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| Certification Vendor: | Salesforce |
|---|---|
| Exam Name: | Salesforce Certified JavaScript Developer I Exam |
| Exam Number: | JS-Dev-101 / JavaScript-Developer-I |
| Real Exam Qty: | 60 scored + up to 5 unscored |
| Available Languages: | Japanese, English |
| Certificate Validity Period: | No expiration; requires regular maintenance via Trailhead modules |
| Related Certifications: | Salesforce Certified Platform Developer I Salesforce Certified Lightning Web Components Specialist |
| Exam Price: | USD 200 (+ applicable taxes) |
| Exam Duration: | 105 minutes |
| Passing Score: | 65% |
| Exam Format: | Multiple-select, Multiple-choice |
| Recommended Training: | Salesforce JavaScript Developer I Certification Prep |
| Exam Registration: | Salesforce Certification Registration |
| Sample Questions: | Salesforce JavaScript-Developer-I Sample Questions |
| Exam Way: | Online proctored or onsite at authorized testing centers |
| Pre Condition: | No formal prerequisites; 1–2 years of JavaScript development experience recommended |
| Official Syllabus URL: | https://trailhead.salesforce.com/credentials/javascript-developer-i |
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NEW QUESTION # 138
Refer to the code:
01 function execute() {
02 return new Promise((resolve, reject) = > reject());
03 }
04 let promise = execute();
05
06 promise
07 .then(() = > console.log( ' Resolved1 ' ))
08 .then(() = > console.log( ' Resolved2 ' ))
09 .then(() = > console.log( ' Resolved3 ' ))
10 .catch(() = > console.log( ' Rejected ' ))
11 .then(() = > console.log( ' Resolved4 ' ));
What is the result when the Promise in the execute function is rejected?
Answer: B
Explanation:
* execute() returns a Promise that immediately calls reject(). So promise starts in a rejected state.
* When a Promise is rejected and you chain .then() calls without rejection handlers, all those .then() callbacks are skipped until a .catch() is encountered:
promise
then(...) // skipped
then(...) // skipped
then(...) // skipped
catch(...) // executed
then(...); // executed after catch
* Execution:
* .then(() = > console.log( ' Resolved1 ' )) is skipped.
* .then(() = > console.log( ' Resolved2 ' )) is skipped.
* .then(() = > console.log( ' Resolved3 ' )) is skipped.
* .catch(() = > console.log( ' Rejected ' )) runs and logs Rejected.
* The .catch() returns a resolved Promise (no explicit return, so undefined), so the next .then() runs:
* .then(() = > console.log( ' Resolved4 ' )) logs Resolved4.
Final output:
Rejected
Resolved4
This matches option D.
NEW QUESTION # 139
Refer to the HTML below:
<div id="main">
<ul>
<li>Leo</li>
<li>Tony</li>
<li>Tiger</li>
</ul>
</div>
Which JavaScript statement results in changing " Tony" to "Mr. T."?
Answer: C
NEW QUESTION # 140
01 function changeValue(obj) {
02 obj.value = obj.value / 2;
03 }
04 const objA = { value: 10 };
05 const objB = objA;
06
07 changeValue(objB);
08 const result = objA.value;
What is the value of result?
Answer: D
Explanation:
* objA is created with:
* { value: 10 }
* objB = objA; Objects in JavaScript are assigned by reference , not copied. So objA and objB refer to the same object in memory.
* The function:
* changeValue(obj) {
* obj.value = obj.value / 2;
* }
When called as changeValue(objB), it updates the value property of the shared object :
value = 10 / 2 # 5
* Therefore:
* objA.value === 5
Because objA and objB refer to the same object, the change is reflected in both.
JavaScript Knowledge References (text-only)
* Objects are stored and passed by reference.
* Mutating an object inside a function mutates the original object.
NEW QUESTION # 141
Universal Containers recently launched its new landing page to host a crowd-funding campaign. The page uses an external library to display some third-party ads. Once the page is fully loaded, it creates more than 50 new HTML items placed randomly inside the DOM, like the one in the code below:
All the elements includes the same ad-library-item class, They are hidden by default, and they are randomly displayed while the user navigates through the page.
Answer: C
NEW QUESTION # 142
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?
Answer: C,D
Explanation:
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 References (no links):
* String() conversion vs .toString()
* String.prototype.concat
* + operator overload: numeric addition vs string concatenation
* Type coercion rules for primitives
NEW QUESTION # 143
......
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