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NEW QUESTION # 74
Refer to the code below:
let timeFunction =() => {
console.log('Timer called.");
};
let timerId = setTimeout (timedFunction, 1000);
Which statement allows a developer to cancel the scheduled timed function?
Answer: B
NEW QUESTION # 75
Refer to the code below:
01 function myFunction(reassign) {
02 let x = 1;
03 var y = 1;
04
05 if (reassign) {
06 let x = 2;
07 var y = 2;
08 console.log(x);
09 console.log(y);
10 }
11
12 console.log(x);
13 console.log(y);
14 }
What is displayed when myFunction(true) is called?
Answer: B
Explanation:
This question tests understanding of let (block scope) and var (function scope) in JavaScript.
Initial declarations in the function:
let x = 1; // line 2
var y = 1; // line 3
Here:
x is declared with let, so it is block-scoped to the function body.
y is declared with var, so it is function-scoped to the entire function.
Inside the if (reassign) block (and since reassign is true, we enter it):
if (reassign) {
let x = 2; // line 6
var y = 2; // line 7
console.log(x); // line 8
console.log(y); // line 9
}
Detailed behavior:
let x = 2; on line 6 creates a new block-scoped variable x that exists only inside the if block. It does not change the outer x declared on line 2.
var y = 2; on line 7 declares y with var again, but var is function-scoped. This effectively reassigns the same y defined on line 3 for the entire function. After this line, y is 2 everywhere in the function.
Now, inside the if block:
console.log(x); (line 8) logs the inner block-scoped x, which is 2.
console.log(y); (line 9) logs y, which is the function-scoped y that was set to 2.
So the first two outputs are:
2
2
After the if block, execution continues:
console.log(x); // line 12
console.log(y); // line 13
Outside the if block:
The block-scoped let x = 2; no longer exists; it was only visible inside the if block.
The outer let x = 1; (line 2) is still in scope and has not been changed.
Thus:
console.log(x); (line 12) logs the outer x, which is still 1.
console.log(y); (line 13) logs y which, due to var y = 2; inside the if, is now 2 for the whole function.
Therefore, when myFunction(true) is called, the output in order is:
2 (inner x in if)
2 (function-scoped y after reassignment)
1 (outer x after if)
2 (function-scoped y remains 2)
This corresponds to:
Answer : B (2 2 1 2)
JavaScript knowledge / study guide reference concepts:
let declarations and block scope
var declarations and function scope
Shadowing of variables with let inside a block
Re-declaration and reassignment of var within a function
Execution order of statements and console output
NEW QUESTION # 76
Given the code below:
01 setTimeout(() => {
02 console.log(1);
03 }, 1100);
04 console.log(2);
05 new Promise((resolve, reject) => {
06 setTimeout(() => {
07 reject(console.log(3));
08 }, 1000);
09 }).catch(() => {
10 console.log(4);
11 });
12 console.log(5);
What is logged to the console?
Answer: D
Explanation:
Comprehensive and Detailed Explanation From JavaScript Knowledge:
We must track synchronous code, setTimeout callbacks (macrotasks), and Promise rejection handling (microtasks).
Step-by-step:
Synchronous code first:
Line 01-03: Schedules a timeout at 1100 ms: logs 1 (later).
Line 04: console.log(2); โ logs 2.
Lines 05-09: Construct a new Promise.
The executor runs immediately.
Inside it, another setTimeout is set for 1000 ms:
setTimeout(() => {
reject(console.log(3));
}, 1000);
Line 09-11: .catch(() => { console.log(4); }) attached to the promise.
Line 12: console.log(5); โ logs 5.
So after all synchronous code, the console has:
2
5
At 1000 ms: inner setTimeout fires
The callback:
() => {
reject(console.log(3));
}
Inside:
console.log(3) runs first, logging 3.
console.log(3) returns undefined.
Then reject(undefined) is called.
So at about 1000 ms, we log:
3
When the promise is rejected:
The .catch handler is scheduled as a microtask.
After the current macrotask (the timeout callback) completes, the microtask queue runs.
Thus .catch(() => { console.log(4); }) runs shortly after, in the same 1000 ms tick.
So immediately after 3, the catch handler logs:
4
Now the logs in order are:
2
5
3
4
At 1100 ms: outer setTimeout fires
The callback from line 01 runs:
console.log(1);
Logs: 1.
Final log order:
2 (line 4, sync)
5 (line 12, sync)
3 (1000 ms timeout, then inside logs before reject)
4 (promise catch microtask after rejection)
1 (1100 ms timeout)
Concatenated: 25341.
Therefore, the correct option is:
Study Guide / Concept Reference (no links):
Event loop: call stack, macrotask queue (timers), microtask queue (promises) setTimeout scheduling and ordering Promise rejection, .catch, and microtasks Evaluation order of function arguments (reject(console.log(3)))
________________________________________
NEW QUESTION # 77
Most accurate tests for:
const arr = Array(5).fill(0);
Answer: B,C
Explanation:
The correct answers are C and D.
The statement:
const arr = Array(5).fill(0);
creates an array with 5 elements, and every element is filled with the value 0.
So the array becomes:
[0, 0, 0, 0, 0]
A professional test should verify both important expectations:
arr.length === 5
and:
each element === 0
Option C is correct because it checks every element in the array:
arr.forEach(e => console.assert(e === 0));
This confirms that each array element was filled with 0.
Option D is correct because it checks the array length:
console.assert(arr.length === 5);
This confirms that the array contains exactly five elements.
Option A is incorrect because the array length is not 0; it is 5.
Option B is incorrect for two reasons. First, arr[5] is outside the valid index range. The valid indexes are:
0, 1, 2, 3, 4
So:
arr[5]
returns:
undefined
Second, arr.length === 0 is false because the length is 5.
Therefore, the verified answers are C and D.
NEW QUESTION # 78
Which two code snippets show working examples of a recursive function?
Answer: A,C
Explanation:
return startNumber;
}
};
(Note: Option D is shown here with corrected syntax: lowercase return and matching parentheses.) Explanation:
Comprehensive and Detailed Explanation From Exact Extract JavaScript knowledge:
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:
Answe r: C, D
Study Guide / Concept Reference (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 # 79
......
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