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問題 #124
A developer is asked to fix some bugs reported by users. To do that, the developer adds a breakpoint for debugging.
01 function Car(maxSpeed, color) {
02 this.maxSpeed = maxSpeed;
03 this.color = color;
04 }
05 let carSpeed = document.getElementById('carSpeed');
06 debugger;
07 let fourWheels = new Car(carSpeed.value, 'red');
When the code execution stops at the breakpoint on line 06, which two types of information are available in the browser console?
答案：B,C
解題說明：
Comprehensive and Detailed Explanation From Exact Extract JavaScript knowledge:
When execution hits the debugger; statement on line 06, JavaScript execution pauses at that point. Most modern browsers (for example, using DevTools) allow you to inspect the current scope, DOM, and various browser APIs in the console.
Let's look at what is available precisely at line 06.
Current code state at the breakpoint:
Lines 01-04 define the Car constructor function.
Line 05 executes:
let carSpeed = document.getElementById('carSpeed');
So at line 06:
carSpeed is a variable containing a reference to a DOM element (the element with id "carSpeed").
Line 07 has not executed yet:
let fourWheels = new Car(carSpeed.value, 'red');
So fourWheels has not been created and is not accessible yet (it is in the temporal dead zone for the let declaration).
Now check each option:
Option A:
"A variable displaying the number of instances created for the Car object" This code does not implement any mechanism to count instances of Car.
There is no static property, global counter, or similar variable tracking the number of Car instances.
At line 06, no instance of Car has even been created yet (new Car(...) is on line 07, which has not run).
Therefore, there is no such variable by default in JavaScript or DevTools.
This option is not available.
Option B:
"The information stored in the window.localStorage property"
At a breakpoint, the console is fully usable to inspect global objects.
window.localStorage is always accessible from the console (assuming standard browser context).
You can type:
window.localStorage
and inspect key/value pairs stored there.
This is independent of the current function or breakpoint line; localStorage is part of the Web Storage API on the window object.
So this information is indeed available in the console at line 06.
Option C:
"The values of the carSpeed and fourWheels variables"
At line 06:
carSpeed has been declared and assigned (line 05), so it is available, and its value (a DOM element) can be inspected.
fourWheels is declared on line 07 with let and has not yet been executed.
Variables declared with let and const are in a temporal dead zone before their declaration line completes.
At line 06, fourWheels is not yet initialized, and attempting to access it would result in a ReferenceError.
Thus, you can inspect carSpeed but not fourWheels. The option explicitly says "the values of the carSpeed and fourWheels variables", which is not correct at this breakpoint, because fourWheels is not available yet.
Option D:
"The style, event listeners and other attributes applied to the carSpeed DOM element" carSpeed holds a reference to a DOM element (document.getElementById('carSpeed')).
In DevTools, you can inspect this element in several ways:
Typing carSpeed in the console.
Inspecting it in the Elements panel.
From the console or Elements panel, you can view:
Its style (inline styles and computed styles).
Event listeners attached to it (using event listener viewer in DevTools).
Other attributes (id, class, etc.).
All of this is accessible at the point where execution is paused.
Therefore, this information is available at the breakpoint.
Conclusion:
B is available (global window.localStorage).
D is available (full inspection of the carSpeed DOM element).
A is not present in this code or environment.
C is partially wrong (only carSpeed exists; fourWheels does not yet).
So the two correct answers are:
Reference of JavaScript knowledge documents or Study Guide (concept names only):
debugger statement and pausing execution
JavaScript execution context and scope at a breakpoint
let declarations and temporal dead zone
Browser DevTools console and inspection of variables
DOM access via document.getElementById and inspection of elements
Web Storage API: window.localStorage
問題 #125
Refer to the code below:
01 const server = require('server');
02 /* Insert code here */
A developer imports a library that creates a web server. Theimported library uses events and callbacks to start the servers Which code should be inserted at the line 03 to set up an event and start the web server ?
答案：A
問題 #126
Refer to the code below:
01 const addBy = ?
02 const addByEight = addBy(8);
03 const sum = addByEight(50);
Which two functions can replace line 01 and return 58 to sum?
答案：A,C
解題說明：
}
Explanation:
We want:
const addByEight = addBy(8);
const sum = addByEight(50);
And we need sum to be 58.
That means:
addBy(8) must return a function.
That returned function, when called with 50, must compute 8 + 50.
So addBy must be a higher-order function that returns another function capturing num1 and later using it with num2 (a closure).
Option A
const addBy = function(num1) {
return function(num2) {
return num1 + num2;
}
}
Step-by-step:
Call addBy(8):
num1 is 8.
The function returns an inner function: function(num2) { return num1 + num2; }.
So addByEight becomes this inner function, with num1 closed over as 8.
Then call addByEight(50):
num2 is 50.
The body computes num1 + num2, i.e., 8 + 50 = 58.
Therefore, with Option A in place:
const addByEight = addBy(8); // returns inner function
const sum = addByEight(50); // 58
So sum is 58. Option A is correct.
Option D (corrected)
const addBy = (num1) => {
return function(num2) {
return num1 + num2;
}
}
This is essentially the same logic expressed with an arrow function for the outer function:
Call addBy(8):
num1 is 8.
