Execution order of multiple setTimeout without delays in angular
up vote
1
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What is the execution order of multiple setTimeout without delays in angular2+?
Example:
setTimeout(() => console.log('1'));
setTimeout(() => console.log('2'));
Questions:
(1) Is 1 always guaranteed to be printed before 2?
(2) If so, what part of the spec says so?
(3) This is called in Angular2+ code. Is angular overriding setTimeout's default behaviour?
How it differs from other questions:
This question (1) asks about the case without delay, and (2) asks it in angular context (in case angular overrides setTimeout).
Non-questions:
I know console.log might (or might not) be async or whatever; this question is not related to that. You can assume console.log is sync.
javascript angular settimeout ngzone
asked Nov 11 at 0:57
Hossam El-Deen
112110
add a comment |
up vote
1
down vote
favorite
What is the execution order of multiple setTimeout without delays in angular2+?
Example:
setTimeout(() => console.log('1'));
setTimeout(() => console.log('2'));
Questions:
(1) Is 1 always guaranteed to be printed before 2?
(2) If so, what part of the spec says so?
(3) This is called in Angular2+ code. Is angular overriding setTimeout's default behaviour?
How it differs from other questions:
This question (1) asks about the case without delay, and (2) asks it in angular context (in case angular overrides setTimeout).
Non-questions:
I know console.log might (or might not) be async or whatever; this question is not related to that. You can assume console.log is sync.
javascript angular settimeout ngzone
asked Nov 11 at 0:57
Hossam El-Deen
112110
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
What is the execution order of multiple setTimeout without delays in angular2+?
Example:
setTimeout(() => console.log('1'));
setTimeout(() => console.log('2'));
Questions:
(1) Is 1 always guaranteed to be printed before 2?
(2) If so, what part of the spec says so?
(3) This is called in Angular2+ code. Is angular overriding setTimeout's default behaviour?
How it differs from other questions:
This question (1) asks about the case without delay, and (2) asks it in angular context (in case angular overrides setTimeout).
Non-questions:
I know console.log might (or might not) be async or whatever; this question is not related to that. You can assume console.log is sync.
javascript angular settimeout ngzone
asked Nov 11 at 0:57
Hossam El-Deen
112110
What is the execution order of multiple setTimeout without delays in angular2+?
Example:
setTimeout(() => console.log('1'));
setTimeout(() => console.log('2'));
Questions:
(1) Is 1 always guaranteed to be printed before 2?
(2) If so, what part of the spec says so?
(3) This is called in Angular2+ code. Is angular overriding setTimeout's default behaviour?
How it differs from other questions:
This question (1) asks about the case without delay, and (2) asks it in angular context (in case angular overrides setTimeout).
Non-questions:
I know console.log might (or might not) be async or whatever; this question is not related to that. You can assume console.log is sync.
javascript angular settimeout ngzone
javascript angular settimeout ngzone
asked Nov 11 at 0:57
Hossam El-Deen
112110
asked Nov 11 at 0:57
Hossam El-Deen
112110
asked Nov 11 at 0:57
Hossam El-Deen
112110
asked Nov 11 at 0:57
Hossam El-Deen
112110
asked Nov 11 at 0:57
Hossam El-Deen
112110
112110
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
1) Yes
2) Step 16 of https://html.spec.whatwg.org/multipage/timers-and-user-prompts.html which says:
Wait until any invocations of this algorithm that had the same method
context, that started before this one, and whose timeout is equal to
or less than this one's, have completed.
3) No, Angular does not override it.
answered Nov 11 at 1:03
user184994
10.8k11526
And for future readers wondering how the exotic step 16 is related: It's because, according to step 14, the rest of steps are parallel. So, actually the example in the original question is similar to the exotic example in step 16, in that the rest of the steps of the 2setTimeoutcalls are indeed running in parallel, but due to step 16, the first will finish first (i.e., executing step 18 & getting to the queue first).
