挂起系统呼叫

Question

因此，我正在使用一个定制的内核模块，我正在为它编写一个python前端。内核模块工作，并向/dev/fb1添加一个框架缓冲区设备文件。我可以很好地读和写。我一直在使用python的mmap模块来映射设备缓冲区，这似乎非常有效。

现在我正在尝试实现numpy，我使用的是numpy的memmap函数，我的假设是，这个函数应该同样工作。问题是，使用numpy的memmap函数打开设备文件会挂起内核(我认为)。

这就是我最初打开文件时所做的

self.surface = np.memmap(dev, dtype=np.uint16, mode='r+', shape=(320,240))

进程挂起，我无法杀死python，除非通过killall python，这可能会使文件资源处于打开状态。任何以后打开文件的任何访问都会无限期挂起，只需执行以下操作

f = open('/dev/fb1', 'r+b')

我在dmesg里找到了这个

[ 1081.480104] INFO: task python2.6:2834 blocked for more than 120 seconds.
[ 1081.480109] "echo 0 > /proc/sys/kernel/hung_task_timeout_secs" disables this message.
[ 1081.480113] python2.6     D 0000000100004eec     0  2834      1 0x00000004
[ 1081.480118]  ffff88020a175db8 0000000000000086 0000000000000000 0000000000015980
[ 1081.480122]  ffff88020a175fd8 0000000000015980 ffff88020a175fd8 ffff88022b69adc0
[ 1081.480127]  0000000000015980 0000000000015980 ffff88020a175fd8 0000000000015980
[ 1081.480131] Call Trace:
[ 1081.480142]  [<ffffffff81049b17>] ? mutex_spin_on_owner+0x97/0xc0
[ 1081.480148]  [<ffffffff81589477>] __mutex_lock_slowpath+0xf7/0x180
[ 1081.480151]  [<ffffffff8158935b>] mutex_lock+0x2b/0x50
[ 1081.480157]  [<ffffffff812f3bcf>] fb_release+0x1f/0x60
[ 1081.480161]  [<ffffffff81154825>] __fput+0xf5/0x210
[ 1081.480164]  [<ffffffff81154965>] fput+0x25/0x30
[ 1081.480168]  [<ffffffff81123d35>] remove_vma+0x45/0x90
[ 1081.480171]  [<ffffffff81126179>] do_munmap+0x309/0x3a0
[ 1081.480174]  [<ffffffff81126266>] sys_munmap+0x56/0x80
[ 1081.480180]  [<ffffffff8100a0f2>] system_call_fastpath+0x16/0x1b
[ 1081.480183] INFO: task ipython:2856 blocked for more than 120 seconds.
[ 1081.480185] "echo 0 > /proc/sys/kernel/hung_task_timeout_secs" disables this message.
[ 1081.480187] ipython       D ffff880256e92018     0  2856   1841 0x00000000
[ 1081.480191]  ffff88022341fb58 0000000000000086 ffffffff81625f10 0000000000015980
[ 1081.480196]  ffff88022341ffd8 0000000000015980 ffff88022341ffd8 ffff88022b7c16e0
[ 1081.480200]  0000000000015980 0000000000015980 ffff88022341ffd8 0000000000015980
[ 1081.480204] Call Trace:
[ 1081.480207]  [<ffffffff81589477>] __mutex_lock_slowpath+0xf7/0x180
[ 1081.480210]  [<ffffffff8158935b>] mutex_lock+0x2b/0x50
[ 1081.480214]  [<ffffffff812f3cd8>] fb_open+0xc8/0x200
[ 1081.480217]  [<ffffffff8115657d>] ? cdev_get+0x2d/0xb0
[ 1081.480221]  [<ffffffff81156e6a>] chrdev_open+0x10a/0x200
[ 1081.480225]  [<ffffffff810878a1>] ? in_group_p+0x31/0x40
[ 1081.480228]  [<ffffffff81156d60>] ? chrdev_open+0x0/0x200
[ 1081.480232]  [<ffffffff811512c5>] __dentry_open+0xe5/0x330
[ 1081.480237]  [<ffffffff81260e4f>] ? security_inode_permission+0x1f/0x30
[ 1081.480240]  [<ffffffff81151624>] nameidata_to_filp+0x54/0x70
[ 1081.480244]  [<ffffffff8115e398>] finish_open+0xe8/0x1d0
[ 1081.480248]  [<ffffffff8116701f>] ? dput+0xdf/0x1b0
[ 1081.480251]  [<ffffffff8115f7f6>] do_last+0x86/0x460
[ 1081.480254]  [<ffffffff81161b2b>] do_filp_open+0x21b/0x660
[ 1081.480259]  [<ffffffff8112117f>] ? handle_mm_fault+0x32f/0x440
[ 1081.480263]  [<ffffffff8116d33a>] ? alloc_fd+0x10a/0x150
[ 1081.480266]  [<ffffffff81151069>] do_sys_open+0x69/0x170
[ 1081.480270]  [<ffffffff811511b0>] sys_open+0x20/0x30
[ 1081.480273]  [<ffffffff8100a0f2>] system_call_fastpath+0x16/0x1b

我想我的问题是，我能手动关闭系统呼叫吗？还是以某种方式解除了互斥锁？还是我完全错过了错误告诉我的一切。

奇怪的是，即使仅仅是memmap调用也会破坏框架缓冲区，并将垃圾写入我的显示器。我猜这只是有点麻烦，不能很好地处理设备文件。

更新

它是来自ps -l的输出。第一个python最初运行numpy memmap调用(至少我相当肯定)。第二个ipython是在第一个进程挂起之后，运行一个简单的普通python打开调用。

0 D  1000  2834     1  0  80   0 - 22101 fb_rel ?        00:00:00 python2.6
0 D  1000  2856  1841  0  80   0 - 15065 fb_ope pts/1    00:00:00 ipython

Answer 1

堆栈跟踪表明，前面的python命令被卡在spinlock中，试图在退出时释放互斥对象(发生了可怕的错误)。POSIX指出，与进入互斥线程的线程不同，您不能将互斥对象从不同线程中释放出来。下一步是找出是什么资源导致互斥锁在退出时被保存。发自：

linux-source-2.6.38/kernel/mutex.c:
     69  * The mutex must later on be released by the same task that
     70  * acquired it. Recursive locking is not allowed. The task
     71  * may not exit without first unlocking the mutex. Also, kernel
     72  * memory where the mutex resides mutex must not be freed with
     73  * the mutex still locked. The mutex must first be initialized
     74  * (or statically defined) before it can be locked. memset()-ing
     75  * the mutex to 0 is not allowed.

您可以首先对python进行stracing，或者在驱动程序上启用dtrace。