Question

我想在一个下雪的周末创建自己的init和一些Linux乐趣。我知道，内核使用rootfs启动，并在一些驱动程序加载和磁盘安装后将流程提供给/ sbin / init。我下载了ubuntu云映像并尝试使用kvm直接进行内核启动，如下所示：

kvm -m 1G -nographic -kernel vmlinuz-3.19.0-32-generic -initrd initrd.img-3.19.0-32-generic -append "console=ttyS0 root=/dev/sda1 rw init=/myinit" -hda mydisk.img

使用trusty-server-cloudimg-amd64-disk1.img（如果你不介意挂在cloud-init上）它可以运行得很好，然后我继续复制它并删除它的内容。

modprobe nbd
qemu-nbd -c /dev/nbd0 mydisk.img 
fdisk -l /dev/nbd0 # confirm partition
mount /dev/nbd0p1 disk/
# Delete all files with myinit.c and myinit

这是我的魔法初学者：

int main(){
    printf("Welcome to my kernel\n");
    printf("Welcome to my kernel\n");
    printf("Welcome to my kernel\n");
    while(1);
}

我用gcc -static myinit.c -o myinit编译它。但是由于我的init，因此发生内核恐慌。我通过将myinit重命名为myinit2并且内核无法找到它来验证它，并且它没有崩溃。我知道编写init不能像上面这么简单，但它需要的步骤是什么？我正在阅读新手的源代码

Begin: Mounting root file system ... Begin: Running /scripts/local-top ... done.
Begin: Running /scripts/local-premount ... [    1.460164] tsc: Refined TSC clocksource calibration: 2394.558 MHz
[    1.866560] input: ImExPS/2 Generic Explorer Mouse as /devices/platform/i8042/serio1/input/input3
done.
[    6.251763] EXT4-fs (sda1): recovery complete
[    6.253623] EXT4-fs (sda1): mounted filesystem with ordered data mode. Opts: (null)
Begin: Running /scripts/local-bottom ... done.
done.
Begin: Running /scripts/init-bottom ... mount: mounting /dev on /root/dev failed: No such file or directory
done.
mount: mounting /sys on /root/sys failed: No such file or directory
mount: mounting /proc on /root/proc failed: No such file or directory
[    6.299404] Kernel panic - not syncing: Attempted to kill init! exitcode=0x00000200
[    6.299404] 
[    6.300013] CPU: 0 PID: 1 Comm: init Not tainted 3.19.0-32-generic #37~14.04.1-Ubuntu
[    6.300013] Hardware name: QEMU Standard PC (i440FX + PIIX, 1996), BIOS Bochs 01/01/2011
[    6.300013]  ffff88003c118700 ffff88003dee7e38 ffffffff817af41b 00000000000017d6
[    6.300013]  ffffffff81a90be8 ffff88003dee7eb8 ffffffff817a925b ffff88003dee8000
[    6.300013]  ffffffff00000010 ffff88003dee7ec8 ffff88003dee7e68 ffffffff81c5ee20
[    6.300013] Call Trace:
[    6.300013]  [<ffffffff817af41b>] dump_stack+0x45/0x57
[    6.300013]  [<ffffffff817a925b>] panic+0xc1/0x1f5
[    6.300013]  [<ffffffff81077b01>] do_exit+0xa11/0xb00
[    6.300013]  [<ffffffff811ec53c>] ? vfs_write+0x15c/0x1f0
[    6.300013]  [<ffffffff81077c7f>] do_group_exit+0x3f/0xa0
[    6.300013]  [<ffffffff81077cf4>] SyS_exit_group+0x14/0x20
[    6.300013]  [<ffffffff817b6dcd>] system_call_fastpath+0x16/0x1b
[    6.300013] Kernel Offset: 0x0 from 0xffffffff81000000 (relocation range: 0xffffffff80000000-0xffffffffbfffffff)
[    6.300013] drm_kms_helper: panic occurred, switching back to text console
[    6.300013] ---[ end Kernel panic - not syncing: Attempted to kill init! exitcode=0x00000200
[    6.300013]

我知道myinit是完全静态的：

# ldd disk/myinit
    not a dynamic executable

所以它不应该依赖别的东西，我想。但是我做错了什么，为什么内核恐慌？ （没有printfs的内核恐慌）

我正在阅读sysvinit源码（它应该比upstart＆amp; systemd＆amp; openrc更简单），但它太长了，但init的主要思想是拥有进程，它也在（1）循环中。< / p>

Answer 1

stdin启动后，stdout，stderr和init可能无法连接。在init程序开始时看到类似于以下内容的序列很常见：

    int onefd = open("/dev/console", O_RDONLY, 0);
    dup2(onefd, 0); // stdin
    int twofd = open("/dev/console", O_RDWR, 0);
    dup2(twofd, 1); // stdout
    dup2(twofd, 2); // stderr

    if (onefd > 2) close(onefd);
    if (twofd > 2) close(twofd);

这可确保stdin，stdout和stderr连接到系统控制台。

编写自己的init可执行文件

1 个答案: