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NEW QUESTION # 65
Refer to the code below:
let inArray = [ [1, 2], [3, 4, 5] ];
Which two statements result in the array [1, 2, 3, 4, 5]?
(With corrected typing errors: usArray → inArray, .. → ....)
Answer: C,D
Explanation:
Comprehensive and Detailed Explanation From Exact Extract JavaScript knowledge:
We start with the array:
let inArray = [ [1, 2], [3, 4, 5] ];
This is an array of two inner arrays:
First element: [1, 2]
Second element: [3, 4, 5]
The desired result is to transform this into a single, flat array:
[1, 2, 3, 4, 5]
This is accomplished by using Array.prototype.concat to concatenate the inner arrays into one new array, optionally combined with the spread syntax (...) or Function.prototype.apply.
Option A: [].concat(...inArray);
Relevant concepts:
Spread syntax (...inArray) expands an iterable into separate arguments.
Array.prototype.concat joins arrays or values into a new array. When you pass arrays as arguments to concat, it flattens them one level into the result.
Step-by-step behavior:
inArray is [ [1, 2], [3, 4, 5] ].
...inArray expands into [1, 2] and [3, 4, 5] as separate arguments.
So [].concat(...inArray) is equivalent to:
[].concat([1, 2], [3, 4, 5]);
concat processes each argument:
For [1, 2], it adds 1 and 2 into the result array.
For [3, 4, 5], it adds 3, 4, and 5 into the result array.
The final outcome is:
[1, 2, 3, 4, 5]
Therefore, Option A correctly produces the desired array.
Option B: [].concat.apply(inArray, []);
Corrected name from usArray to inArray.
Relevant concepts:
Function.prototype.apply(fnThis, argsArray) invokes a function with a specific this value and a list of arguments passed as an array.
Here:
thisArg is inArray.
argsArray is [] (no actual arguments passed).
So the call:
[].concat.apply(inArray, []);
is equivalent to:
Array.prototype.concat.apply(inArray, []);
// i.e. inArray.concat();
Since there are no extra arguments, inArray.concat() simply returns a shallow copy of inArray:
[ [1, 2], [3, 4, 5] ]
This remains an array of arrays and is not flattened into [1, 2, 3, 4, 5]. Therefore, Option B does not produce the required result.
Option C: [].concat::...inArray();
Corrected name from usArray to inArray and .. to ....
This expression uses syntax that is not part of standard JavaScript:
:: (double colon) was proposed in early drafts as a bind operator but is not part of the official ECMAScript standard.
The combination concat::...inArray() is invalid in normal JavaScript engines and cannot be used as a valid way to flatten arrays.
In addition, inArray() would imply calling inArray as a function, but it is an array, which would cause a runtime error.
Hence, Option C is syntactically or semantically invalid in standard JavaScript and does not provide the required result [1, 2, 3, 4, 5].
Option D: [].concat.apply({}, inArray);
Corrected name from usArray to inArray.
Relevant concepts:
Again, Function.prototype.apply is used to call concat with a specific this value and arguments provided as an array.
concat treats its this value as an array-like object but ultimately returns a new array containing concatenated elements.
In this expression:
[].concat.apply({}, inArray);
thisArg is {} (an empty object).
argsArray is inArray, which is [ [1, 2], [3, 4, 5] ].
Using apply, this is interpreted as calling concat like:
Array.prototype.concat.call({}, [1, 2], [3, 4, 5]);
concat then:
Starts from the array-like this (here {} is treated as an empty base).
Takes the first argument [1, 2] and appends its elements 1 and 2 to the result.
Takes the second argument [3, 4, 5] and appends its elements 3, 4, and 5 to the result.
The resulting new array is:
[1, 2, 3, 4, 5]
Therefore, Option D also correctly produces the required array.
Final evaluation of all options:
Option A: Uses spread syntax with concat and correctly flattens one level: [1, 2, 3, 4, 5].
Option B: Equivalent to inArray.concat(), leaving it as [[1, 2], [3, 4, 5]]. Does not flatten the structure.
