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Aspirants must know the exam topics before they start of preparation. Because it will help them to prepare for the below conceptsSalesforce JavaScript-Developer-I Exam will include the following topics:
1. Variable, Types, and Collection: 23%
Scenario based codingVariables Creation and InitializationJSON object understanding
2. Object, Functions, and Classes: 25%
Implementation of different functionsUnderstanding of different modules of JavascriptScope of variables and their execution flow
3. Browser and Events: 17%
Handling and propagation of eventsDevelopment tools of browsersUnderstanding of browser specific APIs
4. Asynchronous Programming: 13%
Asynchrounous programming different conceptsMonitoring and management of different loops
5. Server Side Javascript: 8%
Implementation of Node.jsUnderstanding of Node.js CLI commands
6. Testing: 7%
Unit Testingeffectiveness of different tests
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NEW QUESTION # 53
Refer to the code below:
01 let country = {
02 get capital() {
03 let city = Number( " London " );
04
05 return {
06 cityString: city.toString(),
07 }
08 }
09 }
Which value can a developer expect when referencing country.capital.cityString?
Answer: B
Explanation:
* In the getter:
* let city = Number( " London " );
The Number() constructor attempts to convert the string " London " into a numeric value.
* " London " is not a valid numeric string. When JavaScript attempts numeric conversion of a non- numeric string:
* Number( " London " ) # NaN
* Next line:
* city.toString()
When city is NaN, calling .toString() yields:
NaN.toString() # " NaN "
* The getter returns:
* {
* cityString: " NaN "
* }
* The question asks: Which value can a developer expect? The multiple-choice options include NaN , but not " NaN " .
Given the available choices, the intended correct answer is B (NaN) because the root cause is the Number( " London " ) conversion returning NaN.
JavaScript Knowledge References (text-only)
* Number(nonNumericString) returns NaN.
* NaN.toString() produces " NaN " .
* Getters compute and return their values each time the property is accessed.
NEW QUESTION # 54
A developer wrote the following code:
01 let X = object.value;
02
03 try {
04 handleObjectValue(X);
05 } catch (error) {
06 handleError(error);
07 }
The developer 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 # 55
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: C
Explanation:
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:
The answer: 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 # 56
Universal Containers (UC) just launched anew landing page, but users complain that the website is slow. A developer found some functions any that might cause this problem. To verify this, the developer decides to execute everything and log the time each of these three suspicious functions consumes.
Which function can the developer use to obtain the time spent by every one of the three functions?
Answer: A
NEW QUESTION # 57
Refer to the code declarations below:
let str1 = ' Java ' ;
let str2 = ' Script ' ;
Which three expressions return the string JavaScript?
Answer: B,D,E
Explanation:
The correct answers are A, B, and D .
The two variables are:
let str1 = ' Java ' ;
let str2 = ' Script ' ;
The goal is to combine them into:
JavaScript
Option A is correct because template literals can insert variables directly into a string:
`${str1}${str2}`
This becomes:
`${ ' Java ' }${ ' Script ' }`
Result:
JavaScript
Option B is correct because strings have a concat() method:
str1.concat(str2);
This joins str2 onto the end of str1.
Result:
JavaScript
Option D is correct because the + operator performs string concatenation when both operands are strings:
str1 + str2;
This becomes:
' Java ' + ' Script '
Result:
JavaScript
The incorrect options:
Option C is not valid JavaScript for joining strings. const is used for declaring constants, not concatenating values.
Option E is incorrect because join() is an array method, not a string method. This would only work with an array, for example:
[ ' Java ' , ' Script ' ].join( ' ' );
But str1 is a string, so:
str1.join(str2)
is invalid.
Therefore, the verified answers are A, B, and D .
NEW QUESTION # 58
......
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