Mixing
Swift Async print order?
Clash Royale CLAN TAG#URR8PPP
up vote
7
down vote
favorite
Does this always print in the order of 1 5 2 4 3?
print("1")
DispatchQueue.main.async
print("2")
DispatchQueue.main.async
print(3)
print("4")
print("5")
I feel the answer is no, but I cannot explain it and hope someone could clarify my understanding. Thank you!
swift asynchronous grand-central-dispatch dispatch-async
asked 1 hour ago
Unikorn
4301817
 |Â
show 1 more comment
up vote
7
down vote
favorite
Does this always print in the order of 1 5 2 4 3?
print("1")
DispatchQueue.main.async
print("2")
DispatchQueue.main.async
print(3)
print("4")
print("5")
I feel the answer is no, but I cannot explain it and hope someone could clarify my understanding. Thank you!
swift asynchronous grand-central-dispatch dispatch-async
asked 1 hour ago
Unikorn
4301817
1
Yes, it is obvious for 1 ,5 after that it switches to main queue prints 2, Then you again request main queue task but it will not wait for it and prints 4 and then 3
– Prashant Tukadiya
1 hour ago
3
You need to clarify one thing, whetherprint("1")is executed in the main thread or not.
– OOPer
1 hour ago
1
why would it matter? It would always execute first.
– Unikorn
1 hour ago
I suspect is due to the execution order: print("1") // 1. executes prints 1 DispatchQueue.main.async // 2. switches to main thread print("2") // 4. executes print 3 DispatchQueue.main.async // 5. switches to main thread print(3) // 7. executes print 3 print("4") // 6. executes print 4 print("5") // 3. executes print 5
– Unikorn
44 mins ago
@OOPer, please forgive my ignorant for my earlier remark thinking it will always execute first.
– Unikorn
12 mins ago
 |Â
show 1 more comment
up vote
7
down vote
favorite
up vote
7
down vote
favorite
Does this always print in the order of 1 5 2 4 3?
print("1")
DispatchQueue.main.async
print("2")
DispatchQueue.main.async
print(3)
print("4")
print("5")
I feel the answer is no, but I cannot explain it and hope someone could clarify my understanding. Thank you!
swift asynchronous grand-central-dispatch dispatch-async
asked 1 hour ago
Unikorn
4301817
Does this always print in the order of 1 5 2 4 3?
print("1")
DispatchQueue.main.async
print("2")
DispatchQueue.main.async
print(3)
print("4")
print("5")
I feel the answer is no, but I cannot explain it and hope someone could clarify my understanding. Thank you!
swift asynchronous grand-central-dispatch dispatch-async
swift asynchronous grand-central-dispatch dispatch-async
asked 1 hour ago
Unikorn
4301817
asked 1 hour ago
Unikorn
4301817
edited 1 hour ago
asked 1 hour ago
Unikorn
4301817
asked 1 hour ago
Unikorn
4301817
asked 1 hour ago
Unikorn
4301817
4301817
1
Yes, it is obvious for 1 ,5 after that it switches to main queue prints 2, Then you again request main queue task but it will not wait for it and prints 4 and then 3
– Prashant Tukadiya
1 hour ago
3
You need to clarify one thing, whetherprint("1")is executed in the main thread or not.
– OOPer
1 hour ago
1
why would it matter? It would always execute first.
– Unikorn
1 hour ago
I suspect is due to the execution order: print("1") // 1. executes prints 1 DispatchQueue.main.async // 2. switches to main thread print("2") // 4. executes print 3 DispatchQueue.main.async // 5. switches to main thread print(3) // 7. executes print 3 print("4") // 6. executes print 4 print("5") // 3. executes print 5
– Unikorn
44 mins ago
@OOPer, please forgive my ignorant for my earlier remark thinking it will always execute first.
– Unikorn
12 mins ago
 |Â
show 1 more comment
1
Yes, it is obvious for 1 ,5 after that it switches to main queue prints 2, Then you again request main queue task but it will not wait for it and prints 4 and then 3
– Prashant Tukadiya
1 hour ago
3
You need to clarify one thing, whetherprint("1")is executed in the main thread or not.