– Hossam El-Deen
Nov 11 at 1:23
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
1) Yes
2) Step 16 of https://html.spec.whatwg.org/multipage/timers-and-user-prompts.html which says:
Wait until any invocations of this algorithm that had the same method
context, that started before this one, and whose timeout is equal to
or less than this one's, have completed.
3) No, Angular does not override it.
answered Nov 11 at 1:03
user184994
10.8k11526
And for future readers wondering how the exotic step 16 is related: It's because, according to step 14, the rest of steps are parallel. So, actually the example in the original question is similar to the exotic example in step 16, in that the rest of the steps of the 2setTimeoutcalls are indeed running in parallel, but due to step 16, the first will finish first (i.e., executing step 18 & getting to the queue first).
– Hossam El-Deen
Nov 11 at 1:23
add a comment |
up vote
1
down vote
accepted
1) Yes
2) Step 16 of https://html.spec.whatwg.org/multipage/timers-and-user-prompts.html which says:
Wait until any invocations of this algorithm that had the same method
context, that started before this one, and whose timeout is equal to
or less than this one's, have completed.
3) No, Angular does not override it.
answered Nov 11 at 1:03
user184994
10.8k11526
And for future readers wondering how the exotic step 16 is related: It's because, according to step 14, the rest of steps are parallel. So, actually the example in the original question is similar to the exotic example in step 16, in that the rest of the steps of the 2setTimeoutcalls are indeed running in parallel, but due to step 16, the first will finish first (i.e., executing step 18 & getting to the queue first).
– Hossam El-Deen
Nov 11 at 1:23
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
1) Yes
2) Step 16 of https://html.spec.whatwg.org/multipage/timers-and-user-prompts.html which says:
Wait until any invocations of this algorithm that had the same method
context, that started before this one, and whose timeout is equal to
or less than this one's, have completed.
3) No, Angular does not override it.
answered Nov 11 at 1:03
user184994
10.8k11526
1) Yes
2) Step 16 of https://html.spec.whatwg.org/multipage/timers-and-user-prompts.html which says:
Wait until any invocations of this algorithm that had the same method
context, that started before this one, and whose timeout is equal to
or less than this one's, have completed.
3) No, Angular does not override it.
answered Nov 11 at 1:03
user184994
10.8k11526
answered Nov 11 at 1:03
user184994
10.8k11526
answered Nov 11 at 1:03
user184994
10.8k11526
answered Nov 11 at 1:03
user184994
10.8k11526
10.8k11526
And for future readers wondering how the exotic step 16 is related: It's because, according to step 14, the rest of steps are parallel. So, actually the example in the original question is similar to the exotic example in step 16, in that the rest of the steps of the 2setTimeoutcalls are indeed running in parallel, but due to step 16, the first will finish first (i.e., executing step 18 & getting to the queue first).
– Hossam El-Deen
Nov 11 at 1:23
add a comment |
And for future readers wondering how the exotic step 16 is related: It's because, according to step 14, the rest of steps are parallel. So, actually the example in the original question is similar to the exotic example in step 16, in that the rest of the steps of the 2setTimeoutcalls are indeed running in parallel, but due to step 16, the first will finish first (i.e., executing step 18 & getting to the queue first).
– Hossam El-Deen
Nov 11 at 1:23
And for future readers wondering how the exotic step 16 is related: It's because, according to step 14, the rest of steps are parallel. So, actually the example in the original question is similar to the exotic example in step 16, in that the rest of the steps of the 2
setTimeout calls are indeed running in parallel, but due to step 16, the first will finish first (i.e., executing step 18 & getting to the queue first).– Hossam El-Deen
Nov 11 at 1:23
And for future readers wondering how the exotic step 16 is related: It's because, according to step 14, the rest of steps are parallel. So, actually the example in the original question is similar to the exotic example in step 16, in that the rest of the steps of the 2
setTimeout calls are indeed running in parallel, but due to step 16, the first will finish first (i.e., executing step 18 & getting to the queue first).– Hossam El-Deen
Nov 11 at 1:23
add a comment |
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