Why can't while(flag){} end
code like this:
fun main(args: Array<String>) {
var flag = true
Thread {
Thread.sleep(2000)
println("time over")
flag = false
}.start()
while (flag) {
Thread.sleep(100)
}
println("finish")
}
the run result is :
time over
finish
The program is over
if change to:
fun main(args: Array<String>) {
var flag = true
Thread {
Thread.sleep(2000)
println("time over")
flag = false
}.start()
while (flag) {
//Do nothing
}
println("finish")
}
"finish" cannot be print,The program is stuck. Why is that?
java kotlin
asked Nov 21 '18 at 5:42
Void YoungVoid Young
235
add a comment |
code like this:
fun main(args: Array<String>) {
var flag = true
Thread {
Thread.sleep(2000)
println("time over")
flag = false
}.start()
while (flag) {
Thread.sleep(100)
}
println("finish")
}
the run result is :
time over
finish
The program is over
if change to:
fun main(args: Array<String>) {
var flag = true
Thread {
Thread.sleep(2000)
println("time over")
flag = false
}.start()
while (flag) {
//Do nothing
}
println("finish")
}
"finish" cannot be print,The program is stuck. Why is that?
java kotlin
asked Nov 21 '18 at 5:42
Void YoungVoid Young
235
try marking the flag as "@Volatile var flag = true"
– Abhi
Nov 21 '18 at 6:09
add a comment |
code like this:
fun main(args: Array<String>) {
var flag = true
Thread {
Thread.sleep(2000)
println("time over")
flag = false
}.start()
while (flag) {
Thread.sleep(100)
}
println("finish")
}
the run result is :
time over
finish
The program is over
if change to:
fun main(args: Array<String>) {
var flag = true
Thread {
Thread.sleep(2000)
println("time over")
flag = false
}.start()
while (flag) {
//Do nothing
}
println("finish")
}
"finish" cannot be print,The program is stuck. Why is that?
java kotlin
asked Nov 21 '18 at 5:42
Void YoungVoid Young
235
code like this:
fun main(args: Array<String>) {
var flag = true
Thread {
Thread.sleep(2000)
println("time over")
flag = false
}.start()
while (flag) {
Thread.sleep(100)
}
println("finish")
}
the run result is :
time over
finish
The program is over
if change to:
fun main(args: Array<String>) {
var flag = true
Thread {
Thread.sleep(2000)
println("time over")
flag = false
}.start()
while (flag) {
//Do nothing
}
println("finish")
}
"finish" cannot be print,The program is stuck. Why is that?
java kotlin
java kotlin
asked Nov 21 '18 at 5:42
Void YoungVoid Young
235
asked Nov 21 '18 at 5:42
Void YoungVoid Young
235
asked Nov 21 '18 at 5:42
Void YoungVoid Young
235
asked Nov 21 '18 at 5:42
Void YoungVoid Young
235
asked Nov 21 '18 at 5:42
Void YoungVoid Young
235
235
try marking the flag as "@Volatile var flag = true"
– Abhi
Nov 21 '18 at 6:09
add a comment |
try marking the flag as "@Volatile var flag = true"
– Abhi
Nov 21 '18 at 6:09
try marking the flag as "@Volatile var flag = true"
– Abhi
Nov 21 '18 at 6:09
try marking the flag as "@Volatile var flag = true"
– Abhi
Nov 21 '18 at 6:09
add a comment |
3 Answers
3
active
oldest
votes
Because you are getting a cached version of your flag. Take a look at the volatile keyword.
Basically here you are updating the flag in one thread, but the other is still seeing is own cached version of the flag.
answered Nov 21 '18 at 5:44
alerootaleroot
54.6k20140187
2
Ah, you beat me by 2 seconds for the same answer :)
– PradyumanDixit
Nov 21 '18 at 5:45
Thank you very much for letting me know the use of volatile
– Void Young
Nov 25 '18 at 2:45
add a comment |
The problem is first thread need time to set your flag is false but while loop execute before thread finish that means still your flag is true so that its never get false.
we need to set flow after thread finish then check while condition for that use thread sleep.
var flag = true
Thread {
Thread.sleep(2000)
println("time over")
flag = false
}.start()
Thread.sleep(3000)
while (flag) {
}
println("finish")
answered Nov 21 '18 at 5:49
sasikumarsasikumar
7,48311126
So why the "time over" is also not printing in 2nd case of OP?
