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Salesforce Exam Javascript Developer I Topic 9 Question 64 Discussion

Actual exam question for Salesforce's Javascript Developer I exam
Question #: 64
Topic #: 9
[All Javascript Developer I Questions]

Given the following code:

Counter = 0;

const logCounter = () => {

console.log(counter);

);

logCounter();

setTimeout(logCOunter, 1100);

setInterval(() => {

Counter++

logCounter();

}, 1000);

What is logged by the first four log statements?

Show Suggested Answer Hide Answer
Suggested Answer: B

by Cordelia at Jul 30, 2023, 12:25 AM

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Edwin
1 months ago
Someone should tell the exam writers that their code needs a little more 'set' and a little less 'time-out'.
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Stephaine
5 days ago
A) 0 0 1 2
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Mila
1 months ago
I bet the exam writers were chuckling to themselves when they came up with this one. 'Let's see if they can keep up with our asynchronous antics!'
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Dante
2 months ago
This question is a real head-scratcher! I'm going to go with B) 0 1 2 3 because it just makes the most sense to me. But who knows, maybe the exam writers are trying to trick us with this one.
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Galen
2 months ago
I'm confused. Shouldn't the answer be C) 0 1 1 2? The first logCounter() call logs 0, the setTimeout call logs 1, and the setInterval call increments Counter and logs the new value, but the last log statement should be 2, not 3.
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Brande
14 days ago
B) 0 1 2 3
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Leonida
18 days ago
No, that's not correct. The first logCounter() call logs 0, the setTimeout call logs 0 again, and the setInterval call increments Counter and logs the new value.
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Patria
28 days ago
A) 0 0 1 2
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Kami
2 months ago
I'm going with D) 0 1 2 2. The setInterval call increments Counter and logs the new value, but the last log statement should be 2 since the interval is 1 second and the setTimeout is 1.1 seconds.
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Denna
19 hours ago
I'm not sure about this one, but I think it might be B) 0 1 2 3. The setInterval increments Counter every second, so the log statements should show the increasing values.
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Ryan
2 days ago
I agree with you, it should be D) 0 1 2 2. The setTimeout call logs the initial value of Counter, which is 0. Then the setInterval increments it, resulting in 0 1 2 2.
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Tresa
3 days ago
I'm going with D) 0 1 2 2. The setInterval call increments Counter and logs the new value, but the last log statement should be 2 since the interval is 1 second and the setTimeout is 1.1 seconds.
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Mona
9 days ago
I think it's A) 0 0 1 2. The first log statement logs the initial value of Counter, which is 0. Then the setInterval increments Counter and logs it, so it goes 0 0 1 2.
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Truman
14 days ago
I agree with D) 0 1 2 2. The setTimeout call logs the Counter after the interval.
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Thora
29 days ago
I'm going with D) 0 1 2 2. The setInterval call increments Counter and logs the new value.
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Dewitt
1 months ago
I believe it's B) 0 1 2 3. The Counter is incremented before it is logged.
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Golda
1 months ago
I think it's A) 0 0 1 2. The Counter is logged before it is incremented.
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Michel
2 months ago
That's an interesting point, Matthew. I can see how that could be the case. Let's see what others think.
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Pura
2 months ago
I think the answer is B) 0 1 2 3. The code sets the Counter variable to 0, and then the first logCounter() call logs 0. The setTimeout call logs 0 after 1.1 seconds, and the setInterval call increments Counter and logs the new value every 1 second, so the output should be 0 1 2 3.
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Margret
1 months ago
Finally, the setInterval call increments Counter and logs the new value every 1 second.
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Alex
2 months ago
Then the setTimeout call logs 0 after 1.1 seconds.
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Tabetha
2 months ago
Yes, that's correct. The first log statement logs 0.
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Torie
2 months ago
I think the answer is B) 0 1 2 3.
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Matthew
2 months ago
I disagree, I believe the answer is C) 0 1 1 2 because the logCounter function is called twice before the counter is incremented.
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Michel
3 months ago
I think the answer is B) 0 1 2 3 because the counter starts at 0 and then increments by 1 each time it is logged.
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