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NEW QUESTION # 140
Refer to the code:
01 let car1 = new Promise((_, reject) =>
02 setTimeout(reject, 2000, "Car 1 crashed in"));
03 let car2 = new Promise(resolve =>
04 setTimeout(resolve, 1500, "Car 2 completed"));
05 let car3 = new Promise(resolve =>
06 setTimeout(resolve, 3000, "Car 3 completed"));
07
08 Promise.race([car1, car2, car3])
09 .then(value => {
10 let result = '$(value) the race.';
11 })
12 .catch(err => {
13 console.log("Race is cancelled.", err);
14 });
What is the value of result when Promise.race executes?
Answer: B
Explanation:
Promise.race() returns the result of the first settled promise, whether resolved or rejected.
The promises:
car1 → rejects in 2000 ms
car2 → resolves in 1500 ms
car3 → resolves in 3000 ms
The earliest settled promise is car2 (1500 ms), which resolves with the value:
"Car 2 completed"
Therefore, Promise.race enters the .then() branch with:
value = "Car 2 completed"
Inside the .then():
let result = '$(value) the race.';
This appears to attempt template substitution but uses incorrect syntax.
JavaScript template literals require backticks and ${expression}, for example:
`${value} the race.`
Because the code is incorrect, result is literally the string:
$(value) the race.
However, the question asks:
"What is the value of result when Promise.race executes?"
This refers to the intended value based on which promise wins the race, not the template bug.
The winning value is:
Car 2 completed
So the correct conceptual answer is:
Car 2 completed the race.
JavaScript Knowledge Reference (text-only)
Promise.race() returns the first settled (resolved or rejected) promise.
The earliest resolve here is car2 (1500 ms).
Incorrect template literal syntax does not affect the identity of the winning promise.
NEW QUESTION # 141
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?
Answer: D
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 # 142
Refer to the code snippet below:
Let array = [1, 2, 3, 4, 4, 5, 4, 4];
For (let i =0; i <array.length; i++){
if (array[i] === 4) {
array.splice(i, 1);
}
}
What is the value of the array after the code executes?
Answer: D
NEW QUESTION # 143
let page2stream = "The quick brown fox jumps";
Which three expressions return true for a substring?
Answer: A,B,D
NEW QUESTION # 144
A developer needs to debug a Node.js web server because a runtime error keeps occurring at one of the endpoints.
The developer wants to test the endpoint on a local machine and make the request against a local server to look at the behavior. In the source code, the server.js file will start the server. The developer wants to debug the Node.js server only using the terminal.
Which command can the developer use to open the CLI debugger in their current terminal window?
(With corrected typing errors: node_inspect → node inspect, node_start_inspect → node start inspect.)
Answer: D
Explanation:
Comprehensive and Detailed Explanation From Exact Extract JavaScript knowledge:
Node.js includes a built-in command-line (CLI) debugger. To use it directly in the terminal, the standard command is:
node inspect server.js
This:
Starts server.js under the Node.js inspector.
Opens the CLI debugger right in the terminal window where you ran the command.
Lets you interactively:
Step through code
Set breakpoints
Inspect variables
Continue execution, etc.
Why B is correct:
Option B (corrected to node inspect server.js) is exactly the standard Node.js command to run a script under the CLI debugger in the current terminal.
It does not rely on any external GUI tools.
You stay entirely in the terminal to debug.
Why the other options are incorrect:
A . node start inspect server.js
There is no start subcommand like this in standard Node.js CLI.
This is not a valid Node.js debugging command.
C . node server.js --inspect
This starts Node.js with the Inspector protocol enabled and is typically used to connect external tools like Chrome DevTools or VS Code.
It does not open the CLI debugger in the current terminal. Instead, it opens a debugging port that other tools attach to.
The question specifically says "only using the terminal" and "open the CLI debugger in their current terminal window", which points to node inspect, not --inspect.
D . node -i server.js
-i starts Node in interactive REPL mode, optionally after running a script.
This is for an interactive shell, not the Node debugger.
It does not provide breakpoint/step/next/continue debugging features like the CLI debugger.
Therefore, the only option that opens the Node.js CLI debugger in the current terminal is:
JavaScript / Node.js knowledge / Study Guide references (concept names only, no links):
Node.js CLI debugger: node inspect <script>
Node.js inspector protocol: node --inspect
Difference between CLI debugger and DevTools-based debugging
Node.js command-line options and subcommands
________________________________________
NEW QUESTION # 145
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