question about V4L2_MEMORY_USERPTR on 64bit applications

["Zhang, Ning A" <ning.a.zhang@intel.com>]

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.