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| Certification Vendor: | Salesforce |
|---|---|
| Exam Name: | Salesforce Certified JavaScript Developer I - Multiple Choice Exam (JS-Dev-101) |
| Exam Number: | JS-Dev-101 |
| Related Certifications: | Salesforce Platform Developer II Salesforce Platform Developer I |
| Exam Format: | Multiple Select Questions, Multiple Choice Questions |
| Passing Score: | 65% |
| Real Exam Qty: | 60 (multiple choice) |
| Exam Duration: | 105 minutes |
| Exam Price: | $200 USD |
| Certificate Validity Period: | 1 year (maintenance required annually in Salesforce certification program) |
| Available Languages: | English |
| Recommended Training: | MDN Web Docs - JavaScript Guide Salesforce Trailhead - JavaScript Developer I |
| Exam Registration: | Salesforce Certification Registration Kryterion Webassessor (Exam Delivery) |
| Sample Questions: | Salesforce JS-Dev-101 Sample Questions |
| Exam Way: | Online proctored exam via Salesforce authorized testing platform (e.g., Webassessor) |
| Pre Condition: | No formal prerequisites, but familiarity with JavaScript and web development is recommended. |
| Official Syllabus URL: | https://trailhead.salesforce.com/credentials/javascriptdeveloper |
Salesforce JS-Dev-101 認證試題庫學習資料根據最新的知識點以及輔導資料進行整編，覆蓋面廣，蘊含了眾多最新的 Salesforce 考試知識點。如果你正在準備 JS-Dev-101 考試並且像我一樣急需通過，那 JS-Dev-101 認證試題剛好可以幫助你。因為完善的 JS-Dev-101 學習資料資料覆蓋 Salesforce 考試所有知識點，減少你考試的時間成本和經濟成本，助你輕松通過考試
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問題 #73
Given the code below:
01 const delay = async delay => {
02 return new Promise((resolve, reject) => {
03 console.log(1);
04 setTimeout(resolve, delay);
05 });
06 };
07
08 const callDelay = async () => {
09 console.log(2);
10 const yup = await delay(1000);
11 console.log(3);
12 };
13
14 console.log(4);
15 callDelay();
16 console.log(5);
What is logged to the console?
答案：A
解題說明：
Comprehensive and Detailed Explanation From Exact Extract JavaScript Knowledge:
Execution order:
Top-level code runs synchronously:
Line 14: console.log(4); → logs 4.
Line 15: callDelay(); is called.
Inside callDelay:
Line 9: console.log(2); → logs 2.
Line 10: await delay(1000);:
Calls delay(1000).
Inside delay(1000):
Line 3: console.log(1); → logs 1.
Line 4: setTimeout(resolve, delay); schedules resolve in 1000 ms.
delay returns a pending Promise. await pauses callDelay here and returns control to the event loop.
Back to top-level:
Line 16: console.log(5); → logs 5.
So synchronous log sequence is: 4, 2, 1, 5.
After ~1000 ms:
The setTimeout in delay resolves the Promise.
The await in callDelay resumes.
Line 11: console.log(3); → logs 3.
Final log order: 4 2 1 5 3.
Both A and B show the same sequence; one must be chosen, so A is correct.
Concepts: async/await flow, Promise resolution timing, event loop, and ordering of synchronous vs timer callbacks.
________________________________________
問題 #74
Refer to the code below:
let inArray = [ [1, 2], [3, 4, 5] ];
Which two statements result in the array [1, 2, 3, 4, 5]?
(With corrected typing errors: usArray → inArray, .. → ....)
答案：A,C
解題說明：
We start with the array:
let inArray = [ [1, 2], [3, 4, 5] ];
This is an array of two inner arrays:
First element: [1, 2]
Second element: [3, 4, 5]
The desired result is to transform this into a single, flat array:
[1, 2, 3, 4, 5]
This is accomplished by using Array.prototype.concat to concatenate the inner arrays into one new array, optionally combined with the spread syntax (...) or Function.prototype.apply.
Option A: [].concat(...inArray);
Relevant concepts:
Spread syntax (...inArray) expands an iterable into separate arguments.
Array.prototype.concat joins arrays or values into a new array. When you pass arrays as arguments to concat, it flattens them one level into the result.