– Amit Bera
Nov 21 '18 at 5:57
println("time over") is printing but not printing "finish" in 2nd case
– sasikumar
Nov 21 '18 at 6:02
add a comment |
When the variable flag is not declared volatile, the compiler sometimes (depends on the one used) optimises your while loop code to the following :
if (condition) {
while (true) {
// Do nothing
}
}
Refer this link for more. Here is a similar SO question.
answered Nov 21 '18 at 6:11
PranjalPranjal
1363
Thank you very much for letting me know the use of volatile
– Void Young
Nov 25 '18 at 2:46
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Because you are getting a cached version of your flag. Take a look at the volatile keyword.
Basically here you are updating the flag in one thread, but the other is still seeing is own cached version of the flag.
answered Nov 21 '18 at 5:44
alerootaleroot
54.6k20140187
2
Ah, you beat me by 2 seconds for the same answer :)
– PradyumanDixit
Nov 21 '18 at 5:45
Thank you very much for letting me know the use of volatile
– Void Young
Nov 25 '18 at 2:45
add a comment |
Because you are getting a cached version of your flag. Take a look at the volatile keyword.
Basically here you are updating the flag in one thread, but the other is still seeing is own cached version of the flag.
answered Nov 21 '18 at 5:44
alerootaleroot
54.6k20140187
2
Ah, you beat me by 2 seconds for the same answer :)
– PradyumanDixit
Nov 21 '18 at 5:45
Thank you very much for letting me know the use of volatile
– Void Young
Nov 25 '18 at 2:45
add a comment |
Because you are getting a cached version of your flag. Take a look at the volatile keyword.
Basically here you are updating the flag in one thread, but the other is still seeing is own cached version of the flag.
answered Nov 21 '18 at 5:44
alerootaleroot
54.6k20140187
Because you are getting a cached version of your flag. Take a look at the volatile keyword.
Basically here you are updating the flag in one thread, but the other is still seeing is own cached version of the flag.
answered Nov 21 '18 at 5:44
alerootaleroot
54.6k20140187
answered Nov 21 '18 at 5:44
alerootaleroot
54.6k20140187
answered Nov 21 '18 at 5:44
alerootaleroot
54.6k20140187
answered Nov 21 '18 at 5:44
alerootaleroot
54.6k20140187
54.6k20140187
2
Ah, you beat me by 2 seconds for the same answer :)
– PradyumanDixit
Nov 21 '18 at 5:45
Thank you very much for letting me know the use of volatile
– Void Young
Nov 25 '18 at 2:45
add a comment |
2
Ah, you beat me by 2 seconds for the same answer :)
– PradyumanDixit
Nov 21 '18 at 5:45
Thank you very much for letting me know the use of volatile
– Void Young
Nov 25 '18 at 2:45
2
2
Ah, you beat me by 2 seconds for the same answer :)
– PradyumanDixit
Nov 21 '18 at 5:45
Ah, you beat me by 2 seconds for the same answer :)
– PradyumanDixit
Nov 21 '18 at 5:45
Thank you very much for letting me know the use of volatile
– Void Young
Nov 25 '18 at 2:45
Thank you very much for letting me know the use of volatile
– Void Young
Nov 25 '18 at 2:45
add a comment |
The problem is first thread need time to set your flag is false but while loop execute before thread finish that means still your flag is true so that its never get false.
we need to set flow after thread finish then check while condition for that use thread sleep.
var flag = true
Thread {
Thread.sleep(2000)
println("time over")
flag = false
}.start()
Thread.sleep(3000)
while (flag) {
}
println("finish")
answered Nov 21 '18 at 5:49
sasikumarsasikumar
7,48311126
So why the "time over" is also not printing in 2nd case of OP?
– Amit Bera
Nov 21 '18 at 5:57
println("time over") is printing but not printing "finish" in 2nd case
– sasikumar
Nov 21 '18 at 6:02
add a comment |
The problem is first thread need time to set your flag is false but while loop execute before thread finish that means still your flag is true so that its never get false.
we need to set flow after thread finish then check while condition for that use thread sleep.
var flag = true
Thread {
Thread.sleep(2000)
println("time over")
flag = false
}.start()
Thread.sleep(3000)
while (flag) {
}
println("finish")
answered Nov 21 '18 at 5:49
sasikumarsasikumar
7,48311126
So why the "time over" is also not printing in 2nd case of OP?
