On 4/24/2019 6:37 PM, Al Viro wrote:
On Wed, Apr 24, 2019 at 05:40:40PM +0530, Mukesh Ojha wrote:
Al,

i tried to put traceprintk inside ioctl after fdget and fdput on a simple
call of open  => ioctl => close
in a loop, and multithreaded, presumably?

on /dev/uinput.

           uinput-532   [002] ....    45.312044: SYSC_ioctl: 2     <= f_count
     <After fdget()
           uinput-532   [002] ....    45.312055: SYSC_ioctl: 2
<After fdput()
           uinput-532   [004] ....    45.313766: uinput_open: uinput: 1
           uinput-532   [004] ....    45.313783: SYSC_ioctl: 1
           uinput-532   [004] ....    45.313788: uinput_ioctl_handler:
uinput: uinput_ioctl_handler, 1
           uinput-532   [004] ....    45.313835: SYSC_ioctl: 1
           uinput-532   [004] ....    45.313843: uinput_release: uinput:  0


So while a ioctl is running the f_count is 1, so a fput could be run and do
atomic_long_dec_and_test
this could call release right ?
Look at ksys_ioctl():
int ksys_ioctl(unsigned int fd, unsigned int cmd, unsigned long arg)
{
         int error;
         struct fd f = fdget(fd);
an error or refcount bumped
         if (!f.file)
                 return -EBADF;
not an error, then.  We know that ->release() won't be called
until we drop the reference we've just acquired.
         error = security_file_ioctl(f.file, cmd, arg);
         if (!error)
                 error = do_vfs_ioctl(f.file, fd, cmd, arg);
... and we are done with calling ->ioctl(), so
         fdput(f);
... we drop the reference we'd acquired.

Seeing refcount 1 inside ->ioctl() is possible, all right:

CPU1: ioctl(2) resolves fd to struct file *, refcount 2
CPU2: close(2) rips struct file * from descriptor table and does fput() to drop 
it;
        refcount reaches 1 and fput() is done; no call of ->release() yet.
CPU1: we get arouund to ->ioctl(), where your trace sees refcount 1
CPU1: done with ->ioctl(), drop our reference.  *NOW* refcount gets to 0, and
        ->release() is called.

Thanks for the detail reply, Al

This was my simple program no multithreading just to understand f_counting

int main()
{
        int fd = open("/dev/uinput", O_WRONLY | O_NONBLOCK);
        ioctl(fd, UI_SET_EVBIT, EV_KEY);
        close(fd);
        return 0;
}

           uinput-532   [002] ....    45.312044: SYSC_ioctl: 2   <= f_count >    <After fdget()           uinput-532   [002] ....    45.312055: SYSC_ioctl: 2            <After fdput()           uinput-532   [004] ....    45.313766: uinput_open: uinput: 1   /* This is from the uinput driver uinput_open()*/

  =>>>>                         /* All the above calls happened for the open() in userspace*/

          uinput-532   [004] ....    45.313783: SYSC_ioctl: 1 /* This print is for the trace, i put after fdget */           uinput-532   [004] ....    45.313788: uinput_ioctl_handler: uinput: uinput_ioctl_handler, 1 /* This print is from the uinput_ioctl driver */

          uinput-532   [004] ....    45.313835: SYSC_ioctl: 1 /* This print is for the trace, i put after fdput*/           uinput-532   [004] ....    45.313843: uinput_release: uinput:  0 /* And this is from the close()  */


Should fdget not suppose to increment the f_count here, as it is coming 1 ?
This f_count to one is done at the open, but i have no idea how this  below f_count 2 came before open() for
this simple program.

         uinput-532   [002] ....    45.312044: SYSC_ioctl: 2 <= f_count >    <After fdget()           uinput-532   [002] ....    45.312055: SYSC_ioctl: 2            <After fdput()

-Mukesh

IOW, in your trace fput() has already been run by close(2); having somebody else
do that again while we are in ->ioctl() would be a bug (to start with, where
did they get that struct file * and why wasn't that reference contributing to
struct file refcount?)

In all cases we only call ->release() once all references gone - both
the one(s) in descriptor tables and any transient ones acquired by
fdget(), etc.

I would really like to see a reproducer for the original use-after-free 
report...

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