Returns the inner function function(num2) { return num1 + num2; }.
Call addByEight(50):
num2 is 50.
Computes num1 + num2 → 8 + 50 = 58.
So again:
const addByEight = addBy(8); // inner function with num1 = 8
const sum = addByEight(50); // 58
Option D is also correct.
Thus, A and D are the two functions that satisfy the requirement.
Why B and C are incorrect
Option B:
const addBy = function(num1) {
return num1 * num2;
}
This does not return a function; it returns a value.
addBy(8) returns 8 * num2, but num2 is not defined in this scope, which would cause a ReferenceError.
Also, addByEight would be a number (if num2 existed), not a function, so addByEight(50) would fail.
Option C:
const addBy = (num1) => num1 + num2;
Again, addBy returns a value, not a function.
addBy(8) returns 8 + num2, but num2 is not defined, so this is also invalid due to ReferenceError.
addByEight would be a number (or error), not a function.
Neither B nor C creates the required closure nor returns a function to be called later.
Reference / Study Guide concepts (no links):
Higher-order functions in JavaScript
Closures: inner functions capturing outer variables (num1)
Arrow functions vs function expressions
Returning functions from functions (function factories / currying)
Scope and ReferenceError when a variable is not defined
問題 #127
Refer to the following code block:
01 let array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11];
02 let output = 0;
03
04 for (let num of array) {
05 if (output > 10) {
06 break;
07 }
08 if (num % 2 == 0) {
09 continue;
10 }
11 output += num;
12 }
What is the value of output after the code executes?
答案：D
解題說明：
Comprehensive and Detailed Explanation From Exact Extract JavaScript knowledge:
This code uses:
A for...of loop to iterate over values in array.
break to exit the loop entirely when output > 10.
continue to skip even numbers.
It sums only certain numbers into output.
Let's walk through the loop step by step.
Initial values:
array = [1,2,3,4,5,6,7,8,9,10,11]
output = 0
Loop: for (let num of array) { ... }
First iteration: num = 1
Line 05: if (output > 10) → 0 > 10 is false → no break.
Line 08: if (num % 2 == 0) → 1 % 2 == 1, not 0, so false → no continue.
Line 11: output += num → output = 0 + 1 = 1.
Second iteration: num = 2
output > 10 → 1 > 10 is false → no break.
num % 2 == 0 → 2 % 2 == 0, so true → continue.
Because of continue, line 11 is skipped.
output remains 1.
Third iteration: num = 3
output > 10 → 1 > 10 is false.
num % 2 == 0 → 3 % 2 == 1, false → no continue.
output += num → output = 1 + 3 = 4.
Fourth iteration: num = 4
output > 10 → 4 > 10 is false.
num % 2 == 0 → 4 % 2 == 0, true → continue.
Skip sum; output remains 4.
Fifth iteration: num = 5
output > 10 → 4 > 10 is false.
num % 2 == 0 → 5 % 2 == 1, false.
output += num → output = 4 + 5 = 9.
Sixth iteration: num = 6
output > 10 → 9 > 10 is false.
num % 2 == 0 → 6 % 2 == 0, true → continue.
output remains 9.
Seventh iteration: num = 7
output > 10 → 9 > 10 is false.
num % 2 == 0 → 7 % 2 == 1, false.
output += num → output = 9 + 7 = 16.
Eighth iteration would be num = 8, but:
At the top of the loop body, line 05 is checked again:
if (output > 10) → 16 > 10 is true, so break; is executed.
When break runs:
The loop terminates immediately.
No further iterations (for num = 8, 9, 10, 11) are executed.
Therefore, output stays at 16.
Final value of output after the loop ends is 16.
This matches option A.
Why other options do not match:
B . 25: Would require adding more odd numbers (e.g., 9, 11) after 7, but the loop stops early due to output > 10.
C . 11: Would be smaller; the actual sum of 1 + 3 + 5 + 7 until break is 16.
D . 36: Would require summing many more values (e.g., most or all odd numbers up to 11), but again, the break condition stops the loop once output exceeds 10.
So:
Answe r: A
JavaScript knowledge / Study Guide references (concept names only, no links):
for...of loop over arrays
break statement in loops (terminating a loop early)
continue statement in loops (skipping to the next iteration)
Modulo operator % to test even and odd numbers
Step-by-step execution and control flow in loops
問題 #128
A Developer wrote the following code to test a sum3 function that takes in an array of numbers and returns the sum of the first three number in the array, The test passes:
Let res = sum2([1, 2, 3 ]) ;
console.assert(res === 6 );
Res = sum3([ 1, 2, 3, 4]);
console.assert(res=== 6);
A different developer made changes to the behavior of sum3 to instead sum all of the numbers present in the array. The test passes:
Which two results occur when running the test on the updated sum3 function ?
Choose 2 answers
答案：C,D
問題 #129
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從Google Drive中免費下載最新的PDFExamDumps JS-Dev-101 PDF版考試題庫：https://drive.google.com/open?id=1R8oVsdaTpxYyHmyhYL46QDSUkQKmdNhn