Message ID | 52076172bb8a55305846f6d4dc97bb52@codeaurora.org (mailing list archive) |
---|---|
State | New, archived |
Headers | show |
Series | arm64: break while loop if task had been rescheduled | expand |
Hi, This appears to be a bizarrely formatted reply to Anshuman's questions [1] on the first posting [2] of this patch, and as it stands, it isn't possible to follow. Please follow the usual mailing list ettiquette, and reply inline to questions. I am not going to reply further to this post, but I'll comment on the first post. Thanks, Mark. [1] https://lore.kernel.org/lkml/1558430404-4840-1-git-send-email-tengfeif@codeaurora.org/T/#m415174aacdd100f9386113ed3ae9f427a2255f8a [2] https://lore.kernel.org/lkml/1558430404-4840-1-git-send-email-tengfeif@codeaurora.org/T/#u On Fri, May 24, 2019 at 11:16:16AM +0800, tengfeif@codeaurora.org wrote: > When task isn't current task, this task's state have > chance to be changed during printing this task's > backtrace, so it is possible that task's fp and fp+8 > have the same vaule, so cannot break the while loop. > To fix this issue, we first save the task's state, sp > and fp, then we will get the task's current state, sp > and fp in each while again. we will stop to print > backtrace if we found any of the values are different > than what we saved. > > /********************************answer > question**********************************/ > This is very confusing. IIUC it suggests that while printing > the backtrace for non-current tasks the do/while loop does not > exit because fp and fp+8 might have the same value ? When would > this happen ? Even in that case the commit message here does not > properly match the change in this patch. So > > In our issue, we got fp=pc=0xFFFFFF8025A13BA0, so cannot exit while > loop in dump_basktrace(). > After analyze our issue's dump, we found one task(such as: task A) > is exiting via invoke do_exit() during another task is showing task > A's dumptask. In kernel code, do_exit() and exit_notify are defined > as follows: > void noreturn do_exit(long code) > { > ...... > exit_notify(tsk, group_dead); > ...... > } > static void exit_notify(struct task_struct *tsk, int group_dead) > { > ...... > } > Because of exit_notify() is a static function, so it is inlined to > do_exit() when compile kernel, so we can get partial assembly code > of do_exit() as follows: > …… > { > bool autoreap; > struct task_struct *p, *n; > LIST_HEAD(dead); > > write_lock_irq(&tasklist_lock); > c10: 90000000 adrp x0, 0 <tasklist_lock> > c14: 910003e8 mov x8, sp > c18: 91000000 add x0, x0, #0x0 > */ > static void exit_notify(struct task_struct *tsk, int group_dead) > { > bool autoreap; > struct task_struct *p, *n; > LIST_HEAD(dead); > c1c: a90023e8 stp x8, x8, [sp] > > write_lock_irq(&tasklist_lock); > c20: 94000000 bl 0 <_raw_write_lock_irq> > c24: f9435268 ldr x8, [x19,#1696] > …… > From the code "c14:" and "c1c:", we will find sp's addr value is stored > in sp and sp+8, so sp's vaule equal (sp+8)'s value. > In our issue, there is a chance of fp point sp, so there will be fp=pc=fp's > addr value,so code cannot break from while loop in dump_backtrace(). > > /********************************answer > question**********************************/ > > /********************************answer > question**********************************/ > This patch tries to stop printing the stack for non-current tasks > if their state change while there is one dump_backtrace() trying > to print back trace. Dont we have any lock preventing a task in > this situation (while dumping it's backtrace) from running again > or changing state. > I haven't found any lock preventing a task in this situation, and I think we > shouldn't > prevent task running if this task is scheduled. > /********************************answer > question**********************************/ > > Signed-off-by: Tengfei Fan <tengfeif@codeaurora.org> > --- > arch/arm64/kernel/traps.c | 23 +++++++++++++++++++++++ > 1 file changed, 23 insertions(+) > > diff --git