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Salesforce JavaScript-Developer-I (Salesforce Certified JavaScript Developer I) Exam is a certification exam designed for developers who want to demonstrate their proficiency in developing client-side JavaScript applications using the Salesforce platform. JavaScript-Developer-I exam is highly recommended for those who are interested in pursuing a career in Salesforce development as it is a key requirement for many job positions in the field.
Salesforce JavaScript-Developer-I Certification Exam is a multiple-choice exam that consists of 60 questions. Candidates have 105 minutes to complete the exam, and they must achieve a passing score of 68% to earn the certification. JavaScript-Developer-I exam can be taken online or in person at a testing center.
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NEW QUESTION # 66
A developer has a web server running with Node.js. The command to start the web server is node server.js.
The web server started having
latency issues. Instead of a one second turnaround for web requests, the developer now sees a five second turnaround.
Which command can the web developer run to see what the module is doing during the latency period?
Answer: B
NEW QUESTION # 67
Refer to the code snippet below:
Let array = [1, 2, 3, 4, 4, 5, 4, 4];
For (let i =0; i < array.length; i++)
if (array[i] === 4) {
array.splice(i, 1);
}
}
What is the value of array after the code executes?
Answer: A
Explanation:
NEW QUESTION # 68
Refer to the code below:
console.log(''start);
Promise.resolve('Success') .then(function(value){
console.log('Success');
});
console.log('End');
What is the output after the code executes successfully?
Answer: D
NEW QUESTION # 69
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: A,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:
The answer: A, D
References 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 # 70
Refer to the code below:
Const searchTest = 'Yay! Salesforce is amazing!" ;
Let result1 = searchText.search(/sales/i);
Let result 21 = searchText.search(/sales/i);
console.log(result1);
console.log(result2);
After running this code, which result is displayed on the console?
Answer: A
NEW QUESTION # 71
......
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