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| Certification Vendor: | Salesforce |
|---|---|
| Exam Name: | Salesforce Certified JavaScript Developer I |
| Exam Number: | JS-Dev-101 |
| Exam Duration: | 105 minutes |
| Related Certifications: | Salesforce Certified JavaScript Developer I Lightning Web Components Specialist Superbadge |
| Available Languages: | English |
| Exam Format: | Multiple Select, Multiple Choice |
| Exam Price: | $200 USD |
| Passing Score: | 65% |
| Real Exam Qty: | 60 (plus up to 5 unscored pilot questions) |
| Certificate Validity Period: | Lifetime (No renewal required) |
| Sample Questions: | Salesforce JS-Dev-101 Sample Questions |
| Exam Way: | Online (Proctored) |
| Pre Condition: | Completion of the Lightning Web Components Specialist Superbadge is required to earn the full credential, though it can be completed before or after the exam. |
| Official Syllabus URL: | https://trailhead.salesforce.com/en/credentials/javascriptdeveloperi |
Tech4ExamはIT認定試験のJS-Dev-101問題集を提供して皆さんを助けるウエブサイトです。Tech4Examは先輩の経験を生かして暦年の試験の材料を編集することを通して、最高のJS-Dev-101問題集を作成しました。問題集に含まれているものは実際試験の問題を全部カバーすることができますから、あなたが一回で成功することを保証できます。
| トピック | 出題範囲 |
|---|---|
| トピック 1 |
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| トピック 2 |
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| トピック 3 |
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質問 # 52
Given two expressions var1 and var2, what are two valid ways to return the concatenation of the two expressions and ensure it is data type string?
正解：C、D
解説：
Comprehensive and Detailed Explanation From JavaScript Knowledge:
We want to guarantee that the result is of type string regardless of what var1 and var2 are (numbers, booleans, etc.).
Option A:
String(var1).concat(var2)
String(var1) converts var1 explicitly to a string.
.concat(var2) calls String.prototype.concat, which:
Converts var2 to string if needed.
Returns a new string that is the concatenation of the two.
Result is always a string.
So A is valid.
Option B:
String.concat(var1 + var2)
String.concat (as a static method) does not exist. concat is an instance method on string objects, not on the String constructor.
var1 + var2 might be a number (if both numeric) and used incorrectly here.
So B is invalid.
Option C:
var1 + var2
+ can mean:
String concatenation (if at least one operand is a string).
Numeric addition (if both operands are numbers).
If var1 and var2 are both numbers, var1 + var2 will be a number, not a string.
This does not ensure the result is always a string.
So C is not guaranteed to satisfy the requirement.
Option D:
var1.toString() + var2.toString()
var1.toString() converts var1 to a string (for most primitive/object types).
var2.toString() does the same.
Now both operands of + are strings, so the result is string concatenation.
Result is guaranteed to be a string.
So D is valid.
Therefore, the two valid ways are A and D.
Study Guide / Concept Reference (no links):
String() conversion vs .toString()
String.prototype.concat
+ operator overload: numeric addition vs string concatenation
Type coercion rules for primitives
________________________________________
質問 # 53
Corrected code:
function Person() {
this.firstName = "John";
}
Person.prototype = {
job: x => "Developer"
};
const myFather = new Person();
const result = myFather.firstName + " " + myFather.job();
What is the value of result after line 10 executes?
正解：C
解説：
The verified answer is A.
This constructor function runs when new Person() is called:
function Person() {
this.firstName = "John";
}
So this line:
const myFather = new Person();
creates an object like this:
{
firstName: "John"
}
Then the prototype is assigned a method named job:
Person.prototype = {
job: x => "Developer"
};
Because myFather is created from Person, it can access properties and methods from Person.prototype.
So:
myFather.firstName
returns:
"John"
And:
myFather.job()
returns:
"Developer"
Therefore:
const result = myFather.firstName + " " + myFather.job();
becomes:
const result = "John" + " " + "Developer";
Final value:
"John Developer"
Option B is incorrect because job() returns "Developer", not undefined.
Option C is incorrect because job exists on Person.prototype, so it is callable.
Option D is incorrect because firstName is assigned inside the constructor.
Therefore, the verified answer is A.
質問 # 54
Refer to the code:
01 const exec = (item, delay) =>
02 new Promise(resolve => setTimeout(() => resolve(item), delay));
03
04 async function runParallel() {
05 const [result1, result2, result3] = await Promise.all(
06 [exec('x', '100'), exec('y', '500'), exec('z', '100')]
07 );
08 return `parallel is done: ${result1}${result2}${result3}`;
09 }
Which two statements correctly execute runParallel()?
正解：A、D
解説：
________________________________________
Comprehensive and Detailed Explanation From Exact Extract JavaScript Knowledge Facts about JavaScript Promises:
runParallel() is declared async, which means it always returns a Promise.
Promises are consumed using .then(), .catch(), and .finally().
There is no .done() method in native JavaScript Promises.
The async keyword cannot be placed before a function call (option A is invalid syntax).
Analysis of each option:
A . async runParallel().then(data);
Invalid syntax. async cannot prefix an expression.
B . Valid. Calls the function and attaches a .then() handler.
C . Invalid. .done() is not part of JavaScript Promise API.
D . Valid. Calls runParallel() and chains .then().
Therefore the correct answers are B and D.
________________________________________
JavaScript Knowledge Reference (text-only)
async functions return Promises.
Promises use .then() to retrieve resolved values.
.done() is not part of the standard Promise interface.
質問 # 55
A team at Universal Containers works on a big project and uses Yarn to deal with the project's dependencies. A developer added a dependency to manipulate dates and pushed the updates to the remote repository. The rest of the team complains that the dependency does not get downloaded when they execute yarn.
What could be the reason for this?
正解：B
解説：
If the dependency was added as a dev dependency (yarn add --dev), it goes into devDependencies.
When NODE_ENV=production, Yarn (and npm) typically skip devDependencies on install.
Thus the package doesn't get installed for other team members in a production environment.
YARN_ENV is not a standard variable, --save is npm syntax (and Yarn saves by default), and --add is not how Yarn works (the command itself is yarn add).
________________________________________
質問 # 56
static delay = async delay => {
return new Promise(resolve => {
setTimeout(resolve, delay);
});
};
static asyncCall = async () => {
await delay(1000);
console.log(1);
};
console.log(2);
asyncCall();
console.log(3);
Assume delay and asyncCall are in scope as functions.
What is logged to the console?
正解：B
解説：
The correct answer is D, because JavaScript executes synchronous code first, while the code after await runs later after the Promise resolves.
The execution order is:
console.log(2);
This runs first, so JavaScript logs:
2
Then this line runs:
asyncCall();
The asyncCall() function starts executing. Inside it, JavaScript reaches:
await delay(1000);
The delay(1000) function returns a Promise that resolves after 1000 milliseconds:
return new Promise(resolve => {
setTimeout(resolve, delay);
});
Because await pauses the rest of the asyncCall() function, this line does not run immediately:
console.log(1);
Instead, JavaScript continues executing the remaining synchronous code outside the async function:
console.log(3);
So JavaScript logs:
3
After approximately 1000 milliseconds, the Promise returned by delay(1000) resolves. Then the paused async function continues and runs:
console.log(1);
So JavaScript logs:
1
Final console output:
2
3
1
Important JavaScript concepts involved:
An async function always returns a Promise.
The await keyword pauses execution only inside the async function where it appears.
Code outside the async function continues running normally.
setTimeout() schedules its callback to run later, after the current synchronous code has finished.
Therefore, the verified answer is D. 2 3 1.
質問 # 57
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