spin_lock_irqsave

Question

我在uni处理器机器上面临spin_lock_irqsave的奇怪问题。 我有一个以rx_process命名的关键代码/函数，它由两个函数rx_timeout和rx_callback共享。它位于Linux内核驱动程序omap-serial.c

下

rx_process ( uartNum)
{
//critical section
}

rx_timeout (uartNum)//this is called from softIrq context upon timer expiry
{ 
   rx_process (uartNum);
}
rx_callback(uartNum)//this is called when interrupt handler registers upon interrupt and when 4k buf is full in rx_process function which invokes start_rx_dma which results in this call back function.
{
   rx_process (uartNum);
}

我发现竞争条件发生在这两个函数之间，因此我决定在函数开头和函数结尾处的rx_process中引入spin_lock_irqsave，但它仍然会导致竞争状态并且从时间到时间，我观察到数据丢失和内核恐慌。

rx_process ( uartNum)
{ spin_lock_irqsave(&lock, flags);
 //critical section
 spin_unlock_irqrestore(&lock, flags);
}
rx_timeout (uartNum)//this is called from softIrq context upon timer expiry
{ 
  rx_process (uartNum);
}
rx_callback(uartNum)//this is called when interrupt handler registers upon interrupt and when 4k buf is full in rx_process function which invokes start_rx_dma which results in this call back function
{
 rx_process (uartNum);
}

现在，我将spin_lock_irqsave移动到rx_timeout以及rx_callback函数，我看不到竞争条件。

 rx_process ( uartNum)
    { 
     //critical section
     spin_unlock_irqrestore(&lock, flags);
    }
    rx_timeout (uartNum)//this is called from softIrq context upon timer expiry
    { 
      spin_lock_irqsave(&lock, flags);
      rx_process (uartNum);
      spin_unlock_irqrestore(&lock, flags);
    }
    rx_callback(uartNum)//this is called when interrupt handler registers upon interrupt and when 4k buf is full in rx_process function which invokes start_rx_dma which results in this call back function
    {
     spin_lock_irqsave(&lock, flags);
     rx_process (uartNum);
     spin_unlock_irqrestore(&lock, flags);
    }

如果有人能解释为什么rx_process中使用的spin_lock_irqsave失败并且没有阻止竞争条件，我真的很感激？