How can a C/C++ program put itself into background

问题:

What's the best way for a running C or C++ program that's been launched from the command line to put itself into the background, equivalent to if the user had launched from the unix shell with '&' at the end of the command? (But the user didn't.) It's a GUI app and doesn't need any shell I/O, so there's no reason to tie up the shell after launch. But I want a shell command launch to be auto-backgrounded without the '&' (or on Windows).

Ideally, I want a solution that would work on any of Linux, OS X, and Windows. (Or separate solutions that I can select with #ifdef.) It's ok to assume that this should be done right at the beginning of execution, as opposed to somewhere in the middle.

One solution is to have the main program be a script that launches the real binary, carefully putting it into the background. But it seems unsatisfying to need these coupled shell/binary pairs.

Another solution is to immediately launch another executed version (with 'system' or CreateProcess), with the same command line arguments, but putting the child in the background and then having the parent exit. But this seems clunky compared to the process putting itself into background.

Edited after a few answers: Yes, a fork() (or system(), or CreateProcess on Windows) is one way to sort of do this, that I hinted at in my original question. But all of these solutions make a SECOND process that is backgrounded, and then terminate the original process. I was wondering if there was a way to put the EXISTING process into the background. One difference is that if the app was launched from a script that recorded its process id (perhaps for later killing or other purpose), the newly forked or created process will have a different id and so will not be controllable by any launching script, if you see what I'm getting at.

Edit #2:

fork() isn't a good solution for OS X, where the man page for 'fork' says that it's unsafe if certain frameworks or libraries are being used. I tried it, and my app complains loudly at runtime: "The process has forked and you cannot use this CoreFoundation functionality safely. You MUST exec()."

I was intrigued by daemon(), but when I tried it on OS X, it gave the same error message, so I assume that it's just a fancy wrapper for fork() and has the same restrictions.

Excuse the OS X centrism, it just happens to be the system in front of me at the moment. But I am indeed looking for a solution to all three platforms.

回答1:

My advice: don't do this, at least not under Linux/UNIX.

GUI programs under Linux/UNIX traditionally do not auto-background themselves. While this may occasionally be annoying to newbies, it has a number of advantages:

• Makes it easy to capture standard error in case of core dumps / other problems that need debugging.

• Makes it easy for a shell script to run the program and wait until it's completed.

• Makes it easy for a shell script to run the program in the background and get its process id:

  gui-program &
  pid=$!
  # do something with $pid later, such as check if the program is still running
  

  If your program forks itself, this behavior will break.

"Scriptability" is useful in so many unexpected circumstances, even with GUI programs, that I would hesitate to explicitly break these behaviors.

Windows is another story. AFAIK, Windows programs automatically run in the background--even when invoked from a command shell--unless they explicitly request access to the command window.

回答2:

On Linux, daemon() is what you're looking for, if I understand you correctly.

回答3:

The way it's typically done on Unix-like OSes is to fork() at the beginning and exit from the parent. This won't work on Windows, but is much more elegant than launching another process where forking exists.

回答4:

Three things need doing,

fork
setsid
redirect STDIN, STDOUT and STDERR to /dev/null

This applies to POSIX systems (all the ones you mention claim to be POSIX (but Windows stops at the claiming bit))

回答5:

On UNIX, you need to fork twice in a row and let the parent die.

回答6:

A process cannot put itself into the background, because it isn't the one in charge of background vs. foreground. That would be the shell, which is waiting for process exit. If you launch a process with an ampersand "&" at the end, then the shell does not wait for process exit.

But the only way the process can escape the shell is to fork off another child and then let its original self exit back to the waiting shell.

From the shell, you can background a process with Control-Z, then type "bg".

回答7:

Backgrounding a process is a shell function, not an OS function.

If you want an app to start in the background, the typical trick is to write a shell script to launch it that launches it in the background.

#! /bin/sh
/path/to/myGuiApplication &

回答8:

To followup on your edited question:

I was wondering if there was a way to put the EXISTING process into the background.

