最近，我被一个奇怪的想法所震惊，从 /dev/urandom 获取输入，将相关字符转换为随机整数，然后使用这些整数作为像素的 rgb/x-y 值来绘制到屏幕上。
我做了一些研究（在 StackOverflow 和其他地方），许多人建议您可以直接写入 /dev/fb0，因为它是设备的文件表示形式。不幸的是，这似乎没有产生任何视觉上明显的结果。
我找到了一个来自 QT 教程（不再可用）的示例 C 程序，它使用 mmap 写入缓冲区。程序成功运行，但屏幕上没有任何输出。有趣的是，当我将笔记本电脑置于挂起状态并随后恢复时，我看到图像（红色方块）瞬间闪烁，该图像很早就写入了帧缓冲区。在 Linux 中写入帧缓冲区是否还能用于在屏幕上绘制？理想情况下，我想编写一个 (ba)sh 脚本，但 C 或类似脚本也可以。谢谢！
编辑：这是示例程序......兽医可能看起来很熟悉。
#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 = 100; 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
//location += 4;
} 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;
}
