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Salesforce JS-Dev-101 Exam Syllabus Topics:

Section	Weight	Objectives
Topic 1: Variables, Types, and Collections	23%	- Strings, numbers, dates, arrays and methods - JSON parsing and manipulation - Data types, type coercion, truthy/falsy values - Variable declaration and scope
Topic 2: Server Side JavaScript	8%	- Node.js fundamentals and core modules - Package management and CLI tools
Topic 3: Testing	7%	- Unit test structure and effectiveness - Test coverage and improvement
Topic 4: Browser and Events	17%	- DOM selection and manipulation - Event handling, propagation, listeners - Browser APIs and developer tools
Topic 5: Debugging and Error Handling	7%	- Console usage, breakpoints and debugging techniques - Error types and handling strategies
Topic 6: Objects, Functions, and Classes	25%	- Object creation, properties, prototypes - Function types, scope, closures, arrow functions - ES6 classes, inheritance, modules, decorators
Topic 7: Asynchronous Programming	13%	- Callbacks, promises, async/await - Event loop and execution flow

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Salesforce Certified JavaScript Developer - Multiple Choice Sample Questions (Q146-Q151):

NEW QUESTION # 146
Refer to the following code block (with corrected template literals using backticks):
01 class Animal {
02 constructor(name) {
03 this.name = name;
04 }
05
06 makeSound() {
07 console.log(`${this.name} is making a sound.`);
08 }
09 }
10
11 class Dog extends Animal {
12 constructor(name) {
13 super(name);
14 this.name = name;
15 }
16 makeSound() {
17 console.log(`${this.name} is barking.`);
18 }
19 }
20
21 let myDog = new Dog('Puppy');
22 myDog.makeSound();
What is the console output?

A. > Puppy is barking.
B. > Undefined
C. > Uncaught ReferenceError

Answer: A

Explanation:
Animal class:
Has a constructor that sets this.name.
Has a makeSound method that logs a message using this.name.
Dog class:
Extends Animal.
Its constructor calls super(name) to run the parent constructor, which sets this.name.
It overrides makeSound() to log:
console.log(`${this.name} is barking.`);
Instantiation:
let myDog = new Dog('Puppy');
myDog.makeSound();
new Dog('Puppy'):
Calls Dog constructor with name = 'Puppy'.
Inside, super(name) sets this.name = 'Puppy' in the Animal constructor.
Then this.name = name; in the Dog constructor keeps it as 'Puppy'.
myDog.makeSound():
Calls the overridden makeSound() in Dog.
Logs: "Puppy is barking."
No reference errors, no undefined; the output is:
Puppy is barking.
Study Guide Concepts:
ES6 classes and inheritance (extends)
super() in subclass constructors
Method overriding in subclasses
Template literals: `${this.name} ...`

NEW QUESTION # 147
Refer to the code snippet:
Function getAvailabilityMessage(item) {
If (getAvailability(item)){
Var msg ="Username available";
}
Return msg;
}
A developer writes this code to return a message to user attempting to register a new username. If the username is available, variable.
What is the return value of msg hen getAvailabilityMessage ("newUserName" ) is executed and getAvailability("newUserName") returns false?

A. "Msg is not defined"
B. undefined
C. "Username available"
D. "newUserName"

Answer: B

NEW QUESTION # 148
Refer to the following object.
```javascript
02 const dog = {
02 firstName: 'Beau',
03 lastName: 'Boo',
04 get fullName() {
05 return this.firstName + ' ' + this.lastName;
06 }
07 };
How can a developer access the fullName property for dog?

A. dog.get.fullName
B. dog.function.fullName()
C. dog.fullName()
D. dog.fullName

Answer: D

NEW QUESTION # 149
Refer to the code below:
flag();
function flag() {
console.log('flag');
}
const anotherFlag = () => {
console.log('another flag');
}
anotherFlag();
What is result of the code block?

A. An error is thrown.
B. The console logs 'flag' and 'another flag'.
C. The console logs only 'flag'.
D. The console logs 'flag' and then an error is thrown.

Answer: B

Explanation:
Key points:
Function declarations are hoisted (their definitions are available before the line where they appear).
Function expressions assigned to const or let are not hoisted as callable functions, but here anotherFlag is only used after it is defined.
Step-by-step:
flag(); at the top:
flag is a function declaration defined later; due to hoisting, it is available.
It logs 'flag'.
The declaration:
function flag() {
console.log('flag');
}
is already in effect (hoisted before execution).
const anotherFlag = () => { console.log('another flag'); } defines anotherFlag as an arrow function.
anotherFlag(); is called after its definition; this is valid and logs 'another flag'.
No errors occur. The console output is:
flag
another flag
So option D is correct.
Concepts: function hoisting, difference between function declarations and function expressions, execution order.

NEW QUESTION # 150
A developer wants to use a module named universalContainerslib and then call functions from it. How should a developer import every function from the module and then call the functions foo and bar?

A. import * as lib from '/path/universalContainerslib.js';
lib.foo();
lib.bar();
B. import {foo, bar} from '/path/universalContainerslib.js';
foo();
bar();
C. import * from '/path/universalContainerslib.js';
universalContainerslib.foo();
universalContainerslib.bar();
D. import all from '/path/universalContainerslib.js';
universalContainerslib.foo();
universalContainerslib.bar();

Answer: A

Explanation:
import * as lib from '...' imports all named exports from the module into the namespace object lib.
You then call:
lib.foo();
lib.bar();
Option D correctly imports only foo and bar, not every function. The question explicitly says "import every function... and then call foo and bar", so A best matches.
Options B and C are invalid syntax or reference the wrong identifier.

NEW QUESTION # 151
......

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