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From: "Dan Osborne" <dan DOT osborne AT ramesys DOT com>
To: "Cygwin AT Cygwin. Com" <cygwin AT cygwin DOT com>
Subject: RE: Program exited with code 0303000
Date: Tue, 28 Sep 2004 13:21:51 +0100
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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

add_var(i,var,in_out,apl_tab_size);

However, further digging reveals that my program gets to the throw command
here ...

catch ( RProgReturnException e )
{
   throw;
}

and if I step on throw the debugger thinks I'm at ... gues which line?! yes,
line 3608 of otlv4.h ...

(gdb) s
0x1003b115 in __cxa_rethrow () at otlv4.h:3608
3608              add_var(i,var,in_out,apl_tab_size);

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
main.

Thanks in anticipation and thanks for your input so far,

Dan


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