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| Certification Vendor: | Salesforce |
|---|---|
| Exam Name: | Salesforce Certified JavaScript Developer - Multiple Choice |
| Exam Number: | JS-Dev-101 |
| Exam Price: | USD 200 |
| Available Languages: | English, Japanese |
| Certificate Validity Period: | Does not expire; requires maintenance updates |
| Exam Duration: | 105 minutes |
| Related Certifications: | Salesforce Certified Platform Developer I Salesforce Certified Developer |
| Passing Score: | 65% |
| Real Exam Qty: | 60 scored + up to 5 unscored |
| Exam Format: | Multiple choice, Multiple select |
| Recommended Training: | Trailhead JavaScript Developer I Certification Prep |
| Exam Registration: | Salesforce Certification Registration |
| Sample Questions: | Salesforce JS-Dev-101 Sample Questions |
| Exam Way: | Online proctored or onsite at authorized testing centers |
| Pre Condition: | No formal prerequisites; recommended: strong JavaScript fundamentals, ES6+ experience, web development familiarity |
| Official Syllabus URL: | https://trailheadacademy.salesforce.com/certificate/exam-javascript-dev---JS-Dev-101 |
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NEW QUESTION # 43
A team at Universal Containers works on a big project and uses yarn to manage 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?
Answer: A
Explanation:
In JavaScript server-side development using Node.js, dependency management is typically handled through package managers such as npm or yarn. These tools categorize installed packages into:
dependencies - required for running the application in any environment
devDependencies - required only during development (testing tools, build tools, documentation generators, etc.) When a package is installed using:
yarn add <package> --dev
it is placed under the "devDependencies" section of package.json.
Behavior of Production Mode
Node.js uses the environment variable:
NODE_ENV=production
When this environment variable is set to production, both npm and Yarn follow the standard Node.js convention and skip installing devDependencies. This is done to optimize production builds and reduce deployment size. This is a known and documented behavior in Node.js package management tools.
Therefore, if:
The developer added the date-manipulation library as a dev dependency, and Other team members execute yarn in an environment where NODE_ENV=production is set, then Yarn will not install that dependency because devDependencies are intentionally excluded in production mode.
This explains the behavior described in the question.
Why the Other Options Are Incorrect
Option A:
"YARN_ENV is set to production" is incorrect because Yarn does not use the variable YARN_ENV for dependency installation behavior. Node.js tools use NODE_ENV, not YARN_ENV.
Option B:
This is incorrect because Yarn automatically writes dependencies into package.json. Unlike older npm versions, there is no need for the --save flag.
Option D:
There is no such option as --add. The correct syntax is simply:
yarn add <package>
Missing an option that does not exist cannot be the cause.
JavaScript Knowledge Reference
Node.js uses the environment variable NODE_ENV to determine production or development mode.
Package managers (npm and Yarn) follow the rule that when NODE_ENV=production, only "dependencies" are installed and "devDependencies" are skipped.
Yarn automatically persists installed packages to package.json without requiring --save.
Yarn uses the command yarn add to add dependencies; there is no --add flag.
NEW QUESTION # 44
Refer to the code below:
let inArray = [ [1, 2], [3, 4, 5] ];
Which two statements result in the array [1, 2, 3, 4, 5]?
(With corrected typing errors: usArray → inArray, .. → ....)
Answer: A,B
Explanation:
Comprehensive and Detailed Explanation From Exact Extract JavaScript knowledge:
We start with the array:
let inArray = [ [1, 2], [3, 4, 5] ];
This is an array of two inner arrays:
First element: [1, 2]
Second element: [3, 4, 5]
The desired result is to transform this into a single, flat array:
[1, 2, 3, 4, 5]
This is accomplished by using Array.prototype.concat to concatenate the inner arrays into one new array, optionally combined with the spread syntax (...) or Function.prototype.apply.
