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NEW QUESTION # 49
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: B,C
Explanation:
:
First, recall the fizzbuzz rules:
If n divisible by 3 and not by 5 → return 'fizz'.
If n divisible by 5 and not by 3 → return 'buzz'.
If n divisible by both 3 and 5 → return 'fizzbuzz'.
Otherwise → return '' (empty string).
Evaluate each test:
Option A:
let res = fizzbuzz(true);
console.assert(res === '');
In JavaScript, when using arithmetic operations with true, it is coerced to the number 1. A typical implementation of fizzbuzz would treat the argument as a number and compute:
true % 3 → 1 % 3 → 1
true % 5 → 1 % 5 → 1
So true is effectively treated as 1, which is divisible by neither 3 nor 5. The function should return ''. The assertion res === '' will pass.
This test covers the scenario "neither divisible by 3 nor 5".
Option B:
let res = fizzbuzz(3);
console.assert(res === '');
3 is divisible by 3 but not by 5.
According to fizzbuzz rules, fizzbuzz(3) should return 'fizz'.
This test expects '', so it is incorrect; the assertion would fail in a correct implementation.
Option C:
let res = fizzbuzz(5);
console.assert(res === 'fizz');
5 is divisible by 5 but not by 3.
According to fizzbuzz rules, fizzbuzz(5) should return 'buzz'.
This test expects 'fizz', so it is incorrect.
Option D:
let res = fizzbuzz(15);
console.assert(res === 'fizzbuzz');
15 is divisible by both 3 and 5.
According to fizzbuzz rules, fizzbuzz(15) should return 'fizzbuzz'.
This assertion correctly matches the expected result, so it is a proper test case.
Thus, the two test cases that correctly test valid fizzbuzz behavior are:
Answe r: A, D
Study Guide / Concept Reference (no links):
Modulo operator (%) for divisibility checks
Truthy/number coercion of boolean true in arithmetic (Number(true) === 1) Designing unit tests with console.assert Typical fizzbuzz logic structure and expected outputs
NEW QUESTION # 50
Given the following code:
01 let x = null;
02 console.log(typeof x);
What is the output of line 02?
Answer: A
Explanation:
This question focuses on the typeof operator and how JavaScript represents the type of the value null.
Declaration on line 1:
let x = null;
Here:
x is explicitly assigned the value null.
In JavaScript, null is a special primitive value that represents "no value" or "empty value".
Line 2:
console.log(typeof x);
The typeof operator returns a string indicating the type of the operand. For example:
typeof 1 returns "number"
typeof "hello" returns "string"
typeof true returns "boolean"
typeof undefined returns "undefined"
typeof {} returns "object"
typeof [] returns "object" (arrays are objects)
typeof function() {} returns "function"
For null, due to an early design decision in JavaScript, the result is:
typeof null === "object"
This is a known historical quirk of the language:
Although null is a primitive and conceptually means "no object", typeof null returns the string "object".
Therefore, typeof x when x is null evaluates to "object".
So on line 2, the value printed is:
"object"
This matches option A.
The incorrect options are:
B . "undefined": would be typeof x if x were undefined, not null.
C . "x": typeof returns a string describing the type of the value, not the variable name.
D . "null": although intuitive, JavaScript does not return "null" for typeof null; it returns "object" instead.
Therefore, the correct answer is:
Answe r: A
JavaScript knowledge / study guide reference concepts:
typeof operator and its return values
Primitive types in JavaScript (null, undefined, number, string, boolean, symbol, bigint) Historical behavior: typeof null returning "object" Difference between null and undefined in JavaScript
NEW QUESTION # 51
A developer has an ErrorHandler module that contains multiple functions.
What kind of export should be leveraged so that multiple functions can be used?
Answer: B
NEW QUESTION # 52
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: B,C
Explanation:
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 # 53
Refer to the following object:
01 const cat = {
02 firstName: 'Fancy',
03 lastName: 'Whiskers',
04 get fullName(){
05 return this.firstName + ' ' + this.lastName;
06 }
07 };
How can a developer access the fullName property for cat?
Answer: A
Explanation:
fullName is defined as a getter:
get fullName() { ... }
Getters are accessed like properties, not like functions.
So:
cat.fullName; // "Fancy Whiskers"
No parentheses.
Why others are wrong:
A: cat.fullName() tries to call the returned string as a function.
B and C: These properties (get, function) do not exist; they are misinterpretations of syntax.
NEW QUESTION # 54
......
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