Re: Why dump_stack results different so much?

[Xin Zhao <uszhaoxin@gmail.com>]

Thanks for your reply.

Below is the code that print the kernel calling trace:

/**********************************************************************************/
void show_trace(struct task_struct *task, unsigned long * stack)
{
	unsigned long ebp;

	if (!task)
		task = current;

	if (task == current) {
		/* Grab ebp right from our regs */
		asm ("movl %%ebp, %0" : "=r" (ebp) : );
	} else {
		/* ebp is the last reg pushed by switch_to */
		ebp = *(unsigned long *) task->thread.esp;
	}

	while (1) {
		struct thread_info *context;
		context = (struct thread_info *)
			((unsigned long)stack & (~(THREAD_SIZE - 1)));
		ebp = print_context_stack(context, stack, ebp);
		stack = (unsigned long*)context->previous_esp;
		if (!stack)
			break;
		printk(" =======================\n");
	}
}
/**********************************************************************************/

>From this code, I can see that the show_trace does not scan and guess
the pointers. Instead, it use "previous_esp" to extract the esp and
thus the returning eip. Am I right?

Cheers,
xin




On 7/29/05, bert hubert <bert.hubert@netherlabs.nl> wrote:
> On Fri, Jul 29, 2005 at 04:27:16PM -0400, Xin Zhao wrote:
> > I supprisely noticed that the dump_stack results are quite different!
> > Why did I get the calling traces below our_ssy_open() and above
> > syscall_call()?  Any thought on this? Many thanks!
> 
> This might depend on compiling with frame pointers, or not. I recall that at
> one point, the kernel did a basic scan of addresses that looked like likely
> candidates to have been pointers, and printed those.
> 
> Frame pointers are hailed as improving backtraces. They are in the 'kernel
> hacking' section of the kernel configuration.
> 
> Sorry that I can't be more precise, but try turning on frame pointers.
> 
> Good luck!
> 
> --
> http://www.PowerDNS.com      Open source, database driven DNS Software
> http://netherlabs.nl              Open and Closed source services
>
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