Option A: [].concat(...inArray);
Relevant concepts:
Spread syntax (...inArray) expands an iterable into separate arguments.
Array.prototype.concat joins arrays or values into a new array. When you pass arrays as arguments to concat, it flattens them one level into the result.
Step-by-step behavior:
inArray is [ [1, 2], [3, 4, 5] ].
...inArray expands into [1, 2] and [3, 4, 5] as separate arguments.
So [].concat(...inArray) is equivalent to:
[].concat([1, 2], [3, 4, 5]);
concat processes each argument:
For [1, 2], it adds 1 and 2 into the result array.
For [3, 4, 5], it adds 3, 4, and 5 into the result array.
The final outcome is:
[1, 2, 3, 4, 5]
Therefore, Option A correctly produces the desired array.
Option B: [].concat.apply(inArray, []);
Corrected name from usArray to inArray.
Relevant concepts:
Function.prototype.apply(fnThis, argsArray) invokes a function with a specific this value and a list of arguments passed as an array.
Here:
thisArg is inArray.
argsArray is [] (no actual arguments passed).
So the call:
[].concat.apply(inArray, []);
is equivalent to:
Array.prototype.concat.apply(inArray, []);
// i.e. inArray.concat();
Since there are no extra arguments, inArray.concat() simply returns a shallow copy of inArray:
[ [1, 2], [3, 4, 5] ]
This remains an array of arrays and is not flattened into [1, 2, 3, 4, 5]. Therefore, Option B does not produce the required result.
Option C: [].concat::...inArray();
Corrected name from usArray to inArray and .. to ....
This expression uses syntax that is not part of standard JavaScript:
:: (double colon) was proposed in early drafts as a bind operator but is not part of the official ECMAScript standard.
The combination concat::...inArray() is invalid in normal JavaScript engines and cannot be used as a valid way to flatten arrays.
In addition, inArray() would imply calling inArray as a function, but it is an array, which would cause a runtime error.
Hence, Option C is syntactically or semantically invalid in standard JavaScript and does not provide the required result [1, 2, 3, 4, 5].
Option D: [].concat.apply({}, inArray);
Corrected name from usArray to inArray.
Relevant concepts:
Again, Function.prototype.apply is used to call concat with a specific this value and arguments provided as an array.
concat treats its this value as an array-like object but ultimately returns a new array containing concatenated elements.
In this expression:
[].concat.apply({}, inArray);
thisArg is {} (an empty object).
argsArray is inArray, which is [ [1, 2], [3, 4, 5] ].
Using apply, this is interpreted as calling concat like:
Array.prototype.concat.call({}, [1, 2], [3, 4, 5]);
concat then:
Starts from the array-like this (here {} is treated as an empty base).
Takes the first argument [1, 2] and appends its elements 1 and 2 to the result.
Takes the second argument [3, 4, 5] and appends its elements 3, 4, and 5 to the result.
The resulting new array is:
[1, 2, 3, 4, 5]
Therefore, Option D also correctly produces the required array.
Final evaluation of all options:
Option A: Uses spread syntax with concat and correctly flattens one level: [1, 2, 3, 4, 5].
Option B: Equivalent to inArray.concat(), leaving it as [[1, 2], [3, 4, 5]]. Does not flatten the structure.
Option C: Uses non-standard and invalid syntax; not a valid or correct JavaScript solution.
Option D: Uses apply with concat, passing the inner arrays as arguments and flattening one level to [1, 2, 3, 4, 5].
Thus, the two correct statements are:
Reference of JavaScript knowledge documents or Study Guide (concept names only):
Array.prototype.concat (concatenating and one-level flattening behavior) Spread syntax for arrays (...array) Function.prototype.apply (calling a function with a given this value and arguments list) Array flattening by one level using concat with array arguments Distinction between nested arrays and flat arrays in JavaScript
NEW QUESTION # 45
Refer to the code below:
01 function changeValue(param) {
02 param = 5;
03 }
04 let a = 10;
05 let b = a;
06
07 changeValue(b);
08 const result = a + ' - ' + b;
What is the value of result when the code executes?