– OOPer
1 hour ago
1
why would it matter? It would always execute first.
– Unikorn
1 hour ago
I suspect is due to the execution order: print("1") // 1. executes prints 1 DispatchQueue.main.async // 2. switches to main thread print("2") // 4. executes print 3 DispatchQueue.main.async // 5. switches to main thread print(3) // 7. executes print 3 print("4") // 6. executes print 4 print("5") // 3. executes print 5
– Unikorn
44 mins ago
@OOPer, please forgive my ignorant for my earlier remark thinking it will always execute first.
– Unikorn
12 mins ago
1
1
Yes, it is obvious for 1 ,5 after that it switches to main queue prints 2, Then you again request main queue task but it will not wait for it and prints 4 and then 3
– Prashant Tukadiya
1 hour ago
Yes, it is obvious for 1 ,5 after that it switches to main queue prints 2, Then you again request main queue task but it will not wait for it and prints 4 and then 3
– Prashant Tukadiya
1 hour ago
3
3
You need to clarify one thing, whether
print("1") is executed in the main thread or not.– OOPer
1 hour ago
You need to clarify one thing, whether
print("1") is executed in the main thread or not.– OOPer
1 hour ago
1
1
why would it matter? It would always execute first.
– Unikorn
1 hour ago
why would it matter? It would always execute first.
– Unikorn
1 hour ago
I suspect is due to the execution order: print("1") // 1. executes prints 1 DispatchQueue.main.async // 2. switches to main thread print("2") // 4. executes print 3 DispatchQueue.main.async // 5. switches to main thread print(3) // 7. executes print 3 print("4") // 6. executes print 4 print("5") // 3. executes print 5
– Unikorn
44 mins ago
I suspect is due to the execution order: print("1") // 1. executes prints 1 DispatchQueue.main.async // 2. switches to main thread print("2") // 4. executes print 3 DispatchQueue.main.async // 5. switches to main thread print(3) // 7. executes print 3 print("4") // 6. executes print 4 print("5") // 3. executes print 5
– Unikorn
44 mins ago
@OOPer, please forgive my ignorant for my earlier remark thinking it will always execute first.
– Unikorn
12 mins ago
@OOPer, please forgive my ignorant for my earlier remark thinking it will always execute first.
– Unikorn
12 mins ago
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
6
down vote
It depends on which thread you start the operation.
If you start from the main then, then you get 1, 5, 2, 4, 3
If you start from a background thread, then most of the time you'll get them same result (1, 5, 2, 4, 3), however this is not guaranteed as the background thread can be put to sleep at any time by the OS, and if this happens right before the print(5)call, then 5 will be the last to be printed.
Just a note that if the code from the question is the only one in your app/playground, then you might be surprised by running into partial prints, as the app/playground exits as soon as it hits the print(5) line, before having a chance to execute the async dispatch. To circumvent this, you can make sure that RunLoop.current.run() gets executed on the main thread as the last part of your code.
Here are some diagram that try to illustrate what happens in the main-thread-only scenario, and the one where a background thread is involved:
answered 39 mins ago
Cristik
16.6k114174
2
You get a +1 for the nice pictures.
– rmaddy
24 mins ago
Yeah, +1 for pictures. Thanks!
– Unikorn
9 mins ago
add a comment |Â
up vote
6
down vote
You will always get 1, 2, 4, 3. The 5 will always be after the 1. But where it ends up in relation to the others depends on what queue the whole thing started on.
If this is started from the main queue then 5 will always be between 1 and 2.
Here's why:
This code starts on the main queue. 1 is printed. You then enqueue another block to run asynchronously on the main queue so that block will be run after the current block completes and the main queue gets to the end of the current run loop. The code continues to the next line which is to print 5. The current block ends and the next block on the main queue is run. This is the block of the first call to DispatchQueue.main.async. As this block runs it prints 2 (so now we have 1 5 2). Another block is enqueued to the main queue just like the last one. The current block continues and prints 4 (so now we have 1 5 2 4). The block ends and the next block on the main queue is run. This is the final block we added. That block runs and it prints 3 giving the final output of 1 5 2 4 3.
