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質問 # 75
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.)
正解：D
解説：
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
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質問 # 76
A developer needs to test this function:
01const sum3 = (arr) => (
02if (!arr.length) return 0,
03if (arr.length === 1) return arr[0],
04if (arr.length === 2) return arr[0]+ arr[1],
05 return arr[0] + arr[1] + arr[2],
06 );
Which two assert statements are valid tests for the function?
Choose 2 answers
正解：A、D
質問 # 77
A developer creates a simple webpage with an input field. When a user enters text and clicks the button, the actual value must be displayed in the console:
HTML:
<input type="text" value="Hello" name="input">
<button type="button">Display</button>
JavaScript:
01 const button = document.querySelector('button');
02 button.addEventListener('click', () => {
03 const input = document.querySelector('input');
04 console.log(input.getAttribute('value'));
05 });
When the user clicks the button, the output is always "Hello".
What needs to be done to make this code work as expected?
正解：C
解説：
getAttribute('value')
This returns the initial HTML attribute, not the live, updated value.
Even if the user edits the text, the value attribute remains "Hello" because HTML attributes do not update dynamically.
Input elements have a property value that always reflects the live current text inside the field.
So:
input.value
returns the user-entered value.
Therefore, line 04 must use the property, not the attribute:
console.log(input.value);
This ensures the updated input value is displayed.
JavaScript knowledge references (text-only)
HTML attributes are static and retrieved using getAttribute().
DOM element properties (like value) represent the current live state.
Input value changes update the .value property, not the attribute.
質問 # 78
A developer uses a parsed JSON string to work with user information as in the block below:
01 const userInformation ={
02 " id " : "user-01",
03 "email" : "user01@universalcontainers.demo",
04 "age" : 25
Which two options access the email attribute in the object?
Choose 2 answers
正解：A、B
質問 # 79
Given the code below:
let numValue = 1982;
Which three code segments result in a correct conversion from number to string?
正解：A、B、C
解説：
Comprehensive and Detailed Explanation From JavaScript Knowledge:
We want to convert the number 1982 to a string.
Check each option:
A . numValue.toText()
There is no standard toText() method on numbers.
This will result in TypeError: numValue.toText is not a function.
B . String(numValue);
String() as a function converts its argument to a string.
String(1982) returns "1982".
This is correct.
C . '' + numValue;
'' is a string; + with a string operand performs string concatenation.
'' + 1982 → "1982".
This is a common shorthand for number-to-string conversion.
D . numValue.toString();
Number.prototype.toString() converts the number to its string representation.
1982..toString() or (1982).toString() returns "1982".
For the variable, numValue.toString() is valid: "1982".
E . (String)numValue;
This is not valid JavaScript casting syntax; it is more like a C/Java-style cast.
In JavaScript, that is parsed as a grouping expression (String) and then numValue; it does not convert numValue to a string.
Thus the correct answers are:
B . String(numValue);
C . '' + numValue;
D . numValue.toString();
Relevant concepts: primitive type conversion, String() casting, .toString(), coercion via + with strings.
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質問 # 80
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