Answer: B
Explanation:
We must understand pass-by-value for primitives in JavaScript.
Initial values:
let a = 10;
let b = a; // b gets a copy of the value 10
So:
a is 10
b is 10 (independent copy)
Function call:
changeValue(b);
Function definition:
function changeValue(param) {
param = 5;
}
param receives the value of b, which is 10.
Inside the function, param is a local variable.
param = 5; changes only this local copy.
It does not affect b outside the function.
After the function call:
a is still 10.
b is still 10.
Result:
const result = a + ' - ' + b;
a is 10.
b is 10.
String concatenation: '10 - 10'.
So result is:
"10 - 10"
Therefore, the correct option is:
Study Guide Concepts:
Primitive values (numbers, strings, booleans) are passed by value
Function parameters as local variables
String concatenation with +
Difference between mutating references vs primitives
NEW QUESTION # 46
A developer wants to catch any error that countSheep() may throw and pass it to handleError().
Which implementation is correct?
Answer: C
Explanation:
________________________________________
Comprehensive and Detailed Explanation From Exact Extract JavaScript Knowledge Key JavaScript knowledge:
try...catch catches synchronous errors that occur inside the try block.
Errors thrown asynchronously (e.g., inside a timer callback) cannot be caught by surrounding synchronous try...catch unless the try...catch is inside the callback.
Option A places countSheep() inside the callback but the try...catch is outside the asynchronous function.
However, this is the only syntactically valid answer among the provided choices.
Since the question wants the correct structure based on options provided, A is the only complete and valid try...catch.
Options B, C, and D are syntactically invalid:
B: finally always executes but does not catch the error. Also uses an undefined variable e.
C: Does not follow valid JavaScript grammar.
D: The provided option is incomplete and cannot be correct.
Thus, A is the only valid surrounding structure in the provided choices.
________________________________________
JavaScript Knowledge Reference (text-only)
try...catch syntax must be correctly structured.
finally does not catch errors.
try { } catch (e) { } is valid only when complete and correctly ordered.
NEW QUESTION # 47
JavaScript:
01 function Tiger() {
02 this.type = 'Cat';
03 this.size = 'large';
04 }
05
06 let tony = new Tiger();
07 tony.roar = () => {
08 console.log('They\'re great!');
09 };
10
11 function Lion() {
12 this.type = 'Cat';
13 this.size = 'large';
14 }
15
16 let leo = new Lion();
17 // Insert code here
18 leo.roar();
Which two statements could be inserted at line 17 to enable line 18?
Answer: A,C
Explanation:
________________________________________
Comprehensive and Detailed Explanation From Exact Extract JavaScript Knowledge There are two valid ways to ensure leo.roar() exists:
Directly assign a roar function to leo (Option A).
Assigning a property to an instance object creates a new method on that instance.
leo.roar = () => { console.log('They\'re pretty good!'); };
After this, calling leo.roar() is valid.
Copy tony's properties into leo using Object.assign (Option B).
Object.assign(target, source) copies enumerable own properties from the source object into the target object.
Since tony.roar exists, executing:
Object.assign(leo, tony);
copies roar into leo, making leo.roar() valid.
Why the other answers are incorrect:
Option C:
Object.assign(leo, Tiger) copies properties from the function object Tiger, not from Tiger.prototype and not from a Tiger instance. Tiger (the function) has no roar property, so nothing useful is copied.
Option D:
leo.prototype is undefined because leo is an instance, not a constructor function. Only constructor functions have a .prototype property. This line would cause an error.
________________________________________
JavaScript Knowledge Reference (text-only)
Instances have their own properties and do not contain a .prototype property.
Object.assign(target, source) copies own enumerable properties of the source object.
Assigning a function as a property of an object creates a callable method.
NEW QUESTION # 48
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