Thanks Roland.
But now I am even more confused...
On 09/20, Roland McGrath wrote:
>
> > - PTRACE_SINGLESTEP does user_enable_single_step()
> >
> > - when the tracee returns to user mode, the next instruction
> > causes exception, do_debug()->send_sigtrap() sends SIGTRAP
> >
> > - the tracee notices the signal and reports this SIGTRAP
>
> Correct. But if the tracee is inside the syscall path at the time, then
> it's a little different. TIF_SINGLESTEP being set means it will go to the
> syscall_trace_leave path and send a signal before it actually returns to
> user mode.
I am not sure... I can hardly read entry_64.S, but I am not sure
system_call_fastpath: path calls syscall_trace_leave() on exit...
OK, I added printk to syscall_trace_leave(), please see below.
> > At least. Is it true that X86_EFLAGS_TF should provoke the
> > exception/do_debug?
>
> Yes. You can even set it yourself in userland with:
>
> pushf
> orl $0x100, (%rsp) /* or (%esp) */
> popf
>
> and see the signal.
Yes, this works.
> > But, whatever I did this doesn't work. So, what user_enable_single_step()
> > actually means?
>
> It works very well in the vanilla kernel
> ...
> I tried your prctl hack on the vanilla kernel and a test program does
> indeed get a SIGTRAP. (This is an x86_64 kernel on an Intel chip, trying
> both 32-bit and 64-bit test programs.)
I used the vanilla kernel too. It doesn't work for me. Under KVM at least...
OK, the kernel patch:
--- TTT/kernel/sys.c~TF_FBG 2009-06-17 14:11:26.000000000 +0200
+++ TTT/kernel/sys.c 2009-09-21 20:03:27.000000000 +0200
@@ -1428,6 +1428,12 @@ SYSCALL_DEFINE5(prctl, int, option, unsi
error = 0;
switch (option) {
+ case 666:
+ printk(KERN_INFO "prctl: before=%lX\n",
task_pt_regs(current)->flags);
+ user_enable_single_step(current);
+ printk(KERN_INFO "prctl: after=%lX\n",
task_pt_regs(current)->flags);
+ break;
+
case PR_SET_PDEATHSIG:
if (!valid_signal(arg2)) {
error = -EINVAL;
--- TTT/arch/x86/kernel/ptrace.c~TF_FBG 2009-06-11 14:16:46.000000000
+0200
+++ TTT/arch/x86/kernel/ptrace.c 2009-09-21 20:22:52.000000000
+0200
@@ -1520,6 +1520,8 @@ asmregparm long syscall_trace_enter(stru
asmregparm void syscall_trace_leave(struct pt_regs *regs)
{
+ printk(KERN_INFO "trace_leave enter\n");
+
if (unlikely(current->audit_context))
audit_syscall_exit(AUDITSC_RESULT(regs->ax), regs->ax);
@@ -1543,6 +1545,8 @@ asmregparm void syscall_trace_leave(stru
* system call instruction.
*/
if (test_thread_flag(TIF_SINGLESTEP) &&
- tracehook_consider_fatal_signal(current, SIGTRAP))
+ tracehook_consider_fatal_signal(current, SIGTRAP)) {
+ printk(KERN_INFO "trace_leave sends SIGTRAP\n");
send_sigtrap(current, regs, 0, TRAP_BRKPT);
+ }
}
version:
$ uname -a
Linux myhost 2.6.31-00008-g41b61a1-dirty #2 SMP PREEMPT Mon Sep 21
20:23:05 CEST 2009 x86_64 QEMU Virtual CPU version 0.9.1 GenuineIntel GNU/Linux
The 1ts test:
int main(void)
{
assert(0 == prctl(666));
for (;;)
;
return 0;
}
it doesn't get SIGTRAP, and syscall_trace_leave() is not called.
$ dmesg -c
prctl: before=10246
prctl: after=10346
The same for this test:
int main(void)
{
assert(0 == prctl(666));
pause();
return 0;
}
However,
int main(void)
{
assert(0 == prctl(666));
printf("after prctl\n");
return 0;
}
dies with "Trace/breakpoint trap", dmesg says:
prctl: before=10246
prctl: after=10346
trace_leave enter
I guess "trace_leave enter" corresponds to the next syscall, not to prtcl().
This test
int main(void)
{
printf("before prctl\n");
assert(0 == prctl(666));
printf("after prctl\n");
return 0;
}
gets SIGTRAP too, but printf() succeeds:
before prctl
after prctl
Trace/breakpoint trap
The final test-case:
static inline unsigned long getflags(void)
{
unsigned long flags;
asm volatile("pushf ; pop %0\n"
: "=rm" (flags)
: /* no input */
: "memory");
return flags;
}
int main(void)
{
unsigned long flags;
printf("user_enable_single_step()...\n");
assert(0 == prctl(666));
flags = getflags();
printf("flags=%lX\n", flags);
for (;;)
;
return 0;
}
Result:
user_enable_single_step()...
flags=246
Trace/breakpoint trap
dmesg is the same:
prctl: before=10246
prctl: after=10346
trace_leave enter
Oleg.
