如何通过linux framebuffer在屏幕上显示某些内容？

我发现以下代码旨在在屏幕上绘制正方形.

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

 int main()
 {
     int fbfd = 0;
     struct fb_var_screeninfo vinfo;
     struct fb_fix_screeninfo finfo;
     long int screensize = 0;
     char *fbp = 0;
     int x = 0, y = 0;
     long int location = 0;

 // Open the file for reading and writing
 fbfd = open("/dev/fb0", O_RDWR);
 if (fbfd == -1) {
     perror("Error: cannot open framebuffer device");
     exit(1);
 }
 printf("The framebuffer device was opened successfully.\n");

 // Get fixed screen information
 if (ioctl(fbfd, FBIOGET_FSCREENINFO, &finfo) == -1) {
     perror("Error reading fixed information");
     exit(2);
 }

 // Get variable screen information
 if (ioctl(fbfd, FBIOGET_VSCREENINFO, &vinfo) == -1) {
     perror("Error reading variable information");
     exit(3);
 }

 printf("%dx%d, %dbpp\n", vinfo.xres, vinfo.yres, vinfo.bits_per_pixel);

 // Figure out the size of the screen in bytes
 screensize = vinfo.xres * vinfo.yres * vinfo.bits_per_pixel / 8;

 // Map the device to memory
 fbp = (char *)mmap(0, screensize, PROT_READ | PROT_WRITE, MAP_SHARED,
                    fbfd, 0);
 if ((int)fbp == -1) {
     perror("Error: failed to map framebuffer device to memory");
     exit(4);
 }
 printf("The framebuffer device was mapped to memory successfully.\n");

 x = 300; y = 100;       // Where we are going to put the pixel

 // Figure out where in memory to put the pixel
 for (y = 100; y < 300; y++)
     for (x = 100; x < 300; x++) {

         location = (x+vinfo.xoffset) * (vinfo.bits_per_pixel/8) +
                    (y+vinfo.yoffset) * finfo.line_length;

         if (vinfo.bits_per_pixel == 32) {
             *(fbp + location) = 100;        // Some blue
             *(fbp + location + 1) = 15+(x-100)/2;     // A little green
             *(fbp + location + 2) = 200-(y-100)/5;    // A lot of red
             *(fbp + location + 3) = 0;      // No transparency
         } else  { //assume 16bpp
             int b = 10;
             int g = (x-100)/6;     // A little green
             int r = 31-(y-100)/16;    // A lot of red
             unsigned short int t = r<<11 | g << 5 | b;
             *((unsigned short int*)(fbp + location)) = t;
         }

     }
 munmap(fbp, screensize);
 close(fbfd);
 return 0;
 }

当我运行它时,没有错误,但是不幸的是,没有任何反应,没有任何显示.如何在屏幕上获取图片？我正在研究Ubuntu 14.

解决方法:

运行正常.

帧缓冲程序使用Linux“文本”控制台(它们的作用不只是文本),而不是XWindows,也不是ssh终端会话.

对于大多数目的,帧缓冲不是一个很好的接口.编写可以接管机器的游戏可能没问题.新的Linux桌面程序应使用XWindows兼容的软件.

跑步：

>在XWindows Linux桌面上,按Control Alt F1以获取Linux“文本”控制台. (请勿使用终端窗口).
>使用密码登录
>将程序放入square.c.用gcc square.c -o square之类的程序编译程序.它将给出有关指针/ int转换看起来不错的警告.
> [*]使用sudo su成为root
>运行编译的程序./square

它使阴影变成粉红色的正方形.

或显示错误：如果您不是root用户,则无法打开帧缓冲设备.

[*]切勿以超级用户身份运行您不完全信任的程序

标签：framebuffer,c-3,linux
来源： https://codeday.me/bug/20191119/2039358.html