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| Section | Objectives |
|---|---|
| Topic 1: Testing and Debugging | - Debugging techniques and tools - Unit testing principles |
| Topic 2: Browser and DOM Interaction | - Browser APIs (fetch, storage, timers) - DOM manipulation and traversal - Event handling and propagation |
| Topic 3: Salesforce Platform Integration | - Security considerations in Salesforce JavaScript - Calling Apex methods from JavaScript - Lightning Web Components (LWC) JavaScript usage |
| Topic 4: Asynchronous JavaScript | - Error handling in async workflows - Event loop and concurrency model - Promises and async/await |
| Topic 5: JavaScript Language Fundamentals | - ES6+ syntax and features - Functions, scope, and closures - Data types, variables, and operators |
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NEW QUESTION # 14
Refer to the code below:
After running this code, which result is displayed on the console?
Answer: B
NEW QUESTION # 15
A developer wrote 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 or 5.
Which two test cases will properly test scenarios for the fizzbuss function?
Answer: A,C
NEW QUESTION # 16
At Universal Containers, every team has its own way of copying JavaScript objects. The code snippet shows an implementation from one team:
01 function Person() {
02 this.firstName = " John " ;
03 this.lastName = " Doe " ;
04 this.name = () = > {
05 console.log( ' Hello ${this.firstName} ${this.lastName} ' );
06 }
07 }
08
09 const john = new Person();
10 const dan = JSON.parse(JSON.stringify(john)); // (intended deep copy)
11 dan.firstName = ' Dan ' ;
12 dan.name();
(Original line 10 is logically intended to be JSON.parse(JSON.stringify(john)) to perform a JSON clone.) What is the output of the code execution?
Answer: D
Explanation:
* JSON.stringify(john) converts the john object into a JSON string.
* When you JSON.parse that string back, you get a plain object :
* Only data that can be represented in JSON is preserved (numbers, strings, booleans, arrays, plain objects).
* Functions are not preserved and are dropped.
* So dan is a plain object with properties firstName and lastName, but no name method .
* Therefore, dan.name is undefined, and dan.name() throws:
TypeError: dan.name is not a function
The literal string interpolation inside console.log( ' Hello ${...} ' ) is also wrong (single quotes), but the code never reaches that line.
NEW QUESTION # 17
Given the following code:
01 let x = null;
02 console.log(typeof x);
What is the output of line 02?
Answer: B
Explanation:
This question focuses on the typeof operator and how JavaScript represents the type of the value null.
* Declaration on line 1:
let x = null;
Here:
* x is explicitly assigned the value null.
* In JavaScript, null is a special primitive value that represents "no value" or "empty value".
* Line 2:
console.log(typeof x);
The typeof operator returns a string indicating the type of the operand. For example:
* typeof 1 returns " number "
* typeof " hello " returns " string "
* typeof true returns " boolean "
* typeof undefined returns " undefined "
* typeof {} returns " object "
* typeof [] returns " object " (arrays are objects)
* typeof function() {} returns " function "
For null, due to an early design decision in JavaScript, the result is:
typeof null === " object "
This is a known historical quirk of the language:
* Although null is a primitive and conceptually means "no object", typeof null returns the string " object "
.
Therefore, typeof x when x is null evaluates to " object " .
So on line 2, the value printed is:
" object "
This matches option A.
The incorrect options are:
* B. " undefined " : would be typeof x if x were undefined, not null.
* C. " x " : typeof returns a string describing the type of the value, not the variable name.
* D. " null " : although intuitive, JavaScript does not return " null " for typeof null; it returns " object " instead.
Therefore, the correct answer is:
The answer: A
JavaScript knowledge / study guide reference concepts:
* typeof operator and its return values
* Primitive types in JavaScript (null, undefined, number, string, boolean, symbol, bigint)
* Historical behavior: typeof null returning " object "
* Difference between null and undefined in JavaScript
NEW QUESTION # 18
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: B
Explanation:
Requirement:
* getNextValue() should run only if handleObjectValue(x) does not throw an error.
* If an error occurs, handleError(error) should run, and getNextValue() should be skipped.
Option B:
try {
handleObjectValue(x);
getNextValue();
} catch (error) {
handleError(error);
}
Behavior:
* If handleObjectValue(x) succeeds (no error):
* Execution continues to getNextValue();.
* If handleObjectValue(x) throws:
* Control jumps directly to catch, getNextValue() is skipped.
* handleError(error); is called.
This matches the requirement perfectly.
Why others are incorrect:
* A: } then { is invalid JavaScript syntax.
* C:
try {
handleObjectValue(x);
} catch (error) {
handleError(error);
}
getNextValue();
* getNextValue() is called after the try...catch regardless of whether an error occurred. Not acceptable.
* D: finally always runs:
try {
handleObjectValue(x);
} catch (error) {
handleError(error);
} finally {
getNextValue();
}
* getNextValue() will execute in both success and error cases.
Therefore, the correct approach is B.
Concepts: try/catch control flow, finally semantics, placing code inside vs outside try/catch blocks.
NEW QUESTION # 19
......
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