removes index 3 → [1,2,3,4,5,4,4], then i-- → 2
Next loop, i=3 → 4 again → splice → [1,2,3,5,4,4], i-- → 2
i=3 → 5 (no change)
i=4 → 4 → splice → [1,2,3,5,4], i-- → 3
i=4 → 4 → splice → [1,2,3,5], i-- → 3
Next i=4, array.length is 4 → loop ends.
All 4s removed, final array: [1, 2, 3, 5].
________________________________________
질문 # 100
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Fast2test의Salesforce인증 JS-Dev-101시험덤프 공부가이드는 시장에서 가장 최신버전이자 최고의 품질을 지닌 시험공부자료입니다.IT업계에 종사중이라면 IT자격증취득을 승진이나 연봉협상의 수단으로 간주하고 자격증취득을 공을 들여야 합니다.회사다니면서 공부까지 하려면 몸이 힘들어 스트레스가 많이 쌓인다는것을 헤아려주는Fast2test가 IT인증자격증에 도전하는데 성공하도록Salesforce인증 JS-Dev-101시험대비덤프를 제공해드립니다.
JS-Dev-101시험문제집: https://kr.fast2test.com/JS-Dev-101-premium-file.html
참고: Fast2test에서 Google Drive로 공유하는 무료 2026 Salesforce JS-Dev-101 시험 문제집이 있습니다: https://drive.google.com/open?id=1JPvj2-lvL4VB7RxEynNg909tllWIAPnP