– Amit Bera
Nov 21 '18 at 5:57
println("time over") is printing but not printing "finish" in 2nd case
– sasikumar
Nov 21 '18 at 6:02
add a comment |
The problem is first thread need time to set your flag is false but while loop execute before thread finish that means still your flag is true so that its never get false.
we need to set flow after thread finish then check while condition for that use thread sleep.
var flag = true
Thread {
Thread.sleep(2000)
println("time over")
flag = false
}.start()
Thread.sleep(3000)
while (flag) {
}
println("finish")
answered Nov 21 '18 at 5:49
sasikumarsasikumar
7,48311126
The problem is first thread need time to set your flag is false but while loop execute before thread finish that means still your flag is true so that its never get false.
we need to set flow after thread finish then check while condition for that use thread sleep.
var flag = true
Thread {
Thread.sleep(2000)
println("time over")
flag = false
}.start()
Thread.sleep(3000)
while (flag) {
}
println("finish")
answered Nov 21 '18 at 5:49
sasikumarsasikumar
7,48311126
answered Nov 21 '18 at 5:49
sasikumarsasikumar
7,48311126
answered Nov 21 '18 at 5:49
sasikumarsasikumar
7,48311126
answered Nov 21 '18 at 5:49
sasikumarsasikumar
7,48311126
7,48311126
So why the "time over" is also not printing in 2nd case of OP?
– Amit Bera
Nov 21 '18 at 5:57
println("time over") is printing but not printing "finish" in 2nd case
– sasikumar
Nov 21 '18 at 6:02
add a comment |
So why the "time over" is also not printing in 2nd case of OP?
– Amit Bera
Nov 21 '18 at 5:57
println("time over") is printing but not printing "finish" in 2nd case
– sasikumar
Nov 21 '18 at 6:02
So why the "time over" is also not printing in 2nd case of OP?
– Amit Bera
Nov 21 '18 at 5:57
So why the "time over" is also not printing in 2nd case of OP?
– Amit Bera
Nov 21 '18 at 5:57
println("time over") is printing but not printing "finish" in 2nd case
– sasikumar
Nov 21 '18 at 6:02
println("time over") is printing but not printing "finish" in 2nd case
– sasikumar
Nov 21 '18 at 6:02
add a comment |
When the variable flag is not declared volatile, the compiler sometimes (depends on the one used) optimises your while loop code to the following :
if (condition) {
while (true) {
// Do nothing
}
}
Refer this link for more. Here is a similar SO question.
answered Nov 21 '18 at 6:11
PranjalPranjal
1363
Thank you very much for letting me know the use of volatile
– Void Young
Nov 25 '18 at 2:46
add a comment |
When the variable flag is not declared volatile, the compiler sometimes (depends on the one used) optimises your while loop code to the following :
if (condition) {
while (true) {
// Do nothing
}
}
Refer this link for more. Here is a similar SO question.
answered Nov 21 '18 at 6:11
PranjalPranjal
1363
Thank you very much for letting me know the use of volatile
– Void Young
Nov 25 '18 at 2:46
add a comment |
When the variable flag is not declared volatile, the compiler sometimes (depends on the one used) optimises your while loop code to the following :
if (condition) {
while (true) {
// Do nothing
}
}
Refer this link for more. Here is a similar SO question.
answered Nov 21 '18 at 6:11
PranjalPranjal
1363
When the variable flag is not declared volatile, the compiler sometimes (depends on the one used) optimises your while loop code to the following :
if (condition) {
while (true) {
// Do nothing
}
}
Refer this link for more. Here is a similar SO question.
answered Nov 21 '18 at 6:11
PranjalPranjal
1363
edited Nov 21 '18 at 6:29
answered Nov 21 '18 at 6:11
PranjalPranjal
1363
answered Nov 21 '18 at 6:11
PranjalPranjal
1363
answered Nov 21 '18 at 6:11
PranjalPranjal
1363
1363
Thank you very much for letting me know the use of volatile
– Void Young
Nov 25 '18 at 2:46
add a comment |
Thank you very much for letting me know the use of volatile
– Void Young
Nov 25 '18 at 2:46
Thank you very much for letting me know the use of volatile
– Void Young
Nov 25 '18 at 2:46
Thank you very much for letting me know the use of volatile
– Void Young
Nov 25 '18 at 2:46
add a comment |
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try marking the flag as "@Volatile var flag = true"
– Abhi
Nov 21 '18 at 6:09