If you started on some background queue then 5 can appear anywhere after 1 but for such simple code, the 5 will most likely still appear between 1 and 2 but it can appear anywhere after the 1 in a general case.
Here's why:
This code starts on a background queue. 1 is printed. You then enqueue another block to run asynchronously on the main queue so that block will be run after the current main queue run loop completes. The code continues to the next line which is to print 5. The current block ends. The block added to the main queue is run in parallel to the background queue. Depending on the time of when the block added to the main queue is run in relation to the remaining code on the background queue is run, the output of print("5") can be intermingled with the prints from the main queue. This is why the 5 can appear anywhere after the 1 when started from the background.
But the simple code in the question will likely always give 1 5 2 4 3 even when started on the background because the code is so short and takes so little time.
Here's some code that puts the 5 elsewhere:
func asyncTest()
print("1")
DispatchQueue.main.async
print("2")
DispatchQueue.main.async
print(3)
print("4")
for _ in 0...1000
print("5")
DispatchQueue.global(qos: .background).async
asyncTest()
The existence of the loop causes the 5 to take a little longer before it appears which allows the main queue to get executed some before 5 is printed. Without the loop, the background thread executes too quickly so the 5 appears before 2.
If running this test in a playground, add:
PlaygroundPage.current.needsIndefiniteExecution = true
to the top of the playground (just after the imports). You will also need:
import PlaygroundSupport
answered 1 hour ago
rmaddy
231k27299366
I ran this in playground 10+ times and always get : 1 5 2 4 3
– Unikorn
1 hour ago
Most likely you are starting this on the main queue so that will always give you that result, just as I stated in my answer.
– rmaddy
1 hour ago
1
Can you give an example of a code that could print it in another order?
– Sweeper
1 hour ago
@Sweeper Any code that results in this code being started on any background queue could result in the 5 being elsewhere.
– rmaddy
1 hour ago
1
Please feel free to comment if you down vote and explain why you think this answer is incorrect. I don't bite. Thanks.
– rmaddy
54 mins ago
 |Â
show 5 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
It depends on which thread you start the operation.
If you start from the main then, then you get 1, 5, 2, 4, 3
If you start from a background thread, then most of the time you'll get them same result (1, 5, 2, 4, 3), however this is not guaranteed as the background thread can be put to sleep at any time by the OS, and if this happens right before the print(5)call, then 5 will be the last to be printed.
Just a note that if the code from the question is the only one in your app/playground, then you might be surprised by running into partial prints, as the app/playground exits as soon as it hits the print(5) line, before having a chance to execute the async dispatch. To circumvent this, you can make sure that RunLoop.current.run() gets executed on the main thread as the last part of your code.
Here are some diagram that try to illustrate what happens in the main-thread-only scenario, and the one where a background thread is involved:
answered 39 mins ago
Cristik
16.6k114174
2
You get a +1 for the nice pictures.
– rmaddy
24 mins ago
Yeah, +1 for pictures. Thanks!
– Unikorn
9 mins ago
add a comment |Â
up vote
6
down vote
It depends on which thread you start the operation.
If you start from the main then, then you get 1, 5, 2, 4, 3
If you start from a background thread, then most of the time you'll get them same result (1, 5, 2, 4, 3), however this is not guaranteed as the background thread can be put to sleep at any time by the OS, and if this happens right before the print(5)call, then 5 will be the last to be printed.
Just a note that if the code from the question is the only one in your app/playground, then you might be surprised by running into partial prints, as the app/playground exits as soon as it hits the print(5) line, before having a chance to execute the async dispatch. To circumvent this, you can make sure that RunLoop.current.run() gets executed on the main thread as the last part of your code.
Here are some diagram that try to illustrate what happens in the main-thread-only scenario, and the one where a background thread is involved:
answered 39 mins ago
Cristik
16.6k114174
2
You get a +1 for the nice pictures.
– rmaddy
24 mins ago
Yeah, +1 for pictures. Thanks!
