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RE: Program exited with code 0303000
- From: "Dan Osborne" <dan dot osborne at ramesys dot com>
- To: "Cygwin at Cygwin dot Com" <cygwin at cygwin dot com>
- Date: Tue, 28 Sep 2004 13:21:51 +0100
- Subject: RE: Program exited with code 0303000
Thanks for the reply.
> Since you've got the code up and running in a debugger, you can set
> breakpoints on abort, exit, and the last line of main, then when
> it hits one
> you can see for yourself from which point your code is exiting and the
> reason why it's returning a non-zero exit code.
I was trying to step through the code in gdb and pin this down so setting
breaks on abort and exit sounds useful. However, both b abort and b exit
give me ...
(gdb) b abort
Breakpoint 1 at 0x401552: file otlv4.h, line 3608.
(gdb) b exit
Note: breakpoint 1 also set at pc 0x401552.
Breakpoint 2 at 0x401552: file otlv4.h, line 3608.
But line 3608 of otlv4.h looks like nothing to do with abort or exit
However, further digging reveals that my program gets to the throw command
catch ( RProgReturnException e )
and if I step on throw the debugger thinks I'm at ... gues which line?! yes,
line 3608 of otlv4.h ...
0x1003b115 in __cxa_rethrow () at otlv4.h:3608
and if I step again I get ...
Program exited with code 0303000.
So I'm wondering firstly why gdb seems to have a mismatch between address
and source line number and why that throw didn't get caught in my catch in
Thanks in anticipation and thanks for your input so far,
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