小编典典

通过Linux FrameBuffer将像素绘制到屏幕上

linux

最近,我对一个奇怪的想法感到震惊,他想从/ dev / urandom中获取输入,将相关字符转换为随机整数,然后使用这些整数作为像素rgb /
xy值来绘制到屏幕上。

我已经做过一些研究(在StackOverflow和其他地方),许多建议您可以直接直接写入/ 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,该程序使用mmap写入缓冲区。该程序成功运行,但是再次没有输出到屏幕。有趣的是,当我将笔记本电脑放入Suspend并随后恢复时,我看到了瞬间的图像闪烁(红色方块),该图像早些时候已写入帧缓冲区。在Linux中写帧缓冲区是否可以继续工作以绘画到屏幕上?理想情况下,我想编写一个(ba)sh脚本,但是使用C或类似脚本也可以。谢谢!

编辑:这是示例程序…兽医看似熟悉。

#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;
}

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2020-06-02

共1个答案

小编典典

如果您正在运行X11,则必须通过X11
API才能绘制到屏幕上。在X服务器周围四处乱转(并且,如您所见,通常是行不通的)。它也可能导致崩溃,或者只是一般的显示器损坏。

如果您希望能够在任何地方(控制台和X下)运行,请查看SDL或GGI。如果只关心X11,则可以使用GTK,QT甚至Xlib。有很多很多选择…

2020-06-02