– Unikorn
9 mins ago
add a comment |Â
up vote
6
down vote
up vote
6
down vote
It depends on which thread you start the operation.
If you start from the main then, then you get 1, 5, 2, 4, 3
If you start from a background thread, then most of the time you'll get them same result (1, 5, 2, 4, 3), however this is not guaranteed as the background thread can be put to sleep at any time by the OS, and if this happens right before the print(5)call, then 5 will be the last to be printed.
Just a note that if the code from the question is the only one in your app/playground, then you might be surprised by running into partial prints, as the app/playground exits as soon as it hits the print(5) line, before having a chance to execute the async dispatch. To circumvent this, you can make sure that RunLoop.current.run() gets executed on the main thread as the last part of your code.
Here are some diagram that try to illustrate what happens in the main-thread-only scenario, and the one where a background thread is involved:
answered 39 mins ago
Cristik
16.6k114174
It depends on which thread you start the operation.
If you start from the main then, then you get 1, 5, 2, 4, 3
If you start from a background thread, then most of the time you'll get them same result (1, 5, 2, 4, 3), however this is not guaranteed as the background thread can be put to sleep at any time by the OS, and if this happens right before the print(5)call, then 5 will be the last to be printed.
Just a note that if the code from the question is the only one in your app/playground, then you might be surprised by running into partial prints, as the app/playground exits as soon as it hits the print(5) line, before having a chance to execute the async dispatch. To circumvent this, you can make sure that RunLoop.current.run() gets executed on the main thread as the last part of your code.
Here are some diagram that try to illustrate what happens in the main-thread-only scenario, and the one where a background thread is involved:
answered 39 mins ago
Cristik
16.6k114174
edited 26 mins ago
answered 39 mins ago
Cristik
16.6k114174
answered 39 mins ago
Cristik
16.6k114174
answered 39 mins ago
Cristik
16.6k114174
16.6k114174
2
You get a +1 for the nice pictures.
– rmaddy
24 mins ago
Yeah, +1 for pictures. Thanks!
– Unikorn
9 mins ago
add a comment |Â
2
You get a +1 for the nice pictures.
– rmaddy
24 mins ago
Yeah, +1 for pictures. Thanks!
– Unikorn
9 mins ago
2
2
You get a +1 for the nice pictures.
– rmaddy
24 mins ago
You get a +1 for the nice pictures.
– rmaddy
24 mins ago
Yeah, +1 for pictures. Thanks!
– Unikorn
9 mins ago
Yeah, +1 for pictures. Thanks!
– Unikorn
9 mins ago
add a comment |Â
up vote
6
down vote
You will always get 1, 2, 4, 3. The 5 will always be after the 1. But where it ends up in relation to the others depends on what queue the whole thing started on.
If this is started from the main queue then 5 will always be between 1 and 2.
Here's why:
This code starts on the main queue. 1 is printed. You then enqueue another block to run asynchronously on the main queue so that block will be run after the current block completes and the main queue gets to the end of the current run loop. The code continues to the next line which is to print 5. The current block ends and the next block on the main queue is run. This is the block of the first call to DispatchQueue.main.async. As this block runs it prints 2 (so now we have 1 5 2). Another block is enqueued to the main queue just like the last one. The current block continues and prints 4 (so now we have 1 5 2 4). The block ends and the next block on the main queue is run. This is the final block we added. That block runs and it prints 3 giving the final output of 1 5 2 4 3.
If you started on some background queue then 5 can appear anywhere after 1 but for such simple code, the 5 will most likely still appear between 1 and 2 but it can appear anywhere after the 1 in a general case.
Here's why:
This code starts on a background queue. 1 is printed. You then enqueue another block to run asynchronously on the main queue so that block will be run after the current main queue run loop completes. The code continues to the next line which is to print 5. The current block ends. The block added to the main queue is run in parallel to the background queue. Depending on the time of when the block added to the main queue is run in relation to the remaining code on the background queue is run, the output of print("5") can be intermingled with the prints from the main queue. This is why the 5 can appear anywhere after the 1 when started from the background.
But the simple code in the question will likely always give 1 5 2 4 3 even when started on the background because the code is so short and takes so little time.
