RE: question about V4L2_MEMORY_USERPTR on 64bit applications

[tbhardwa]

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.


å 2018-10-12äç 11:04 +0800ïZhang Ningåéï
> 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.