a/arch/arm64/kernel/traps.c b/arch/arm64/kernel/traps.c > index 2975598..9df6e02 100644 > --- a/arch/arm64/kernel/traps.c > +++ b/arch/arm64/kernel/traps.c > @@ -103,6 +103,9 @@ void dump_backtrace(struct pt_regs *regs, struct > task_struct *tsk) > { > struct stackframe frame; > int skip = 0; > + long cur_state = 0; > + unsigned long cur_sp = 0; > + unsigned long cur_fp = 0; > > pr_debug("%s(regs = %p tsk = %p)\n", __func__, regs, tsk); > > @@ -127,6 +130,9 @@ void dump_backtrace(struct pt_regs *regs, struct > task_struct *tsk) > */ > frame.fp = thread_saved_fp(tsk); > frame.pc = thread_saved_pc(tsk); > + cur_state = tsk->state; > + cur_sp = thread_saved_sp(tsk); > + cur_fp = frame.fp; > > /********************************answer > question**********************************/ > Should 'saved_state|sp|fp' instead as its applicable to non-current > tasks only. > 'saved_state|sp|fp' only applies to non-current tasks. > > /********************************answer > question**********************************/ > > } > #ifdef CONFIG_FUNCTION_GRAPH_TRACER > frame.graph = 0; > @@ -134,6 +140,23 @@ void dump_backtrace(struct pt_regs *regs, struct > task_struct *tsk) > > printk("Call trace:\n"); > do { > + if (tsk != current && (cur_state != tsk->state > + /* > + * We would not be printing backtrace for the task > + * that has changed state from "saved" state to ohter > + * state before hitting the do-while loop but after > + * saving the current state. If task's current state > + * not equal the "saved" state, then we may print > + * wrong call trace or end up in infinite while loop > + * if *(fp) and *(fp+8) are same. While the situation > + * should be stoped once we found the task's state > + * is changed, so we detect the task's current state, > + * sp and fp in each while. > + */ > + || cur_sp != thread_saved_sp(tsk) > + || cur_fp != thread_saved_fp(tsk))) { > > /********************************answer > question**********************************/ > Why does any of these three mismatches detect the problematic transition > not just the state ? > 1. we can use "cur_state != tsk->state" prevent printing backtrace if the > task's > state is changed after "saved" task's state. > 2. we can use "cur_sp != thread_saved_sp(tsk)" and "cur_fp != > thread_saved_fp(tsk)" > prevent printing backtrace if the task's state is changed before "saved" > task's > state. Because the value of "thread_saved_sp(tsk)" and > "thread_saved_fp(tsk)" > will not equal "saved" sp(cur_sp) and fp(cur_fp). > /********************************answer > question**********************************/
diff --git a/arch/arm64/kernel/traps.c b/arch/arm64/kernel/traps.c index 2975598..9df6e02 100644 --- a/arch/arm64/kernel/traps.c +++ b/arch/arm64/kernel/traps.c @@ -103,6 +103,9 @@ void dump_backtrace(struct pt_regs *regs, struct task_struct *tsk) { struct stackframe frame; int skip = 0; + long cur_state = 0; + unsigned long cur_sp = 0; + unsigned long cur_fp = 0; pr_debug("%s(regs = %p tsk = %p)\n", __func__, regs, tsk);
When task isn't current task, this task's state have chance to be changed during printing this task's backtrace, so it is possible that task's fp and fp+8 have the same vaule, so cannot break the while loop. To fix this issue, we first save the task's state, sp and fp, then we will get the task's current state, sp and fp in each while again. we will stop to print backtrace if we found any of the values are different than what we saved. /********************************answer question**********************************/ This is very confusing. IIUC it suggests that while printing the backtrace for non-current tasks the do/while loop does not exit because fp and fp+8 might have the same value ? When would this happen ? Even in that case the commit message here does not properly match the change in this patch. In our issue, we got fp=pc=0xFFFFFF8025A13BA0, so cannot exit while loop in dump_basktrace(). After analyze our issue's dump, we found one task(such as: task A) is exiting via invoke