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質問 # 105
Refer to the code:
01 let car1 = new Promise((_, reject) =>
02 setTimeout(reject, 2000, "Car 1 crashed in"));
03 let car2 = new Promise(resolve =>
04 setTimeout(resolve, 1500, "Car 2 completed"));
05 let car3 = new Promise(resolve =>
06 setTimeout(resolve, 3000, "Car 3 completed"));
07
08 Promise.race([car1, car2, car3])
09 .then(value => {
10 let result = '$(value) the race.';
11 })
12 .catch(err => {
13 console.log("Race is cancelled.", err);
14 });
What is the value of result when Promise.race executes?
正解：C
解説：
________________________________________
Comprehensive and Detailed Explanation From Exact Extract JavaScript Knowledge Promise.race() returns the result of the first settled promise, whether resolved or rejected.
The promises:
car1 → rejects in 2000 ms
car2 → resolves in 1500 ms
car3 → resolves in 3000 ms
The earliest settled promise is car2 (1500 ms), which resolves with the value:
"Car 2 completed"
Therefore, Promise.race enters the .then() branch with:
value = "Car 2 completed"
Inside the .then():
let result = '$(value) the race.';
This appears to attempt template substitution but uses incorrect syntax.
JavaScript template literals require backticks and ${expression}, for example:
`${value} the race.`
Because the code is incorrect, result is literally the string:
$(value) the race.
However, the question asks:
"What is the value of result when Promise.race executes?"
This refers to the intended value based on which promise wins the race, not the template bug.
The winning value is:
Car 2 completed
So the correct conceptual answer is:
Car 2 completed the race.
________________________________________
JavaScript Knowledge Reference (text-only)
Promise.race() returns the first settled (resolved or rejected) promise.
The earliest resolve here is car2 (1500 ms).
Incorrect template literal syntax does not affect the identity of the winning promise.
質問 # 106
Which two code snippets show working examples of a recursive function?
正解：B、C
解説：
return startNumber;
}
};
(Note: Option D is shown here with corrected syntax: lowercase return and matching parentheses.) Explanation:
A recursive function is a function that calls itself and has a base case to terminate the recursion.
Evaluate each option:
Option A:
const sumToTen = numVar => {
if (numVar < 0)
return;
return sumToTen(numVar + 1);
};
This function calls itself: sumToTen(numVar + 1) - so it is recursive.
However, the base condition is if (numVar < 0) return;.
If you call sumToTen(0):
numVar < 0 is false, so it calls sumToTen(1), then sumToTen(2), and so on, incrementing forever.
There is no condition to stop the recursion when numVar increases; it will eventually cause a stack overflow.
This code does not represent a properly working recursive function with a valid termination for increasing values and is not a good example of correct recursion.
Option B:
function factorial(numVar) {
if (numVar < 0) return;
if (numVar === 0) return 1;
return numVar - 1;
}
This function does not call itself anywhere.
It has conditional returns, but there is no recursive call such as factorial(numVar - 1).
Therefore, it is not recursive at all.
Option C:
const factorial = numVar => {
if (numVar < 0) return;
if (numVar === 0) return 1;
return numVar * factorial(numVar - 1);
};
This is a classic recursive factorial implementation.
It calls itself with a smaller argument: factorial(numVar - 1).
Base cases:
If numVar < 0, it simply returns (could be treated as invalid input).
If numVar === 0, it returns 1, which is the mathematical definition of 0! (zero factorial).
For positive integers, it correctly multiplies numVar by factorial(numVar - 1) until it reaches the base case.
This is a correct and working recursive function.
Option D (corrected):
let countingDown = function(startNumber) {
if (startNumber > 0) {
console.log(startNumber);
return countingDown(startNumber - 1);
} else {
return startNumber;
}
};
This function also calls itself: countingDown(startNumber - 1).
Base case:
When startNumber is not greater than 0 (i.e., 0 or negative), it returns startNumber and stops recursing.
For example, countingDown(3) would:
Log 3, call countingDown(2)
Log 2, call countingDown(1)
Log 1, call countingDown(0)
At 0, it hits the else branch and returns 0, ending the recursion.
This is a valid working recursive function structure (once syntax is corrected).
Therefore, the snippets that show working recursive functions are:
Answe r: C, D
Study Guide / Concept Reference (no links):
Definition of recursion: a function calling itself
Base case vs recursive step
Recursive factorial implementation
Recursive countdown example
Importance of a terminating condition to avoid infinite recursion
質問 # 107
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?
正解：B、D
解説：
}
Explanation:
Comprehensive and Detailed Explanation From Exact Extract JavaScript knowledge:
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
________________________________________
質問 # 108
Correct implementation of try...catch for countsDeep():
正解：C
解説：
The correct answer is B because countsDeep() is executed inside the callback function passed to setTimeout(), and the try...catch block is also placed inside that same callback.
In JavaScript, setTimeout() schedules a function to run later. The outer code finishes first, and the callback runs asynchronously after the delay. Because of this, a try...catch block placed outside setTimeout() cannot catch errors thrown later inside the callback.
Correct logic:
setTimeout(function() {
try {
countsDeep();
} catch (e) {
handleError(e);
}
}, 1000);
Here, when countsDeep() runs after 1000 milliseconds, any error thrown by countsDeep() happens inside the try block. Therefore, the catch (e) block can catch that error and pass it to handleError(e).
Why the other options are incorrect:
A is incorrect because the syntax is invalid JavaScript. A valid try...catch structure must be:
try {
// code
} catch (e) {
// handle error
}
Option A incorrectly writes:
} handleError (e){
catch(e);
}
That is not valid try...catch syntax.
C is incorrect because the try...catch surrounds only the call to setTimeout(), not the later execution of countsDeep(). If countsDeep() throws an error after the timer expires, the outer catch block will not catch it.
D is incorrect for the same reason as C. In the original question, D also had a typing error: it used countSheep() instead of countsDeep(). Even after correcting that typing error, D is still incorrect because the try...catch is outside the asynchronous callback.
質問 # 109
Refer to the following code:
function test (val) {
If (val === undefined) {
return 'Undefined values!' ;
}
if (val === null) {
return 'Null value! ';
}
return val;
}
Let x;
test(x);
What is returned by the function call on line 13?
正解：D
質問 # 110
......
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