Option C: Uses non-standard and invalid syntax; not a valid or correct JavaScript solution.
Option D: Uses apply with concat, passing the inner arrays as arguments and flattening one level to [1, 2, 3, 4, 5].
Thus, the two correct statements are:
Reference of JavaScript knowledge documents or Study Guide (concept names only):
Array.prototype.concat (concatenating and one-level flattening behavior) Spread syntax for arrays (...array) Function.prototype.apply (calling a function with a given this value and arguments list) Array flattening by one level using concat with array arguments Distinction between nested arrays and flat arrays in JavaScript
NEW QUESTION # 66
A developer has a fizzbuzz function that, when passed in a number, returns the following:
'fizz' if the number is divisible by 3.
'buzz' if the number is divisible by 5.
'fizzbuzz' if the number is divisible by both 3 and 5.
Empty string if the number is divisible by neither 3 nor 5.
Which two test cases properly test scenarios for the fizzbuzz function?
Answer: A,B
Explanation:
The correct answers are B and D.
A strong test case checks whether the actual output matches the expected behavior of the function.
For Option B:
let res = fizzbuzz(15);
console.assert(res === 'fizzbuzz');
The number 15 is divisible by both 3 and 5:
15 % 3 === 0
15 % 5 === 0
So the correct return value is:
'fizzbuzz'
That makes option B a valid test.
For Option D:
let res = fizzbuzz(Infinity);
console.assert(res === '');
Infinity is not a normal finite divisible number. In JavaScript:
Infinity % 3
Infinity % 5
produce NaN, and comparisons such as:
Infinity % 3 === 0
are false.
So a typical fizzbuzz implementation would not treat Infinity as divisible by 3 or 5, meaning it should return:
''
That makes option D a valid test for the "neither divisible by 3 nor 5" path.
Now compare the incorrect options:
Option A is wrong because:
fizzbuzz(3)
should return:
'fizz'
not:
'buzz'
Option C is wrong because the function specification says the function returns strings such as 'fizz', 'buzz', 'fizzbuzz', or ''. It does not say the function returns numeric NaN.
So this assertion is not appropriate:
console.assert(isNaN(res));
Therefore, the verified answers are B and D.
NEW QUESTION # 67
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: A
NEW QUESTION # 68
Refer to the code below:
01 const myFunction = arr => {
02 return arr.reduce((result, current) => {
03 return result + current;
04 }, 10);
05 }
What is the output of this function when called with an empty array?
Answer: D
Explanation:
Comprehensive and Detailed Explanation From JavaScript Knowledge:
We call:
myFunction([]);
Inside:
arr.reduce((result, current) => {
return result + current;
}, 10);
Key points about Array.prototype.reduce:
Signature: array.reduce(callback, initialValue)
If initialValue is provided and the array is empty, reduce:
Does not call the callback at all.
Simply returns initialValue.
Here:
arr is [] (empty).
initialValue is 10.
So:
No iterations of the callback happen (no elements to process).
The return value is the initial value: 10.
So the actual output is:
myFunction([]) === 10;
Among the options:
A: 0 - incorrect, because the initial value is 10, not 0.
B: Throws an error - reduce throws only if the array is empty and there is no initialValue. Here we have an initial value, so no error.
C: NaN - there is no arithmetic with undefined or invalid values; we just return 10.
D: Returns 5 - the numeric value given is wrong; the correct value is 10.
Given the logic, the correct conceptual result is 10. The option text "Returns 5" is almost certainly a typo for "Returns 10". Since the letter that is intended to represent the correct behavior is D, we keep:
Study Guide / Concept Reference (no links):
Array.prototype.reduce behavior with initialValue
Behavior of reduce on an empty array with and without initialValue
Return value when no iterations run
________________________________________
NEW QUESTION # 69
Given the following code:
Counter = 0;
const logCounter = () => {
console.log(counter);
);
logCounter();
setTimeout(logCOunter, 1100);
setInterval(() => {
Counter++
logCounter();
}, 1000);
What is logged by the first four log statements?
Answer: B
NEW QUESTION # 70
......
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