do_exit() during another task is showing task A's dumptask. In kernel code, do_exit() and exit_notify are defined as follows: void noreturn do_exit(long code) { ...... exit_notify(tsk, group_dead); ...... } static void exit_notify(struct task_struct *tsk, int group_dead) { ...... } Because of exit_notify() is a static function, so it is inlined to do_exit() when compile kernel, so we can get partial assembly code of do_exit() as follows: …… { bool autoreap; struct task_struct *p, *n; LIST_HEAD(dead); write_lock_irq(&tasklist_lock); c10: 90000000 adrp x0, 0 <tasklist_lock> c14: 910003e8 mov x8, sp c18: 91000000 add x0, x0, #0x0 */ static void exit_notify(struct task_struct *tsk, int group_dead) { bool autoreap; struct task_struct *p, *n; LIST_HEAD(dead); c1c: a90023e8 stp x8, x8, [sp] write_lock_irq(&tasklist_lock); c20: 94000000 bl 0 <_raw_write_lock_irq> c24: f9435268 ldr x8, [x19,#1696] …… From the code "c14:" and "c1c:", we will find sp's addr value is stored in sp and sp+8, so sp's vaule equal (sp+8)'s value. In our issue, there is a chance of fp point sp, so there will be fp=pc=fp's addr value,so code cannot break from while loop in dump_backtrace(). /********************************answer question**********************************/ /********************************answer question**********************************/ This patch tries to stop printing the stack for non-current tasks if their state change while there is one dump_backtrace() trying to print back trace. Dont we have any lock preventing a task in this situation (while dumping it's backtrace) from running again or changing state. I haven't found any lock preventing a task in this situation, and I think we shouldn't prevent task running if this task is scheduled. /********************************answer question**********************************/ Signed-off-by: Tengfei Fan <tengfeif@codeaurora.org> --- arch/arm64/kernel/traps.c | 23 +++++++++++++++++++++++ 1 file changed, 23 insertions(+) @@ -127,6 +130,9 @@ void dump_backtrace(struct pt_regs *regs, struct task_struct *tsk) */ frame.fp = thread_saved_fp(tsk); frame.pc = thread_saved_pc(tsk); + cur_state = tsk->state; + cur_sp = thread_saved_sp(tsk); + cur_fp = frame.fp; /********************************answer question**********************************/ Should 'saved_state|sp|fp' instead as its applicable to non-current tasks only. 'saved_state|sp|fp' only applies to non-current tasks. /********************************answer question**********************************/ } #ifdef CONFIG_FUNCTION_GRAPH_TRACER frame.graph = 0; @@ -134,6 +140,23 @@ void dump_backtrace(struct pt_regs *regs, struct task_struct *tsk) printk("Call trace:\n"); do { + if (tsk != current && (cur_state != tsk->state + /* + * We would not be printing backtrace for the task + * that has changed state from "saved" state to ohter + * state before hitting the do-while loop but after + * saving the current state. If task's current state + * not equal the "saved" state, then we may print + * wrong call trace or end up in infinite while loop + * if *(fp) and *(fp+8) are same. While the situation + * should be stoped once we found the task's state + * is changed, so we detect the task's current state, + * sp and fp in each while. + */ + || cur_sp != thread_saved_sp(tsk) + || cur_fp != thread_saved_fp(tsk))) { /********************************answer question**********************************/ Why does any of these three mismatches detect the problematic transition not just the state ? 1. we can use "cur_state != tsk->state" prevent printing backtrace if the task's state is changed after "saved" task's state. 2. we can use "cur_sp != thread_saved_sp(tsk)" and "cur_fp != thread_saved_fp(tsk)" prevent printing backtrace if the task's state is changed before "saved" task's state. Because the value of "thread_saved_sp(tsk)" and "thread_saved_fp(tsk)" will not equal "saved" sp(cur_sp) and fp(cur_fp). /********************************answer question**********************************/