Here's some code that puts the 5 elsewhere:
func asyncTest()
print("1")
DispatchQueue.main.async
print("2")
DispatchQueue.main.async
print(3)
print("4")
for _ in 0...1000
print("5")
DispatchQueue.global(qos: .background).async
asyncTest()
The existence of the loop causes the 5 to take a little longer before it appears which allows the main queue to get executed some before 5 is printed. Without the loop, the background thread executes too quickly so the 5 appears before 2.
If running this test in a playground, add:
PlaygroundPage.current.needsIndefiniteExecution = true
to the top of the playground (just after the imports). You will also need:
import PlaygroundSupport
answered 1 hour ago
rmaddy
231k27299366
I ran this in playground 10+ times and always get : 1 5 2 4 3
– Unikorn
1 hour ago
Most likely you are starting this on the main queue so that will always give you that result, just as I stated in my answer.
– rmaddy
1 hour ago
1
Can you give an example of a code that could print it in another order?
– Sweeper
1 hour ago
@Sweeper Any code that results in this code being started on any background queue could result in the 5 being elsewhere.
– rmaddy
1 hour ago
1
Please feel free to comment if you down vote and explain why you think this answer is incorrect. I don't bite. Thanks.
– rmaddy
54 mins ago
 |Â
show 5 more comments
up vote
6
down vote
You will always get 1, 2, 4, 3. The 5 will always be after the 1. But where it ends up in relation to the others depends on what queue the whole thing started on.
If this is started from the main queue then 5 will always be between 1 and 2.
Here's why:
This code starts on the main queue. 1 is printed. You then enqueue another block to run asynchronously on the main queue so that block will be run after the current block completes and the main queue gets to the end of the current run loop. The code continues to the next line which is to print 5. The current block ends and the next block on the main queue is run. This is the block of the first call to DispatchQueue.main.async. As this block runs it prints 2 (so now we have 1 5 2). Another block is enqueued to the main queue just like the last one. The current block continues and prints 4 (so now we have 1 5 2 4). The block ends and the next block on the main queue is run. This is the final block we added. That block runs and it prints 3 giving the final output of 1 5 2 4 3.
If you started on some background queue then 5 can appear anywhere after 1 but for such simple code, the 5 will most likely still appear between 1 and 2 but it can appear anywhere after the 1 in a general case.
Here's why:
This code starts on a background queue. 1 is printed. You then enqueue another block to run asynchronously on the main queue so that block will be run after the current main queue run loop completes. The code continues to the next line which is to print 5. The current block ends. The block added to the main queue is run in parallel to the background queue. Depending on the time of when the block added to the main queue is run in relation to the remaining code on the background queue is run, the output of print("5") can be intermingled with the prints from the main queue. This is why the 5 can appear anywhere after the 1 when started from the background.
But the simple code in the question will likely always give 1 5 2 4 3 even when started on the background because the code is so short and takes so little time.
Here's some code that puts the 5 elsewhere:
func asyncTest()
print("1")
DispatchQueue.main.async
print("2")
DispatchQueue.main.async
print(3)
print("4")
for _ in 0...1000
print("5")
DispatchQueue.global(qos: .background).async
asyncTest()
The existence of the loop causes the 5 to take a little longer before it appears which allows the main queue to get executed some before 5 is printed. Without the loop, the background thread executes too quickly so the 5 appears before 2.
If running this test in a playground, add:
PlaygroundPage.current.needsIndefiniteExecution = true
to the top of the playground (just after the imports). You will also need:
import PlaygroundSupport
answered 1 hour ago
rmaddy
231k27299366
I ran this in playground 10+ times and always get : 1 5 2 4 3
– Unikorn
1 hour ago
Most likely you are starting this on the main queue so that will always give you that result, just as I stated in my answer.
– rmaddy
1 hour ago
1
Can you give an example of a code that could print it in another order?
– Sweeper
1 hour ago
@Sweeper Any code that results in this code being started on any background queue could result in the 5 being elsewhere.
