Free Salesforce JavaScript-Developer-I Practice Exam with Questions & Answers | Set: 3

The correct answer is A , not D.

Inside myFunction(), the declaration:

var b = 1;

is hoisted to the top of the function scope, but only the declaration is hoisted, not the assignment. So internally, JavaScript treats the function approximately like this:

function myFunction() {

var b;

a = a + b;

b = 1;

}

Now examine this line:

a = a + b;

Before JavaScript can assign a value to a, it must first evaluate the right-hand side:

a + b

At this point:

b

exists locally because var b was hoisted, so its value is:

undefined

However:

a

has not been declared anywhere. Reading an undeclared variable causes a:

ReferenceError

So this line throws an error before assignment happens:

a = a + b;

Because the error occurs on that line, the next line inside the function is never executed:

var b = 1;

Also, because the error is not handled with try...catch, the program stops before reaching:

console.log(a);

console.log(b);

Therefore, the accurate statement is:

Line 02 throws a reference error, therefore line 03 is never executed.

So the verified answer is A .

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); → 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); → 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:

Answer: A

Study Guide / Concept References (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)))

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:

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

From earlier pattern (cars/cats race):

One promise resolves the fastest with " Car 2 completed " (or analogous " Cat 2 completes " ).

Promise.race resolves with the first settled promise’s value.

In .then, value is " Car 2 completed " and:

let result = `${value} the race.`;

// " Car 2 completed the race. "

Thus result is " Car 2 completed the race. " → option A.

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:

Answer: D

Study Guide / Concept References (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

Comprehensive and Detailed Explanation:

Errors thrown inside setTimeout callbacks occur asynchronously, in a different tick of the event loop. A try...catch around setTimeout itself (as in D) cannot catch errors thrown later in the scheduled callback.

To catch errors from countSheep() when it’s called asynchronously, the try...catch must wrap the call inside the timeout callback:

setTimeout(function() {

try {

countSheep();

} catch (e) {

handleError(e);

}

}, 1000);

B uses finally incorrectly and references e out of scope.

C is not valid JavaScript syntax.

D’s catch can only handle synchronous errors thrown before setTimeout returns, not those thrown inside the callback.

In JavaScript, ES modules (files imported using import / export) have special semantics:

All code inside a module is automatically executed in strict mode .

This is specified by the ECMAScript standard and does not require the explicit " use strict " ; directive.

Effects:

You do not need to write " use strict " ; at the top of a module file.

You cannot turn strict mode off for module code.

All module top-level code is treated as strict.

Now evaluate each option:

Option A:

Add the statement use non-strict; before any other statements in the module to enable notstrict mode.

There is no use non-strict; directive in JavaScript.

The only directive recognized by engines is " use strict " ;.

You cannot explicitly enable a “non-strict” mode with such a string.

Option B:

Imported modules are in strict mode whether you declare them as such or not.

This matches the language specification.

ES modules are always in strict mode, automatically.

This is true.

Option C:

Add the statement use strict = false; before any other statements in the module to enable notstrict mode.

use strict = false; is not a directive; it is parsed as a normal expression, and does not affect strict mode.

There is no supported way to disable strict mode in modules.

Option D:

A developer can only reference notStrict() functions from the imported module.

There is no special notStrict() restriction in modules.

Functions exported/imported in modules are simply subject to strict mode rules like the rest of the module’s code.

Therefore, the only correct statement is:

Answer: B

Study Guide / Concept References (no links):

ES modules (import / export)

Automatic strict mode in modules

" use strict " ; directive for scripts vs modules

Inability to disable strict mode in ES module code

setInterval(fn, delay, arg1, arg2, ...) calls fn every delay ms, passing extra arguments.

To call isDeg every 60 seconds with " cat " :

setInterval(isDeg, 60000, ' cat ' );

Why others are wrong:

A/D: isDeg( ' cat ' ) is called immediately, and its return value is scheduled, not the function.

C: setTimeout runs once after 60 seconds, not repeatedly.

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 .

Key details:

var has function scope , not block scope.

Declaration var msg is hoisted to the top of the function, but initialization only happens if the if condition is true.

Effectively, the function behaves like:

function getAvailableilityMessage(item) {

var msg; // hoisted declaration

if (getAvailableility(item)) {

msg = " Username available " ;

}

return msg;

}

Now, given:

getAvailableility( " newUserName " ) returns false.

Execution:

The if condition is false, so the body does not execute.

Therefore msg is declared but never assigned .

In JavaScript, an uninitialized variable that has been declared with var has the value undefined.

Thus:

return msg; // returns undefined

Why other options are wrong:

A: " newUserName " – msg never receives the parameter value; it ' s only set to " Username available " inside the if, which does not run.

B: " msg is not defined " – That kind of error occurs if msg were never declared. Here it is declared via var, so it is defined but undefined.

D: " Username available " – This would require the if branch to run, which it does not when getAvailableility(...) is false.

So the return value is:

undefined

Study Guide Concepts:

var hoisting and function scope

Uninitialized variables default to undefined

Control flow and conditional initialization