68 lines
2.9 KiB
Plaintext
68 lines
2.9 KiB
Plaintext
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Chinese translated version of Documentation/io_ordering.txt
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If you have any comment or update to the content, please contact the
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original document maintainer directly. However, if you have a problem
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communicating in English you can also ask the Chinese maintainer for
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help. Contact the Chinese maintainer if this translation is outdated
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or if there is a problem with the translation.
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Chinese maintainer: Lin Yongting <linyongting@gmail.com>
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---------------------------------------------------------------------
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Documentation/io_ordering.txt 的中文翻译
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如果想评论或更新本文的内容，请直接联系原文档的维护者。如果你使用英文
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交流有困难的话，也可以向中文版维护者求助。如果本翻译更新不及时或者翻
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译存在问题，请联系中文版维护者。
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中文版维护者： 林永听 Lin Yongting <linyongting@gmail.com>
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中文版翻译者： 林永听 Lin Yongting <linyongting@gmail.com>
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中文版校译者： 林永听 Lin Yongting <linyongting@gmail.com>
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以下为正文
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---------------------------------------------------------------------
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在某些平台上，所谓的内存映射I/O是弱顺序。在这些平台上，驱动开发者有责任
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保证I/O内存映射地址的写操作按程序图意的顺序达到设备。通常读取一个“安全”
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设备寄存器或桥寄存器，触发IO芯片清刷未处理的写操作到达设备后才处理读操作，
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而达到保证目的。驱动程序通常在spinlock保护的临界区退出之前使用这种技术。
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这也可以保证后面的写操作只在前面的写操作之后到达设备（这非常类似于内存
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屏障操作，mb()，不过仅适用于I/O）。
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假设一个设备驱动程的具体例子：
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...
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CPU A: spin_lock_irqsave(&dev_lock, flags)
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CPU A: val = readl(my_status);
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CPU A: ...
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CPU A: writel(newval, ring_ptr);
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CPU A: spin_unlock_irqrestore(&dev_lock, flags)
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...
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CPU B: spin_lock_irqsave(&dev_lock, flags)
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CPU B: val = readl(my_status);
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CPU B: ...
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CPU B: writel(newval2, ring_ptr);
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CPU B: spin_unlock_irqrestore(&dev_lock, flags)
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...
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上述例子中，设备可能会先接收到newval2的值，然后接收到newval的值，问题就
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发生了。不过很容易通过下面方法来修复：
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...
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CPU A: spin_lock_irqsave(&dev_lock, flags)
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CPU A: val = readl(my_status);
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CPU A: ...
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CPU A: writel(newval, ring_ptr);
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CPU A: (void)readl(safe_register); /* 配置寄存器？*/
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CPU A: spin_unlock_irqrestore(&dev_lock, flags)
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...
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CPU B: spin_lock_irqsave(&dev_lock, flags)
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CPU B: val = readl(my_status);
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CPU B: ...
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CPU B: writel(newval2, ring_ptr);
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CPU B: (void)readl(safe_register); /* 配置寄存器？*/
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CPU B: spin_unlock_irqrestore(&dev_lock, flags)
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在解决方案中，读取safe_register寄存器，触发IO芯片清刷未处理的写操作，
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再处理后面的读操作，防止引发数据不一致问题。
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