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質問 # 148
Given the code below:
01 const delay = async delay => {
02 return new Promise((resolve, reject) => {
03 console.log(1);
04 setTimeout(resolve, delay);
05 });
06 };
07
08 const callDelay = async () => {
09 console.log(2);
10 const yup = await delay(1000);
11 console.log(3);
12 };
13
14 console.log(4);
15 callDelay();
16 console.log(5);
What is logged to the console?
正解：A
解説：
Comprehensive and Detailed Explanation From Exact Extract JavaScript Knowledge:
Execution order:
Top-level code runs synchronously:
Line 14: console.log(4); → logs 4.
Line 15: callDelay(); is called.
Inside callDelay:
Line 9: console.log(2); → logs 2.
Line 10: await delay(1000);:
Calls delay(1000).
Inside delay(1000):
Line 3: console.log(1); → logs 1.
Line 4: setTimeout(resolve, delay); schedules resolve in 1000 ms.
delay returns a pending Promise. await pauses callDelay here and returns control to the event loop.
Back to top-level:
Line 16: console.log(5); → logs 5.
So synchronous log sequence is: 4, 2, 1, 5.
After ~1000 ms:
The setTimeout in delay resolves the Promise.
The await in callDelay resumes.
Line 11: console.log(3); → logs 3.
Final log order: 4 2 1 5 3.
Both A and B show the same sequence; one must be chosen, so A is correct.
Concepts: async/await flow, Promise resolution timing, event loop, and ordering of synchronous vs timer callbacks.
________________________________________
質問 # 149
Refer to the following object.
How can a developer access the fullName property for dog?
正解：A
質問 # 150
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:
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 # 151
Refer to the code below:
Const resolveAfterMilliseconds = (ms) => Promise.resolve (
setTimeout ((=> console.log(ms), ms ));
Const aPromise = await resolveAfterMilliseconds(500);
Const bPromise = await resolveAfterMilliseconds(500);
Await aPromise, wait bPromise;
What is the result of running line 05?
正解：C
質問 # 152
Refer to the code below:
01 let timedFunction = () => {
02 console.log('Timer called.');
03 };
04
05 let timerId = setInterval(timedFunction, 1000);
Which statement allows a developer to cancel the scheduled timed function?
正解：A
解説：
Comprehensive and Detailed Explanation From JavaScript Knowledge:
The code:
let timerId = setInterval(timedFunction, 1000);
setInterval schedules timedFunction to run every 1000 ms.
It returns an interval ID (here stored in timerId), which is used to cancel the interval later.
To cancel:
Use clearInterval(timerId);
This is the standard browser (and Node.js) API:
let id = setInterval(fn, delay);
clearInterval(id); stops future executions of that interval.
Check other options:
B . removeInterval(timerId);
There is no removeInterval function in the standard JavaScript timer API.
C . removeInterval(timedFunction);
Again, no such function; and timers are cancelled by ID, not by the callback function reference.
D . clearInterval(timedFunction);
clearInterval expects the ID returned by setInterval, not the callback function.
Passing the function does not cancel the timer.
Therefore, the correct statement is:
Answe r: A
Study Guide / Concept Reference (no links):
Timer APIs: setInterval and clearInterval
Relationship between timer ID and cancellation
Difference between interval ID and callback function
Basic async timing patterns in JavaScript
質問 # 153
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