linux下如何实现简单刷屏

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本文分享了一款用于检测LCD屏幕是否存在坏点的简单刷屏程序。通过逐个填充屏幕像素来观察是否有颜色不变化的部分，以此判断LCD屏幕的好坏。

今天为了测试我的LCD有没有坏点写了一个简单的刷屏程序，效果还不错，分析给大家。代码如下：

#include <unistd.h>
#include <stdio.h>
#include <fcntl.h>
#include <linux/fb.h>
#include <sys/mman.h>
#include <string.h>
#include <stdlib.h>
#include <unistd.h>

int main(int argc,char *argv[])
{
int fbfd = 0;
struct fb_var_screeninfo vinfo;
struct fb_fix_screeninfo finfo;
long int screensize = 0;
char *fbp = 0;
char inbuf[240*320*2];
    int i = 0;
/*open /dev/fb0 */
fbfd = open("/dev/fb0",O_RDWR);
if(fbfd<=0)
{
  printf("Error:cannot open framebuffer devices./n");
  return -1;
}
printf("The framebuffer devices was opened successfully./n");
/*get framebuffer infomation*/
if(ioctl(fbfd,FBIOGET_FSCREENINFO,&finfo))
{
  printf("Error: reanding fixed informatin./n");
  return -1;
}
    if(ioctl(fbfd,FBIOGET_VSCREENINFO,&vinfo))
     {
   printf("Error: reanding var informatin./n");
   return -1;
}

/*set screensiez*/
screensize = vinfo.xres*vinfo.yres*vinfo.bits_per_pixel/8;
    /*mmap*/
fbp = (char*)mmap(0,screensize,PROT_READ|PROT_WRITE,MAP_SHARED,fbfd,0);
if((int)fbp == -1)
{

   printf("Error: failed to map framebuffer device to memory./n");
   return -4;
}

printf(" The framebuffer device was mapped to memory sucessfully./n");
memcpy(inbuf,fbp,screensize);//保存刷屏前显存中的数据
     for(i = 0;i<0xff;i++)
{
   memset(fbp,i,screensize);
   sleep(1);
}
   memcpy(fbp,inbuf,screensize);//将保存的显存数据复原回去
munmap(fbp,screensize);
printf("The framebuffer device was munmaped successfully./n");
close(fbfd);
printf("The framebuffer device was closed successfully./n");
return 0;