– rmaddy
1 hour ago
1
Please feel free to comment if you down vote and explain why you think this answer is incorrect. I don't bite. Thanks.
– rmaddy
54 mins ago
 |Â
show 5 more comments
up vote
6
down vote
up vote
6
down vote
You will always get 1, 2, 4, 3. The 5 will always be after the 1. But where it ends up in relation to the others depends on what queue the whole thing started on.
If this is started from the main queue then 5 will always be between 1 and 2.
Here's why:
This code starts on the main queue. 1 is printed. You then enqueue another block to run asynchronously on the main queue so that block will be run after the current block completes and the main queue gets to the end of the current run loop. The code continues to the next line which is to print 5. The current block ends and the next block on the main queue is run. This is the block of the first call to DispatchQueue.main.async. As this block runs it prints 2 (so now we have 1 5 2). Another block is enqueued to the main queue just like the last one. The current block continues and prints 4 (so now we have 1 5 2 4). The block ends and the next block on the main queue is run. This is the final block we added. That block runs and it prints 3 giving the final output of 1 5 2 4 3.
If you started on some background queue then 5 can appear anywhere after 1 but for such simple code, the 5 will most likely still appear between 1 and 2 but it can appear anywhere after the 1 in a general case.
Here's why:
This code starts on a background queue. 1 is printed. You then enqueue another block to run asynchronously on the main queue so that block will be run after the current main queue run loop completes. The code continues to the next line which is to print 5. The current block ends. The block added to the main queue is run in parallel to the background queue. Depending on the time of when the block added to the main queue is run in relation to the remaining code on the background queue is run, the output of print("5") can be intermingled with the prints from the main queue. This is why the 5 can appear anywhere after the 1 when started from the background.
But the simple code in the question will likely always give 1 5 2 4 3 even when started on the background because the code is so short and takes so little time.
Here's some code that puts the 5 elsewhere:
func asyncTest()
print("1")
DispatchQueue.main.async
print("2")
DispatchQueue.main.async
print(3)
print("4")
for _ in 0...1000
print("5")
DispatchQueue.global(qos: .background).async
asyncTest()
The existence of the loop causes the 5 to take a little longer before it appears which allows the main queue to get executed some before 5 is printed. Without the loop, the background thread executes too quickly so the 5 appears before 2.
If running this test in a playground, add:
PlaygroundPage.current.needsIndefiniteExecution = true
to the top of the playground (just after the imports). You will also need:
import PlaygroundSupport
answered 1 hour ago
rmaddy
231k27299366
You will always get 1, 2, 4, 3. The 5 will always be after the 1. But where it ends up in relation to the others depends on what queue the whole thing started on.
If this is started from the main queue then 5 will always be between 1 and 2.
Here's why:
This code starts on the main queue. 1 is printed. You then enqueue another block to run asynchronously on the main queue so that block will be run after the current block completes and the main queue gets to the end of the current run loop. The code continues to the next line which is to print 5. The current block ends and the next block on the main queue is run. This is the block of the first call to DispatchQueue.main.async. As this block runs it prints 2 (so now we have 1 5 2). Another block is enqueued to the main queue just like the last one. The current block continues and prints 4 (so now we have 1 5 2 4). The block ends and the next block on the main queue is run. This is the final block we added. That block runs and it prints 3 giving the final output of 1 5 2 4 3.
If you started on some background queue then 5 can appear anywhere after 1 but for such simple code, the 5 will most likely still appear between 1 and 2 but it can appear anywhere after the 1 in a general case.
Here's why:
This code starts on a background queue. 1 is printed. You then enqueue another block to run asynchronously on the main queue so that block will be run after the current main queue run loop completes. The code continues to the next line which is to print 5. The current block ends. The block added to the main queue is run in parallel to the background queue. Depending on the time of when the block added to the main queue is run in relation to the remaining code on the background queue is run, the output of print("5") can be intermingled with the prints from the main queue. This is why the 5 can appear anywhere after the 1 when started from the background.
But the simple code in the question will likely always give 1 5 2 4 3 even when started on the background because the code is so short and takes so little time.
