Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
690 views
in Technique[技术] by (71.8m points)

jvm - Behavior of a Java process in case of OutOfMemoryError

What would be the behavior of a Java program on getting OutOfMemoryError. Is there any defined behavior? Will the process crash or would it go into wait/sleep state?

Update: if I am not handling it in my code?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

And OutOfMemoryError is handled like any other exception:

  • If it is caught, then nothing more happens.
  • If it is not caught, then either the threads or the threads groups uncaught exception handler handles it. This pretty much always leads to the thread being stopped.

However there are two factors that are not really there in other exceptions:

  • OutOfMemoryError is an Error and not an Exception. This means that it's very unlikely to be caught anywhere: You should not try to catch an Error generally (with very few exceptions) and it's not usually done, so the chances of it being handled are rather low.
  • When an OutOfMemoryError happens and no object become eligible for GC because of that, then you'll still have little memory left and chances are that you'll run into the exact same problem again later on.

And if the thread this happens to is the only non-daemon thread (often, but not necessarily, that's the main thread, that executes the main method), then that thread getting killed results in the whole JVM shutting down (which is often perceived as "a crash").

So the tl;dr is: It will probably kill the thread, and if the memory-issue is not solved, then this can happen to more and more threads.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...