Step-by-step behavior:
inArray is [ [1, 2], [3, 4, 5] ].
...inArray expands into [1, 2] and [3, 4, 5] as separate arguments.
So [].concat(...inArray) is equivalent to:
[].concat([1, 2], [3, 4, 5]);
concat processes each argument:
For [1, 2], it adds 1 and 2 into the result array.
For [3, 4, 5], it adds 3, 4, and 5 into the result array.
The final outcome is:
[1, 2, 3, 4, 5]
Therefore, Option A correctly produces the desired array.
Option B: [].concat.apply(inArray, []);
Corrected name from usArray to inArray.
Relevant concepts:
Function.prototype.apply(fnThis, argsArray) invokes a function with a specific this value and a list of arguments passed as an array.
Here:
thisArg is inArray.
argsArray is [] (no actual arguments passed).
So the call:
[].concat.apply(inArray, []);
is equivalent to:
Array.prototype.concat.apply(inArray, []);
// i.e. inArray.concat();
Since there are no extra arguments, inArray.concat() simply returns a shallow copy of inArray:
[ [1, 2], [3, 4, 5] ]
This remains an array of arrays and is not flattened into [1, 2, 3, 4, 5]. Therefore, Option B does not produce the required result.
Option C: [].concat::...inArray();
Corrected name from usArray to inArray and .. to ....
This expression uses syntax that is not part of standard JavaScript:
:: (double colon) was proposed in early drafts as a bind operator but is not part of the official ECMAScript standard.
The combination concat::...inArray() is invalid in normal JavaScript engines and cannot be used as a valid way to flatten arrays.
In addition, inArray() would imply calling inArray as a function, but it is an array, which would cause a runtime error.
Hence, Option C is syntactically or semantically invalid in standard JavaScript and does not provide the required result [1, 2, 3, 4, 5].
Option D: [].concat.apply({}, inArray);
Corrected name from usArray to inArray.
Relevant concepts:
Again, Function.prototype.apply is used to call concat with a specific this value and arguments provided as an array.
concat treats its this value as an array-like object but ultimately returns a new array containing concatenated elements.
In this expression:
[].concat.apply({}, inArray);
thisArg is {} (an empty object).
argsArray is inArray, which is [ [1, 2], [3, 4, 5] ].
Using apply, this is interpreted as calling concat like:
Array.prototype.concat.call({}, [1, 2], [3, 4, 5]);
concat then:
Starts from the array-like this (here {} is treated as an empty base).
Takes the first argument [1, 2] and appends its elements 1 and 2 to the result.
Takes the second argument [3, 4, 5] and appends its elements 3, 4, and 5 to the result.
The resulting new array is:
[1, 2, 3, 4, 5]
Therefore, Option D also correctly produces the required array.
Final evaluation of all options:
Option A: Uses spread syntax with concat and correctly flattens one level: [1, 2, 3, 4, 5].
Option B: Equivalent to inArray.concat(), leaving it as [[1, 2], [3, 4, 5]]. Does not flatten the structure.
Option C: Uses non-standard and invalid syntax; not a valid or correct JavaScript solution.
Option D: Uses apply with concat, passing the inner arrays as arguments and flattening one level to [1, 2, 3, 4, 5].
Thus, the two correct statements are:
Reference of JavaScript knowledge documents or Study Guide (concept names only):
Array.prototype.concat (concatenating and one-level flattening behavior) Spread syntax for arrays (...array) Function.prototype.apply (calling a function with a given this value and arguments list) Array flattening by one level using concat with array arguments Distinction between nested arrays and flat arrays in JavaScript
問題 #75
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(`${this.name} is barking.`);
18 }
19 }
20
21 let myDog = new Dog('Puppy');
22 myDog.makeSound();
What is the console output?
答案：B
解題說明：
Animal class:
Has a constructor that sets this.name.
Has a makeSound method that logs a message using this.name.
Dog class:
Extends Animal.
Its constructor calls super(name) to run the parent constructor, which sets this.name.
It overrides makeSound() to log:
console.log(`${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} ...`
問題 #76
What are two unique features of fat-arrow functions compared to normal function definitions?
答案：B,D
解題說明：
________________________________________
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.
問題 #77
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?
答案：B
解題說明：
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.
問題 #78
......
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