Here's some code that puts the 5 elsewhere:
func asyncTest()
print("1")
DispatchQueue.main.async
print("2")
DispatchQueue.main.async
print(3)
print("4")
for _ in 0...1000
print("5")
DispatchQueue.global(qos: .background).async
asyncTest()
The existence of the loop causes the 5 to take a little longer before it appears which allows the main queue to get executed some before 5 is printed. Without the loop, the background thread executes too quickly so the 5 appears before 2.
If running this test in a playground, add:
PlaygroundPage.current.needsIndefiniteExecution = true
to the top of the playground (just after the imports). You will also need:
import PlaygroundSupport
answered 1 hour ago
rmaddy
231k27299366
edited 25 mins ago
answered 1 hour ago
rmaddy
231k27299366
answered 1 hour ago
rmaddy
231k27299366
answered 1 hour ago
rmaddy
231k27299366
231k27299366
I ran this in playground 10+ times and always get : 1 5 2 4 3
– Unikorn
1 hour ago
Most likely you are starting this on the main queue so that will always give you that result, just as I stated in my answer.
– rmaddy
1 hour ago
1
Can you give an example of a code that could print it in another order?
– Sweeper
1 hour ago
@Sweeper Any code that results in this code being started on any background queue could result in the 5 being elsewhere.
– rmaddy
1 hour ago
1
Please feel free to comment if you down vote and explain why you think this answer is incorrect. I don't bite. Thanks.
– rmaddy
54 mins ago
 |Â
show 5 more comments
I ran this in playground 10+ times and always get : 1 5 2 4 3
– Unikorn
1 hour ago
Most likely you are starting this on the main queue so that will always give you that result, just as I stated in my answer.
– rmaddy
1 hour ago
1
Can you give an example of a code that could print it in another order?
– Sweeper
1 hour ago
@Sweeper Any code that results in this code being started on any background queue could result in the 5 being elsewhere.
– rmaddy
1 hour ago
1
Please feel free to comment if you down vote and explain why you think this answer is incorrect. I don't bite. Thanks.
– rmaddy
54 mins ago
I ran this in playground 10+ times and always get : 1 5 2 4 3
– Unikorn
1 hour ago
I ran this in playground 10+ times and always get : 1 5 2 4 3
– Unikorn
1 hour ago
Most likely you are starting this on the main queue so that will always give you that result, just as I stated in my answer.
– rmaddy
1 hour ago
Most likely you are starting this on the main queue so that will always give you that result, just as I stated in my answer.
– rmaddy
1 hour ago
1
1
Can you give an example of a code that could print it in another order?
– Sweeper
1 hour ago
Can you give an example of a code that could print it in another order?
– Sweeper
1 hour ago
@Sweeper Any code that results in this code being started on any background queue could result in the 5 being elsewhere.
– rmaddy
1 hour ago
@Sweeper Any code that results in this code being started on any background queue could result in the 5 being elsewhere.
– rmaddy
1 hour ago
1
1
Please feel free to comment if you down vote and explain why you think this answer is incorrect. I don't bite. Thanks.
– rmaddy
54 mins ago
Please feel free to comment if you down vote and explain why you think this answer is incorrect. I don't bite. Thanks.
– rmaddy
54 mins ago
 |Â
show 5 more comments
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1
Yes, it is obvious for 1 ,5 after that it switches to main queue prints 2, Then you again request main queue task but it will not wait for it and prints 4 and then 3
– Prashant Tukadiya
1 hour ago
3
You need to clarify one thing, whether
print("1")is executed in the main thread or not.– OOPer
1 hour ago
1
why would it matter? It would always execute first.
– Unikorn
1 hour ago
I suspect is due to the execution order: print("1") // 1. executes prints 1 DispatchQueue.main.async // 2. switches to main thread print("2") // 4. executes print 3 DispatchQueue.main.async // 5. switches to main thread print(3) // 7. executes print 3 print("4") // 6. executes print 4 print("5") // 3. executes print 5
– Unikorn
44 mins ago
@OOPer, please forgive my ignorant for my earlier remark thinking it will always execute first.
– Unikorn
12 mins ago