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我在单处理器机器上遇到了 spin_lock_irqsave 的奇怪问题。我有一段以 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);
}

我发现这两个函数之间发生了竞态条件，因此我决定在函数开始和函数结束时在 rx_process 中引入 spin_lock_irqsave，但它仍然会导致竞态条件，并且不时，我观察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，我没有看到竞争条件。

 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在 rx_process 中使用的 spin_lock_irqsave 失败并且没有防止竞争条件，我将非常感激？

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0 回答 0