英文:
C start a process without seeing the input or output
问题
我正在尝试编写一个初始化系统，我想要在不涉及输入输出的情况下启动一个进程。换句话说，我不想看到输出，也不想输入。我目前使用了suckless的sinit中的以下函数 -
void spawn(char *const argv[]) {
switch (fork()) {
case 0:
sigprocmask(SIG_UNBLOCK, &set, NULL);
setsid();
execvp(argv[0], argv);
perror("execvp");
_exit(1);
break;
case -1:
perror("fork");
break;
default:
break;
}
}
但是，如果我启动一个进程（以top作为测试），它不会在后台运行。我该如何做到这一点？
英文:
I'm trying to write an init system, and I want to spawn a process without the IO. meaning, I don't want to see output or for it to take input. I currently use the function from suckless's sinit -
void spawn(char *const argv[]) {
switch (fork()) {
case 0:
sigprocmask(SIG_UNBLOCK, &set, NULL);
setsid();
execvp(argv[0], argv);
perror("execvp");
_exit(1);
break;
case -1:
perror("fork");
break;
default:
break;
}
}
but if I start a process (used top as a test), it doesn't run "in the background". how can I do that?
答案1
得分: 2
以下是翻译好的部分：
所有进程都期望从它们的父进程继承文件描述符 0、1 和 2，分别作为标准输入、标准输出和标准错误。
传统的方法涉及将它们重定向到 /dev/null。所有输出都会消失，而任何试图从标准输入读取的尝试都会立即显示文件结束指示。
将它们重定向到 /dev/null 包括关闭它们，并在它们的位置打开 /dev/null。
void spawn(char *const argv[]) {
switch (fork()) {
case 0:
sigprocmask(SIG_UNBLOCK, &set, NULL);
setsid();
close(0);
close(1);
close(2);
open("/dev/null", O_RDONLY);
open("/dev/null", O_WRONLY);
open("/dev/null", O_wRONLY);
execvp(argv[0], argv);
鲁棒性建议要仔细检查每个 open() 是否成功，如果打开 /dev/null 失败，您将有更多问题需要担心。
英文:
All processes expect to inherit file descriptors 0, 1, and 2 from their parent process, as standard input, output and error, respectively.
The traditional approach involves redirecting them to /dev/null. All output disappears, and any attempts to read from standard input result in an immediately end-of-file indication.
And redirecting them to /dev/null consists of closing them, and opening /dev/null in their place.
void spawn(char *const argv[]) {
switch (fork()) {
case 0:
sigprocmask(SIG_UNBLOCK, &set, NULL);
setsid();
close(0);
close(1);
close(2);
open("/dev/null", O_RDONLY);
open("/dev/null", O_WRONLY);
open("/dev/null", O_wRONLY);
execvp(argv[0], argv);
Robustness suggests meticulously checking if each open() succeeded, or not. But, if opening /dev/null fails you'll have more problems to worry about.
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