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Desktop practice test software, and web-based practice test software. All three VCEEngine JS-Dev-101 practice test questions formats are easy to use and compatible with all devices and operating systems. The VCEEngine JS-Dev-101 desktop practice test software and web-based practice test software both are the JS-Dev-101 Practice copyright. While practicing on copyright copyright Certified JavaScript Developer - Multiple Choice practice test software you will experience the real-timecopyright Certified JavaScript Developer - Multiple Choice JS-Dev-101 copyright environment for preparation. This will help you to understand the pattern of final JS-Dev-101 copyright questions and answers.
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copyright Certified JavaScript Developer - Multiple Choice Sample Questions (Q86-Q91):
NEW QUESTION # 86
Given the code below:
01 setCurrentUrl();
02 console.log("The current URL is: " + url);
03
04 function setCurrentUrl() {
05 url = window.location.href;
06 }
What happens when the code executes?
- A. The url variable has global scope and line 02 executes correctly.
- B. The url variable has global scope and line 02 throws an error.
- C. The url variable has local scope and line 02 throws an error.
- D. The url variable has local scope and line 02 executes correctly.
Answer: A
Explanation:
Inside setCurrentUrl, url is assigned without var, let, or const:
url = window.location.href;
In non-strict mode, this implicitly creates a global variable url on window.
Execution order:
setCurrentUrl(); creates/sets global url to window.location.href.
console.log("The current URL is: " + url); has access to url in global scope and prints correctly.
Thus:
url has global scope.
Line 02 runs without error and logs a valid string.
So B is correct.
NEW QUESTION # 87
A developer has a web server running with Node.js. The command to start the web server is node server.js. The web server started having latency issues. Instead of a one second turnaround for web requests, the developer now sees a five second turnaround.
Which command can the web developer run to see what the module is doing during the latency period?
- A. DEBUG=true node server.js
- B. DEBUG=http, https node server.js
- C. NODE_DEBUG=http,https node server.js
- D. NODE_DEBUG=true node server.js
Answer: A
NEW QUESTION # 88
Refer to the code below:
class Student {
constructor(name) {
this._name = name;
}
displayGrade() {
console.log(`${this._name} got 70% on test.`);
}
}
class GraduateStudent extends Student {
constructor(name) {
super(name);
this._name = "Graduate Student " + name;
}
displayGrade() {
console.log(`${this._name} got 100% on test.`);
}
}
let student = new GraduateStudent("Jane");
student.displayGrade();
What is the console output?
- A. Uncaught ReferenceError
- B. Graduate Student Jane got 100% on test.
- C. Jackie got 70% on test.
- D. Better student Jackie got 70% on test.
Answer: B
Explanation:
The correct answer is C, after correcting the option text to match the actual code.
The object is created here:
let student = new GraduateStudent("Jane");
Because GraduateStudent extends Student, its constructor runs:
constructor(name) {
super(name);
this._name = "Graduate Student " + name;
}
The call to:
super(name);
runs the parent Student constructor first and temporarily sets:
this._name = "Jane";
Then this line in the child constructor overwrites that value:
this._name = "Graduate Student " + name;
So the final value of this._name becomes:
"Graduate Student Jane"
Next, this line executes:
student.displayGrade();
Since student is an instance of GraduateStudent, JavaScript uses the overridden displayGrade() method from GraduateStudent, not the parent method from Student.
The executed method is:
displayGrade() {
console.log(`${this._name} got 100% on test.`);
}
Therefore, the console output is:
Graduate Student Jane got 100% on test.
Important correction: the original option C said something like "Better student Jackie got 100% on test.", but the actual code uses "Graduate Student " and the name "Jane". The verified corrected answer remains C.
NEW QUESTION # 89
Corrected code:
let a = "*";
let b = "**";
// x = 3;
console.log(a);
What is displayed when the code executes?
- A. ReferenceError: a is not defined
- B. null
- C. undefined
- D. *
Answer: D
Explanation:
The correct answer is B because variable a is declared and initialized before it is printed.
let a = "*";
This creates a block-scoped variable named a and assigns it the string value:
"*"
The line:
// x = 3;
is a comment. JavaScript ignores comments during execution, so this line has no effect.
Then this line runs:
console.log(a);
Since a already exists and contains "*", the console displays:
* Option A is incorrect because a is defined.
Option C is incorrect because a has an assigned value, so it is not undefined.
Option D is incorrect because a was never assigned null.
Therefore, the verified answer is B.
NEW QUESTION # 90
Given the code below:
01 setCurrentUrl();
02 console.log("The current URL is: " + url);
03
04 function setCurrentUrl() {
05 url = window.location.href;
06 }
What happens when the code executes?
- A. The url variable has global scope and line 02 executes correctly.
- B. The url variable has global scope and line 02 throws an error.
- C. The url variable has local scope and line 02 throws an error.
- D. The url variable has local scope and line 02 executes correctly.
Answer: A
Explanation:
Inside setCurrentUrl, url is assigned without var, let, or const:
url = window.location.href;
In non-strict mode, this implicitly creates a global variable url on window.
Execution order:
setCurrentUrl(); creates/sets global url to window.location.href.
console.log("The current URL is: " + url); has access to url in global scope and prints correctly.
Thus:
url has global scope.
Line 02 runs without error and logs a valid string.
So B is correct.
________________________________________
NEW QUESTION # 91
......
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