android/linux usb 摄像头获取一张图像

usb uvc 摄像头测试程序:获取一张图片

代码是网上找的，能编译过

#include <stdio.h>#include <stdlib.h>#include <string.h>#include <assert.h> #include <getopt.h>             #include <fcntl.h>             #include <unistd.h>#include <errno.h>#include <malloc.h>#include <sys/stat.h>#include <sys/types.h>#include <sys/time.h>#include <sys/mman.h>#include <sys/ioctl.h>#include <asm/types.h>         #include <linux/videodev2.h>#define CLEAR(x) memset (&(x), 0, sizeof (x))struct buffer {        void *                  start;        size_t                  length;};static char *           dev_name        = "/dev/video0";//摄像头设备名static int              fd              = -1;struct buffer *         buffers         = NULL;static unsigned int     n_buffers       = 0;FILE *file_fd;static unsigned long file_length;static unsigned char *file_name;////////////////////////////////////////////////////////获取一帧数据//////////////////////////////////////////////////////static int read_frame (void){struct v4l2_buffer buf;unsigned int i;CLEAR (buf);buf.type = V4L2_BUF_TYPE_VIDEO_CAPTURE;buf.memory = V4L2_MEMORY_MMAP;ioctl (fd, VIDIOC_DQBUF, &buf); //出列采集的帧缓冲assert (buf.index < n_buffers);printf ("buf.index dq is %d,\n",buf.index);fwrite(buffers[buf.index].start, buffers[buf.index].length, 1, file_fd); //将其写入文件中  ioctl (fd, VIDIOC_QBUF, &buf); //再将其入列return 1;}int main (int argc,char ** argv){struct v4l2_capability cap; struct v4l2_format fmt;unsigned int i;enum v4l2_buf_type type;file_fd = fopen("test-mmap.jpg", "w");//图片文件名fd = open (dev_name, O_RDWR /* required */ | O_NONBLOCK, 0);//打开设备ioctl (fd, VIDIOC_QUERYCAP, &cap);//获取摄像头参数CLEAR (fmt);fmt.type = V4L2_BUF_TYPE_VIDEO_CAPTURE;fmt.fmt.pix.width       = 640; fmt.fmt.pix.height      = 480;fmt.fmt.pix.pixelformat = V4L2_PIX_FMT_YUYV;fmt.fmt.pix.field       = V4L2_FIELD_INTERLACED;ioctl (fd, VIDIOC_S_FMT, &fmt); //设置图像格式file_length = fmt.fmt.pix.bytesperline * fmt.fmt.pix.height; //计算图片大小struct v4l2_requestbuffers req;CLEAR (req);req.count = 4;req.type = V4L2_BUF_TYPE_VIDEO_CAPTURE;req.memory = V4L2_MEMORY_MMAP;ioctl (fd, VIDIOC_REQBUFS, &req); //申请缓冲，count是申请的数量if (req.count < 2)printf("Insufficient buffer memory\n");buffers = calloc (req.count, sizeof (*buffers));//内存中建立对应空间for (n_buffers = 0; n_buffers < req.count; ++n_buffers) {struct v4l2_buffer buf;   //驱动中的一帧CLEAR (buf);buf.type = V4L2_BUF_TYPE_VIDEO_CAPTURE;buf.memory = V4L2_MEMORY_MMAP;buf.index = n_buffers;if (-1 == ioctl (fd, VIDIOC_QUERYBUF, &buf)) //映射用户空间printf ("VIDIOC_QUERYBUF error\n");buffers[n_buffers].length = buf.length;buffers[n_buffers].start =mmap (NULL /* start anywhere */,    //通过mmap建立映射关系buf.length,PROT_READ | PROT_WRITE /* required */,MAP_SHARED /* recommended */,fd, buf.m.offset);if (MAP_FAILED == buffers[n_buffers].start)printf ("mmap failed\n");}for (i = 0; i < n_buffers; ++i) {struct v4l2_buffer buf;CLEAR (buf);buf.type = V4L2_BUF_TYPE_VIDEO_CAPTURE;buf.memory = V4L2_MEMORY_MMAP;buf.index = i;if (-1 == ioctl (fd, VIDIOC_QBUF, &buf))//申请到的缓冲进入列队printf ("VIDIOC_QBUF failed\n");}            type = V4L2_BUF_TYPE_VIDEO_CAPTURE;if (-1 == ioctl (fd, VIDIOC_STREAMON, &type)) //开始捕捉图像数据printf ("VIDIOC_STREAMON failed\n");for (;;) //这一段涉及到异步IO{fd_set fds;struct timeval tv;int r;FD_ZERO (&fds);//将指定的文件描述符集清空FD_SET (fd, &fds);//在文件描述符集合中增加一个新的文件描述符/* Timeout. */tv.tv_sec = 3;tv.tv_usec = 0;r = select (fd + 1, &fds, NULL, NULL, &tv);//判断是否可读（即摄像头是否准备好），tv是定时if (-1 == r) {if (EINTR == errno)continue;printf ("select err\n");        }        if (0 == r) {fprintf (stderr, "select timeout\n");exit (EXIT_FAILURE);        }if (read_frame ())//如果可读，执行read_frame ()函数，并跳出循环break;}unmap:for (i = 0; i < n_buffers; ++i)if (-1 == munmap (buffers[i].start, buffers[i].length))printf ("munmap error");close (fd);fclose (file_fd);exit (EXIT_SUCCESS);return 0;}

Linux上直接 arm-linux-gcc videoTest.c -o videoTest ，然后运行./videoTest，此目录下就会生成test-mmap.jpeg
 如果是在andorid上， 编译时加上-static : arm-linux-gcc videoTest.c -o videoText -static

我测试的时候是在android4.0 (Linux kernel2.6.34)，试了两个usb uvc 摄像头，都生成了图片test-mmap.jpeg 大小为600K,但打不开。

程序会随机出现select timeout的错误，原因还在寻找