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NEW QUESTION # 127
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?
Answer: C
NEW QUESTION # 128
A class was written to represent regular items and sale items. Code:
01 let regItem = new Item( ' Scarf ' , 55);
02 let saleItem = new SaleItem( ' Shirt ' , 80, .1);
03 Item.prototype.description = function() { return ' This is a ' + this.name; }
04 console.log(regItem.description());
05 console.log(saleItem.description());
06
07 SaleItem.prototype.description = function() { return ' This is a discounted ' + this.name; }
08 console.log(regItem.description());
09 console.log(saleItem.description());
What is the output?
Answer: D
Explanation:
* At line 03, the developer assigns:
Item.prototype.description = function() { return ' This is a ' + this.name; } This affects all objects whose prototype chain includes Item.prototype .
* regItem inherits from Item # gets this method.
* saleItem, as an instance of SaleItem, also inherits Item.prototype (since SaleItem uses prototype inheritance from Item), so it also has this method at this moment.
Outputs at lines 04 and 05:
* regItem.description() # " This is a Scarf "
* saleItem.description() # " This is a Shirt "
* At line 07, the developer overrides the method on SaleItem.prototype :
SaleItem.prototype.description = function() {
return ' This is a discounted ' + this.name;
}
From this point:
* regItem still uses the Item.prototype version
* saleItem uses the overridden SaleItem.prototype version
Outputs at lines 08 and 09:
* regItem.description() # " This is a Scarf "
* saleItem.description() # " This is a discounted Shirt "
Combining all results:
This is a Scarf
This is a Shirt
This is a Scarf
This is a discounted Shirt
This matches option B .
JavaScript Knowledge References (text-only)
* Objects created from constructor functions use prototype chaining.
* Overriding a subclass prototype method does not affect the parent class prototype.
* Instances inherit the most specific version of the method on their prototype chain.
NEW QUESTION # 129
Refer to the following code that imports a module named utils:
import (foo, bar) from '/path/Utils.js';
foo() ;
bar() ;
Which two implementations of Utils.js export foo and bar such that the code above runs without error?
Choose 2 answers
Answer: C,D
NEW QUESTION # 130
A developer copied a JavaScript object:
How does the developer access dan's forstName,lastName? Choose 2 answers
Answer: A,B
NEW QUESTION # 131
Which two code snippets show working examples of a recursive function?
Answer: A,B
Explanation:
A recursive function is a function that calls itself and has a base case to terminate the recursion.
Evaluate each option:
Option A:
const sumToTen = numVar = > {
if (numVar < 0)
return;
return sumToTen(numVar + 1);
};
* This function calls itself: sumToTen(numVar + 1) - so it is recursive.
* However, the base condition is if (numVar < 0) return;.
* If you call sumToTen(0):
* numVar < 0 is false, so it calls sumToTen(1), then sumToTen(2), and so on, incrementing forever.
* There is no condition to stop the recursion when numVar increases; it will eventually cause a stack overflow.
* This code does not represent a properly working recursive function with a valid termination for increasing values and is not a good example of correct recursion.
Option B:
function factorial(numVar) {
if (numVar < 0) return;
if (numVar === 0) return 1;
return numVar - 1;
}
* This function does not call itself anywhere.
* It has conditional returns, but there is no recursive call such as factorial(numVar - 1).
* Therefore, it is not recursive at all.
Option C:
const factorial = numVar = > {
if (numVar < 0) return;
if (numVar === 0) return 1;
return numVar * factorial(numVar - 1);
};
* This is a classic recursive factorial implementation.
* It calls itself with a smaller argument: factorial(numVar - 1).
* Base cases:
* If numVar < 0, it simply returns (could be treated as invalid input).
* If numVar === 0, it returns 1, which is the mathematical definition of 0! (zero factorial).
* For positive integers, it correctly multiplies numVar by factorial(numVar - 1) until it reaches the base case.
This is a correct and working recursive function.
Option D (corrected):
let countingDown = function(startNumber) {
if (startNumber > 0) {
console.log(startNumber);
return countingDown(startNumber - 1);
} else {
return startNumber;
}
};
* This function also calls itself: countingDown(startNumber - 1).
* Base case:
* When startNumber is not greater than 0 (i.e., 0 or negative), it returns startNumber and stops recursing.
* For example, countingDown(3) would:
* Log 3, call countingDown(2)
* Log 2, call countingDown(1)
* Log 1, call countingDown(0)
* At 0, it hits the else branch and returns 0, ending the recursion.
This is a valid working recursive function structure (once syntax is corrected).
Therefore, the snippets that show working recursive functions are:
The answer: C, D
Study Guide / Concept References (no links):
* Definition of recursion: a function calling itself
* Base case vs recursive step
* Recursive factorial implementation
* Recursive countdown example
* Importance of a terminating condition to avoid infinite recursion
NEW QUESTION # 132
......
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