question about V4L2_MEMORY_USERPTR on 64bit applications
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.
Hi, Ning,
unsigned long userptr; <<<--- this is a 32bit addr.
I think it's wrong here,for LP64 data modelmachine(unix-like systems), the actual size ofdata type 'unsigned long'is 8(64bits value)whichis equal to pointer.
Quoted 37 lines · Ning A
Hi,
I have question about
V4L2_MEMORY_USERPTRon 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.
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.
Quoted 37 lines · 在 2018-10-12五的 11:04 +0800，Zhang Ning写道：
Hi,
I have question about
V4L2_MEMORY_USERPTRon 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.
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: [email protected] <[email protected]> On Behalf Of Zhang, Ning A
Sent: Friday, October 12, 2018 10:03 AM
To: [email protected]; [email protected]
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.
Quoted 37 lines · 在 2018-10-12五的 11:04 +0800，Zhang Ning写道：
Hi,
I have question about
V4L2_MEMORY_USERPTRon 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.
Quoted 1 line · On Fri, Oct 12, 2018 at 10:52:51AM +0530, [email protected] wrote:
(in 64 bit userspace long is 32-bit)
Not on Linux.
$ cat /tmp/c.c
int main(void)
{
return sizeof(long);
}
$ gcc /tmp/c.c
$ ./a.out; echo $?
8
$ file ./a.out
./a.out: ELF 64-bit LSB pie executable, 64-bit PowerPC or cisco 7500, version 1 (SYSV), dynamically linked, interpreter /lib64/ld64.so.2, for GNU/Linux 3.10.0, BuildID[sha1]=dad78822e741b7900dc7568222822d5d63c31a6c, not stripped