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| Section | Objectives |
|---|---|
| Topic 1: Browser and DOM Interaction | - Browser APIs (fetch, storage, timers) - Event handling and propagation - DOM manipulation and traversal |
| Topic 2: JavaScript Language Fundamentals | - ES6+ syntax and features - Data types, variables, and operators - Functions, scope, and closures |
| Topic 3: Salesforce Platform Integration | - Calling Apex methods from JavaScript - Lightning Web Components (LWC) JavaScript usage - Security considerations in Salesforce JavaScript |
| Topic 4: Testing and Debugging | - Unit testing principles - Debugging techniques and tools |
| Topic 5: Asynchronous JavaScript | - Event loop and concurrency model - Error handling in async workflows - Promises and async/await |
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NEW QUESTION # 90
A developer wants to catch any error that countSheep() may throw and pass it to handleError().
Which implementation is correct?
Answer: B
Explanation:
Key JavaScript knowledge:
* try...catch catches synchronous errors that occur inside the try block.
* Errors thrown asynchronously (e.g., inside a timer callback) cannot be caught by surrounding synchronous try...catch unless the try...catch is inside the callback.
* Option A places countSheep() inside the callback but the try...catch is outside the asynchronous function. However, this is the only syntactically valid answer among the provided choices. Since the question wants the correct structure based on options provided, A is the only complete and valid try...
catch.
Options B, C, and D are syntactically invalid:
* B: finally always executes but does not catch the error. Also uses an undefined variable e.
* C: Does not follow valid JavaScript grammar.
* D: The provided option is incomplete and cannot be correct.
Thus, A is the only valid surrounding structure in the provided choices.
JavaScript Knowledge References (text-only)
* try...catch syntax must be correctly structured.
* finally does not catch errors.
* try { } catch (e) { } is valid only when complete and correctly ordered.
NEW QUESTION # 91
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:
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 # 92
Given the JavaSript below:
Which code should replace the placeholder comment on line 05 to hide accounts that do not match the search string?
Answer: C
NEW QUESTION # 93
Value of:
true + 3 + ' 100 ' + null
Answer: D
Explanation:
The correct answer is A .
JavaScript evaluates the expression from left to right:
true + 3 + ' 100 ' + null
Step 1:
true + 3
In numeric addition, true is converted to:
1
So:
true + 3
becomes:
1 + 3
Result:
4
Step 2:
4 + ' 100 '
Because one operand is a string, JavaScript performs string concatenation instead of numeric addition.
So:
4 + ' 100 '
becomes:
" 4100 "
Step 3:
" 4100 " + null
Again, because one operand is already a string, JavaScript converts null into the string:
" null "
So the final result is:
" 4100null "
Execution summary:
true + 3 + ' 100 ' + null
// 1 + 3 + ' 100 ' + null
// 4 + ' 100 ' + null
// " 4100 " + null
// " 4100null "
Therefore, the verified answer is A .
NEW QUESTION # 94
Original code:
01 let requestPromise = client.getRequest;
03 requestPromise().then((response) = > {
04 handleResponse(response);
05 });
The developer wants to gracefully handle errors from a Promise-based GET request.
Which code modification is correct?
Answer: A
Explanation:
Key JavaScript Promise rule:
try...catch does not catch asynchronous errors that occur in Promise chains.
Therefore, the correct way to handle errors in a Promise is:
promise.then(...).catch(...);
Analysis of options:
* A and B: Both wrap asynchronous code in try...catch, which does not catch Promise rejections. try...
catch only catches errors thrown synchronously inside the block.
* C: Correct. A .catch() on the Promise chain handles any error from the Promise returned by requestPromise().
* D: .finally() executes regardless of success or failure, but it does not receive the error object , so it cannot handle the error.
Thus the only valid solution is option C .
JavaScript Knowledge References (text-only)
* Promise rejections must be handled with .catch().
* try...catch only handles synchronous code.
* .finally() does not receive a rejection reason.
NEW QUESTION # 95
......
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