What does it mean when a solution outputs Runtime Error SIGABRT in Java?
1 Like
If you show us your code, we could probably find the error.
I am also getting this error in one of the problems. Also its not a Throwable object. So how to find the reason behind it?
I’m not sure but I think it might be because of the memory limit. But I don’t know why the admins don’t confirm that that could be one of the reasons, since a problem has a time limit and a memory limit, so accordingly we have TLE and maybe a SIGABRT???
SIGABRT Error simply means abnormal termination of your program. The JVM raises this signal whenever it detects a JVM fault. It might be because of the extreme memory limit. The official Codechef response says, “A SIGABRT is caused if your program aborted due to a fatal error. This can also be caused if you are using an assert() which fails or an abort().”
May be my experience could help u. I was programming in c for double link list. i created node temp. using malloc i was making new node. but in the middle of program i freed the temp and this error poped up. May be u also doing smthing like that. But i have no knowledge about java.