On 10/30/2014 07:26 AM, Daniel Thompson wrote:
> On 29/10/14 18:14, Stephen Boyd wrote:
>> The handle_rx() path calls uart_handle_sysrq_char() with the port
>> lock held. This causes a spinlock recursion. Release and
>> reacquire the lock here to avoid this.
>>
>> BUG: spinlock recursion on CPU#0, swapper/0
>> lock: msm_uart_ports+0x1e0/0x2d0, .magic: dead4ead, .owner: swapper/0, .owner_cpu: 0
>> CPU: 0 PID: 0 Comm: swapper Not tainted 3.17.0-rc7-00012-gb38ee8265941 #69
>> [<c0013964>] (unwind_backtrace) from [<c0011f74>] (show_stack+0x10/0x14)
>> [<c0011f74>] (show_stack) from [<c004ed1c>] (do_raw_spin_lock+0x11c/0x13c)
>> [<c004ed1c>] (do_raw_spin_lock) from [<c02d44c0>] (msm_console_write+0x78/0x188)
>> [<c02d44c0>] (msm_console_write) from [<c0052880>] (call_console_drivers.constprop.22+0xb4/0x144)
>> [<c0052880>] (call_console_drivers.constprop.22) from [<c0053570>] (console_unlock+0x27c/0x4ac)
>> [<c0053570>] (console_unlock) from [<c0053bb4>] (vprintk_emit+0x1f4/0x5a8)
>> [<c0053bb4>] (vprintk_emit) from [<c04ad0ac>] (printk+0x30/0x40)
>> [<c04ad0ac>] (printk) from [<c02c2990>] (__handle_sysrq+0x58/0x1b8)
>> [<c02c2990>] (__handle_sysrq) from [<c02d41b0>] (msm_irq+0x694/0x6f8)
>> [<c02d41b0>] (msm_irq) from [<c0055740>] (handle_irq_event_percpu+0x58/0x270)
>> [<c0055740>] (handle_irq_event_percpu) from [<c0055994>] (handle_irq_event+0x3c/0x5c)
>> [<c0055994>] (handle_irq_event) from [<c0057e84>] (handle_level_irq+0x9c/0x138)
>> [<c0057e84>] (handle_level_irq) from [<c005509c>] (generic_handle_irq+0x24/0x38)
>> [<c005509c>] (generic_handle_irq) from [<c000f730>] (handle_IRQ+0x44/0xb0)
>> [<c000f730>] (handle_IRQ) from [<c0008518>] (msm_vic_handle_irq+0x44/0x64)
>> [<c0008518>] (msm_vic_handle_irq) from [<c04b5ac4>] (__irq_svc+0x44/0x7c)
>> Exception stack(0xc0719f68 to 0xc0719fb0)
>> 9f60: 00000001 00000001 00000000 c0722938 c0718000 c0769acc
>> 9f80: 00000000 c0720098 c0769305 4117b362 c0769acc 00000000 01000000 c0719fb0
>> 9fa0: c004cab0 c000f880 20000013 ffffffff
>> [<c04b5ac4>] (__irq_svc) from [<c000f880>] (arch_cpu_idle+0x20/0x30)
>> [<c000f880>] (arch_cpu_idle) from [<c004691c>] (cpu_startup_entry+0xf4/0x23c)
>> [<c004691c>] (cpu_startup_entry) from [<c06d8b70>] (start_kernel+0x32c/0x394)
>>
>> Cc: Frank Rowand <frank.rowand@xxxxxxxxxxxxxx>
>> Cc: Daniel Thompson <daniel.thompson@xxxxxxxxxx>
>> Signed-off-by: Stephen Boyd <sboyd@xxxxxxxxxxxxxx>
>> ---
>> drivers/tty/serial/msm_serial.c | 6 +++++-
>> 1 file changed, 5 insertions(+), 1 deletion(-)
>>
>> diff --git a/drivers/tty/serial/msm_serial.c b/drivers/tty/serial/msm_serial.c
>> index 4b6c78331a64..cedcc36762a2 100644
>> --- a/drivers/tty/serial/msm_serial.c
>> +++ b/drivers/tty/serial/msm_serial.c
>> @@ -174,6 +174,7 @@ static void handle_rx(struct uart_port *port)
>> while ((sr = msm_read(port, UART_SR)) & UART_SR_RX_READY) {
>> unsigned int c;
>> char flag = TTY_NORMAL;
>> + int sysrq;
>>
>> c = msm_read(port, UART_RF);
>>
>> @@ -195,7 +196,10 @@ static void handle_rx(struct uart_port *port)
>> else if (sr & UART_SR_PAR_FRAME_ERR)
>> flag = TTY_FRAME;
>>
>> - if (!uart_handle_sysrq_char(port, c))
>> + spin_unlock(&port->lock);
>> + sysrq = uart_handle_sysrq_char(port, c);
>> + spin_lock(&port->lock);
>> + if (!sysrq)
>> tty_insert_flip_char(tport, c, flag);
>
> Does tty_insert_flip_char() need the port to be locked?
No.
It does not serialize internally, so concurrent use will blow up, but
that doesn't look possible here.
Can bad things happen if a well-timed set_termios() happens while the
lock is dropped?
Regards,
Peter Hurley
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