On Thu, Jun 24, 2021 at 05:37:38PM +0530, Saubhik Mukherjee wrote:
On 6/23/21 12:46 PM, Johan Hovold wrote:
On Wed, Jun 23, 2021 at 11:06:53AM +0530, Saubhik Mukherjee wrote:
On 6/17/21 4:52 PM, Greg KH wrote:
On Thu, Jun 17, 2021 at 04:34:43PM +0530, Saubhik Mukherjee wrote:
Suppose the driver is registered and a UART port is added. Once an
application opens the port, owl_uart_startup is called which registers
the interrupt handler owl_uart_irq.
We could have the following race condition:
When device is removed, owl_uart_remove is called, which calls
uart_remove_one_port, which calls owl_uart_release_port, which writes
NULL to port->membase. At this point parallely, an interrupt could be
handled by owl_uart_irq which reads port->membase.
This is because it is possible to remove device without closing a port.
Thus, we need to check it and call owl_uart_shutdown in owl_uart_remove.
No, this makes no sense at all. The port is deregistered and hung up by
uart_remove_one_port() (and the interrupt line is consequently disabled
by the driver) before it is released so this can never happen.
Thanks for the reply. I am not sure I understand. I could not find any
interrupt disabling in owl_uart_remove. Could you point out where/how is
the interrupt line is disabled before releasing the port?
The interrupt line is disabled by owl_uart_shutdown(), which is called
when uart_remove_one_port() hangs up an open tty. And as I mentioned
this happens after deregistering the port (so no new opens) and before
releasing the port.
Johan