64 bit kernel have 64 bit long(not 32 bit), which is not the case with userspace (in 64 bit userspace long is 32-bit). Probably thig got you confused.
-----Original Message-----
From: linux-media-owner@xxxxxxxxxxxxxxx <linux-media-owner@xxxxxxxxxxxxxxx> On Behalf Of Zhang, Ning A
Sent: Friday, October 12, 2018 10:03 AM
To: linux-kernel@xxxxxxxxxxxxxxx; linux-media@xxxxxxxxxxxxxxx
Subject: Re: question about V4L2_MEMORY_USERPTR on 64bit applications
sorry for wrong question, I really meet memory address truncated issue, when use V4L2 kernel APIs.
in a kernel thread created by kernel_thread() I vm_mmap a shmem_file to addr: 00007ffff7fa8000 and queue it to V4L2, after dequeue it, and I find the address is truncated to 00000000f7fa8000
I use __u64 {aka long long unsigned int} to save address, and I find userptr is unsigned long, wrongly think it as "data truncated"
and a lot of __u32 in this structure.
everything works fine, but I still don't understand why high 32bit be 0..
BR.
Ning.
在 2018-10-12五的 11:04 +0800，Zhang Ning写道：
> Hi,
>
> I have question about V4L2_MEMORY_USERPTR on 64bit applications.
>
> struct v4l2_buffer {
> __u32 index;
> __u32 type;
> __u32 bytesused;
> __u32 flags;
> __u32 field;
> struct timeval timestamp;
> struct v4l2_timecode timecode;
> __u32 sequence;
>
> /* memory location */
> __u32 memory;
> union {
> __u32 offset;
> unsigned long userptr; <<<--- this is a 32bit addr.
> struct v4l2_plane *planes;
> __s32 fd;
> } m;
> __u32 length;
> __u32 reserved2;
> __u32 reserved;
> };
>
> when use a 64bit application, memory from malloc is 64bit address.
> memory from GPU (eg, intel i915) are also 64bit address.
>
> when use these kind of memory as V4L2_MEMORY_USERPTR, address will be
> truncated into 32bit.
>
> this would be error, but actually not. I really don't understand.
>
> BR.
> Ning.
