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| Section | Objectives |
|---|---|
| Topic 1: Asynchronous Programming | - API Calls and Data Fetching - Callbacks, Promises, and Async/Await |
| Topic 2: Object-Oriented JavaScript | - Objects and Prototypes - Classes and Inheritance |
| Topic 3: Salesforce Platform Integration | - Working with Salesforce Data Models - Lightning Web Components (LWC) Basics |
| Topic 4: Browser and DOM Interaction | - DOM Manipulation - Events and Event Handling |
| Topic 5: JavaScript Fundamentals | - Control Flow and Error Handling - Variables, Data Types, and Operators - Functions and Scope |
>> JS-Dev-101 Intereactive Testing Engine <<
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NEW QUESTION # 124
A developer has a fizzbuzz function that, when passed in a number, returns the following:
'fizz' if the number is divisible by 3.
'buzz' if the number is divisible by 5.
'fizzbuzz' if the number is divisible by both 3 and 5.
Empty string '' if the number is divisible by neither 3 nor 5.
Which two test cases properly test scenarios for the fizzbuzz function?
Answer: C,D
Explanation:
Comprehensive and Detailed Explanation From Exact Extract JavaScript knowledge:
First, recall the fizzbuzz rules:
If n divisible by 3 and not by 5 → return 'fizz'.
If n divisible by 5 and not by 3 → return 'buzz'.
If n divisible by both 3 and 5 → return 'fizzbuzz'.
Otherwise → return '' (empty string).
Evaluate each test:
Option A:
let res = fizzbuzz(true);
console.assert(res === '');
In JavaScript, when using arithmetic operations with true, it is coerced to the number 1. A typical implementation of fizzbuzz would treat the argument as a number and compute:
true % 3 → 1 % 3 → 1
true % 5 → 1 % 5 → 1
So true is effectively treated as 1, which is divisible by neither 3 nor 5. The function should return ''. The assertion res === '' will pass.
This test covers the scenario "neither divisible by 3 nor 5".
Option B:
let res = fizzbuzz(3);
console.assert(res === '');
3 is divisible by 3 but not by 5.
According to fizzbuzz rules, fizzbuzz(3) should return 'fizz'.
This test expects '', so it is incorrect; the assertion would fail in a correct implementation.
Option C:
let res = fizzbuzz(5);
console.assert(res === 'fizz');
5 is divisible by 5 but not by 3.
According to fizzbuzz rules, fizzbuzz(5) should return 'buzz'.
This test expects 'fizz', so it is incorrect.
Option D:
let res = fizzbuzz(15);
console.assert(res === 'fizzbuzz');
15 is divisible by both 3 and 5.
According to fizzbuzz rules, fizzbuzz(15) should return 'fizzbuzz'.
This assertion correctly matches the expected result, so it is a proper test case.
Thus, the two test cases that correctly test valid fizzbuzz behavior are:
Study Guide / Concept Reference (no links):
Modulo operator (%) for divisibility checks
Truthy/number coercion of boolean true in arithmetic (Number(true) === 1) Designing unit tests with console.assert Typical fizzbuzz logic structure and expected outputs
________________________________________
NEW QUESTION # 125
Refer to the following code block:
What is the console output?
Answer: D
NEW QUESTION # 126
A developer executes:
document.cookie;
document.cookie = 'key=John Smith';
What is the behavior?
Answer: B
Explanation:
________________________________________
Comprehensive and Detailed Explanation From Exact Extract JavaScript Knowledge 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 # 127
Correct implementation of try...catch for countsDeep():
Answer: B
Explanation:
Errors thrown inside a setTimeout callback are asynchronous.
A try...catch around setTimeout (options C and D) can't catch errors thrown inside the callback later.
You must put the try...catch inside the timeout callback (option B).
A is nonsense syntax, D also calls countSheep() instead of countsDeep().
NEW QUESTION # 128
Correct implementation of try...catch for countsDeep():
Answer: B
Explanation:
The correct answer is B because countsDeep() is executed inside the callback function passed to setTimeout(), and the try...catch block is also placed inside that same callback.
In JavaScript, setTimeout() schedules a function to run later. The outer code finishes first, and the callback runs asynchronously after the delay. Because of this, a try...catch block placed outside setTimeout() cannot catch errors thrown later inside the callback.
Correct logic:
setTimeout(function() {
try {
countsDeep();
} catch (e) {
handleError(e);
}
}, 1000);
Here, when countsDeep() runs after 1000 milliseconds, any error thrown by countsDeep() happens inside the try block. Therefore, the catch (e) block can catch that error and pass it to handleError(e).
Why the other options are incorrect:
A is incorrect because the syntax is invalid JavaScript. A valid try...catch structure must be:
try {
// code
} catch (e) {
// handle error
}
Option A incorrectly writes:
} handleError (e){
catch(e);
}
That is not valid try...catch syntax.
C is incorrect because the try...catch surrounds only the call to setTimeout(), not the later execution of countsDeep(). If countsDeep() throws an error after the timer expires, the outer catch block will not catch it.
D is incorrect for the same reason as C. In the original question, D also had a typing error: it used countSheep() instead of countsDeep(). Even after correcting that typing error, D is still incorrect because the try...catch is outside the asynchronous callback.
NEW QUESTION # 129
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