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| Certification Vendor: | Salesforce |
|---|---|
| Exam Name: | Salesforce Certified JavaScript Developer - Multiple Choice |
| Exam Number: | JS-Dev-101 |
| Passing Score: | 65% |
| Exam Format: | Multiple choice, Multiple select |
| Available Languages: | English, Japanese |
| Related Certifications: | Salesforce Certified Developer Salesforce Certified Platform Developer I |
| Certificate Validity Period: | Does not expire; requires maintenance updates |
| Exam Duration: | 105 minutes |
| Real Exam Qty: | 60 scored + up to 5 unscored |
| Exam Price: | USD 200 |
| 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 # 144
Refer to the code below:
```html
<html>
<body>
<div id="logo">Hello Logo!</div>
<button id="test">Click me</button>
</body>
<script>
function printMessage() {
console.log('This is a test message');
}
let el = document.getElementById('test');
el.addEventListener("click", printMessage, false);
</script>
</html>
Answer: A
NEW QUESTION # 145
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 # 146
Referto the following code:
Which two statement could be inserted at line 17 to enable the function call on line 18? Choose 2 answers
Answer: B,D
NEW QUESTION # 147
Refer to the code below:
01 function myFunction(reassign) {
02 let x = 1;
03 var y = 1;
04
05 if (reassign) {
06 let x = 2;
07 var y = 2;
08 console.log(x);
09 console.log(y);
10 }
11
12 console.log(x);
13 console.log(y);
14 }
What is displayed when myFunction(true) is called?
Answer: B
Explanation:
This question tests understanding of let (block scope) and var (function scope) in JavaScript.
Initial declarations in the function:
let x = 1; // line 2
var y = 1; // line 3
Here:
x is declared with let, so it is block-scoped to the function body.
y is declared with var, so it is function-scoped to the entire function.
Inside the if (reassign) block (and since reassign is true, we enter it):
if (reassign) {
let x = 2; // line 6
var y = 2; // line 7
console.log(x); // line 8
console.log(y); // line 9
}
Detailed behavior:
let x = 2; on line 6 creates a new block-scoped variable x that exists only inside the if block. It does not change the outer x declared on line 2.
var y = 2; on line 7 declares y with var again, but var is function-scoped. This effectively reassigns the same y defined on line 3 for the entire function. After this line, y is 2 everywhere in the function.
Now, inside the if block:
console.log(x); (line 8) logs the inner block-scoped x, which is 2.
console.log(y); (line 9) logs y, which is the function-scoped y that was set to 2.
So the first two outputs are:
2
2
After the if block, execution continues:
console.log(x); // line 12
console.log(y); // line 13
Outside the if block:
The block-scoped let x = 2; no longer exists; it was only visible inside the if block.
The outer let x = 1; (line 2) is still in scope and has not been changed.
Thus:
console.log(x); (line 12) logs the outer x, which is still 1.
console.log(y); (line 13) logs y which, due to var y = 2; inside the if, is now 2 for the whole function.
Therefore, when myFunction(true) is called, the output in order is:
2 (inner x in if)
2 (function-scoped y after reassignment)
1 (outer x after if)
2 (function-scoped y remains 2)
This corresponds to:
Answer : B (2 2 1 2)
JavaScript knowledge / study guide reference concepts:
let declarations and block scope
var declarations and function scope
Shadowing of variables with let inside a block
Re-declaration and reassignment of var within a function
Execution order of statements and console output
NEW QUESTION # 148
A developer has a formatName function that takes two arguments, firstName and lastName and returns a string. They want to schedule the function to run once after five seconds.
What is the correct syntax toschedule this function?
Answer: C
NEW QUESTION # 149
......
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