BONUS!!! Download part of PassTorrent JS-Dev-101 dumps for free: https://drive.google.com/open?id=1WWJSDl7R6N98r4dplf0azlhdn2U3Cn3x
As we all know, the examination fees about JS-Dev-101 exam test is too expensive, so many IT candidates want to get the most valid and useful JS-Dev-101 study material and expect to pass the actual test at first attempt. PassTorrent provide you with the latest JS-Dev-101 exam prep study material which can ensure you 100% pass. The quality & service of JS-Dev-101 exam dumps will give you a good shopping experience. The quality and quantities are controlled by strict standards. PassTorrent has IT experts handling the latest IT information so as to adjust the outline for the exam dumps at the first time, thus to ensure the Salesforce JS-Dev-101 training exam cram shown front of you is the latest and most relevant.
| Topic | Details |
|---|---|
| Topic 1 |
|
| Topic 2 |
|
| Topic 3 |
|
| Topic 4 |
|
Like other Salesforce examinations, the JS-Dev-101 exam preparation calls for a strong preparation and precise JS-Dev-101 practice material. Finding original and latest 121 exam questions however, is a difficult process. Candidates require assistance finding the JS-Dev-101 updated questions. It will be hard for applicants to pass the Salesforce JS-Dev-101 exam on their first try if Salesforce Certified JavaScript Developer - Multiple Choice questions they have are not real and updated.
NEW QUESTION # 89
What are two unique features of fat-arrow functions compared to normal function definitions?
Answer: A,D
Explanation:
________________________________________
Comprehensive and Detailed Explanation From Exact Extract JavaScript Knowledge Arrow functions have two defining characteristics:
Implicit return when written as a single expression:
const fn = () => 5; // returns 5 automatically
This is unique to arrow functions → A is correct.
Lexical this binding:
Arrow functions do not create their own this.
Instead, they use the this value from the surrounding scope → B is correct.
Incorrect options:
C is false: There is no special argument called parentThis.
D is the opposite of arrow function behavior:
Normal functions create their own this; arrow functions do not.
________________________________________
JavaScript Knowledge Reference (text-only)
Arrow functions have lexical this binding.
Arrow functions support implicit return for single expressions.
NEW QUESTION # 90
A developer executes:
document.cookie;
document.cookie = 'key=John Smith';
What is the behavior?
Answer: C
Explanation:
document.cookie retrieves the cookie string.
Setting document.cookie = 'key=John Smith' adds or updates only that one cookie.
Important details:
Writing to document.cookie does not overwrite all cookies.
The browser does not require URL encoding, though it is recommended. Non-encoded spaces are allowed and automatically handled.
document.cookies does not exist; the correct property is document.cookie.
Therefore, the correct behavior:
Cookies are read.
A new cookie key=John Smith is set.
Existing cookies remain.
JavaScript Knowledge Reference (text-only)
Writing document.cookie appends or updates cookies, not replaces them.
document.cookie is both getter and setter for cookie strings.
NEW QUESTION # 91
A developer wrote the following code:
01 let X = object.value;
02
03 try {
04 handleObjectValue(X);
05 } catch (error) {
06 handleError(error);
07 }
Thedeveloper has a getNextValue function to execute after handleObjectValue(), but does not want to execute getNextValue() if an error occurs.
How can the developer change the code to ensure this behavior?
Answer: A
NEW QUESTION # 92
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: C
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 # 93
Refer to the following code block:
01 let array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11];
02 let output = 0;
03
04 for (let num of array) {
05 if (output > 10) {
06 break;
07 }
08 if (num % 2 == 0) {
09 continue;
10 }
11 output += num;
12 }
What is the value of output after the code executes?
Answer: D
Explanation:
Comprehensive and Detailed Explanation From Exact Extract JavaScript knowledge:
This code uses:
A for...of loop to iterate over values in array.
break to exit the loop entirely when output > 10.
continue to skip even numbers.
It sums only certain numbers into output.
Let's walk through the loop step by step.
Initial values:
array = [1,2,3,4,5,6,7,8,9,10,11]
output = 0
Loop: for (let num of array) { ... }
First iteration: num = 1
Line 05: if (output > 10) → 0 > 10 is false → no break.
Line 08: if (num % 2 == 0) → 1 % 2 == 1, not 0, so false → no continue.
Line 11: output += num → output = 0 + 1 = 1.
Second iteration: num = 2
output > 10 → 1 > 10 is false → no break.
num % 2 == 0 → 2 % 2 == 0, so true → continue.
Because of continue, line 11 is skipped.
output remains 1.
Third iteration: num = 3
output > 10 → 1 > 10 is false.
num % 2 == 0 → 3 % 2 == 1, false → no continue.
output += num → output = 1 + 3 = 4.
Fourth iteration: num = 4
output > 10 → 4 > 10 is false.
num % 2 == 0 → 4 % 2 == 0, true → continue.
Skip sum; output remains 4.
Fifth iteration: num = 5
output > 10 → 4 > 10 is false.
num % 2 == 0 → 5 % 2 == 1, false.
output += num → output = 4 + 5 = 9.
Sixth iteration: num = 6
output > 10 → 9 > 10 is false.
num % 2 == 0 → 6 % 2 == 0, true → continue.
output remains 9.
Seventh iteration: num = 7
output > 10 → 9 > 10 is false.
num % 2 == 0 → 7 % 2 == 1, false.
output += num → output = 9 + 7 = 16.
Eighth iteration would be num = 8, but:
At the top of the loop body, line 05 is checked again:
if (output > 10) → 16 > 10 is true, so break; is executed.
When break runs:
The loop terminates immediately.
No further iterations (for num = 8, 9, 10, 11) are executed.
Therefore, output stays at 16.
Final value of output after the loop ends is 16.
This matches option A.
Why other options do not match:
B . 25: Would require adding more odd numbers (e.g., 9, 11) after 7, but the loop stops early due to output > 10.
C . 11: Would be smaller; the actual sum of 1 + 3 + 5 + 7 until break is 16.
D . 36: Would require summing many more values (e.g., most or all odd numbers up to 11), but again, the break condition stops the loop once output exceeds 10.
So:
Answe r: A
JavaScript knowledge / Study Guide references (concept names only, no links):
for...of loop over arrays
break statement in loops (terminating a loop early)
continue statement in loops (skipping to the next iteration)
Modulo operator % to test even and odd numbers
Step-by-step execution and control flow in loops
NEW QUESTION # 94
......
Because of not having appropriate review methods and review materials, or not grasping the rule of the questions, so many candidates eventually failed to pass the JS-Dev-101 exam even if they have devoted much effort. At this moment, we sincerely recommend our JS-Dev-101 Exam Materials to you, which will be your best companion on the way to preparing for the exam. And with high pass rate as 98% to 100%, you will be bound to pass the exam as long as you choose our JS-Dev-101 praparation questions.
JS-Dev-101 Latest Learning Material: https://www.passtorrent.com/JS-Dev-101-latest-torrent.html
BONUS!!! Download part of PassTorrent JS-Dev-101 dumps for free: https://drive.google.com/open?id=1WWJSDl7R6N98r4dplf0azlhdn2U3Cn3x