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When I call abort() in an if statement, it causes a segmentation fault, if I do exit(0), it exits fine. But I need to use abort(), can anyone provide the details of what abort() exactly does? I googled it, but I don't much about what this function actually does to abort the program
If it helps, I get an error saying
"6 [sig]" along with something else like cored dumped corrupted stack.
Do you have a signal handler installed for SIGABRT, or are you possibly using a library of functions which might have installed a signal handler for abort? (Which, by the way, is 6).
I don't know what exactly a signal handler is. Neither did I install such a thing, so I'm guessing it's to do with the latter (library functions)
Can you post your code?
I will post a part of it because it's assignment work and I don't want someone else here to copy it, if they happen to be in my class.
I ran the gdb earlier and it indicated a problem was at this part:
Of course it didn't indicate which line so I had no idea.Code:AR node;
AR head = NULL;
AR tail;
AR next;
int N = 0; //line number-node number
/*PROTOTYPE DECLARATIONS*/
AR makeNode(void);
AR traverseE(AR ar); // function for traversing to the end of the list
int main (int argc, char *argv[]) {
This is where I call the abort function:
Essentially N is the number of nodes and if a position entered by user is greater than that or less than 0, just abort.Code:if((pos1 > N) || (pos1 < 0) || (pos2 > N) || (pos2 < 0)) {
printf("Error position out of range\n");
abort();
}
??? Can't tell from that.
Quote:
Originally Posted by Todd Burch
Well do you know exactly how the abort function works? If I knew that I could probably figure out what exactly it's doing before it exits.
What platform are you using?
I just read up on abort(). It's doing just what it is supposed to do.
It is terminating your program, and sending the SIGABRT signal to the caller of your program, and the caller of your program is taking a core dump just like it is supposed to.
I suspect you'll need to find a new mechanism to exit clean if you want to exit clean.
You can verify this by writing a small "hello world" program and aborting instead of returning.
That's what the description on the man page says.Quote:
The abort() function causes abnormal process termination to
occur, unless the signal SIGABRT is being caught and the
signal handler does not return. The abnormal termination
processing includes at least the effect of fclose(3C) on all
open streams and message catalogue descriptors, and the
default actions defined for SIGABRT. The SIGABRT signal is
sent to the calling process as if by means of the raise(3C)
function with the argument SIGABRT.
The status made available to wait(3C) or waitpid(3C) by
abort will be that of a process terminated by the SIGABRT
signal. abort will override blocking or ignoring the
SIGABRT signal.
And the Usage says:
I guess that makes it clear to you why your code is dumping core.Quote:
Catching the signal is intended to provide the application
writer with a portable means to abort processing, free from
possible interference from any implementation-provided
library functions. If SIGABRT is neither caught nor
ignored, and the current directory is writable, a core dump
may be produced.
I'd recommend writing a signal handler to work with SIGABRT (6) for a clean exit or use other means exiting your code.
of clean up and exit.
No it's not clear to me exactly. Though I understand there is some kind of stray signal so the streams haven't been completely closed. I think I'm going to try to fix up my code instead of writing a signal handler, we haven't learn such a thing and I don't think the lecturer intended for such a thing. It was meant to be a simple clean exit.
Abort was not made for a simple and clean exit. It's more meant if something fatal happens in the program and it needs to be terminated directly without allowing the flow of the program to continue.