In a Unix-like OS, there really is not a way to do this that I know of. The shell is blocked because it is executing one of the variants of a wait() call, waiting for the child process to exit. There is not a way for the child process to remain running but somehow cause the shell's wait() to return with a "please stop watching me" status. The reason you have the child fork and exit the original is so the shell will return from wait().

回答9:

Here is some pseudocode for Linux/UNIX:

initialization_code()
if(failure) exit(1)
if( fork() > 0 ) exit(0)
setsid()
setup_signal_handlers()
for(fd=0; fd<NOFILE; fd++) close(fd)
open("/dev/null", O_RDONLY)
open("/dev/null", O_WRONLY)
open("/dev/null", o_WRONLY)
chdir("/")

And congratulations, your program continues as an independent "daemonized" process without a controlling TTY and without any standard input or output.

Now, in Windows you simply build your program as a Win32 application with WinMain() instead of main(), and it runs without a console automatically. If you want to run as a service, you'll have to look that up because I've never written one and I don't really know how they work.

回答10:

You edited your question, but you may still be missing the point that your question is a syntax error of sorts -- if the process wasn't put in the background to begin with and you want the PID to stay the same, you can't ignore the fact that the program which started the process is waiting on that PID and that is pretty much the definition of being in the foreground.

I think you need to think about why you want to both put something in the background and keep the PID the same. I suggest you probably don't need both of those constraints.

回答11:

As others mentioned, fork() is how to do it on *nix. You can get fork() on Windows by using MingW or Cygwin libraries. But those will require you to switch to using GCC as your compiler.

In pure Windows world, you'd use CreateProcess (or one of its derivatives CreateProcessAsUser, CreateProcessWithLogonW).

回答12:

The most common way of doing this under Linux is via forking. The same should work on Mac, as for Windows I'm not 100% sure but I believe they have something similar.

Basically what happens is the process splits itself into two processes, and then the original one exits (returning control to the shell or whatever), and the second process continues to run in the background.

回答13:

I'm not sure about Windows, but on UNIX-like systems, you can fork() then setsid() the forked process to move it into a new process group that is not connected to a terminal.

回答14:

Under Windows, the closing thing you're going to get to fork() is loading your program as a Windows service, I think.

Here is a link to an intro article on Windows services... CodeProject: Simple Windows Service Sample

回答15:

So, as you say, just fork()ing will not do the trick. What you must do is fork() and then re-exec(), as this code sample does:

#include stdio.h>
#include <unistd.h>
#include <string.h>

#include <CoreFoundation/CoreFoundation.h>

int main(int argc, char **argv)
{
    int i, j;

    for (i=1; i<argc; i++)
        if (strcmp(argv[i], "--daemon") == 0)
        {
            for (j = i+1; j<argc; j++)
                argv[j-1] = argv[j];

            argv[argc - 1] = NULL;

            if (fork()) return 0;

            execv(argv[0], argv);

            return 0;
        }


    sleep(1);

    CFRunLoopRun();

    CFStringRef hello = CFSTR("Hello, world!");

    printf("str: %s\n", CFStringGetCStringPtr(hello, CFStringGetFastestEncoding(hello)));

    return 0;
}

The loop is to check for a --daemon argument, and if it is present, remove it before re-execing so an infinite loop is avoided.

I don't think this will work if the binary is put into the path because argv[0] is not necessarily a full path, so it will need to be modified.

回答16:

/**Deamonize*/

pid_t pid;
pid = fork(); /**father makes a little deamon(son)*/
if(pid>0)
exit(0); /**father dies*/
while(1){
printf("Hello I'm your little deamon %d\n",pid); /**The child deamon goes on*/
sleep(1)
}

/** try 'nohup' in linux(usage: nohup <command> &) */

回答17:

In Unix, I have learned to do that using fork(). If you want to put a running process into the background, fork it twice.

回答18:

I was trying the solution.

Only one fork is needed from the parent process.

The most important point is that, after fork, the parent process must die by calling _exit(0); and NOT by calling exit(0);

When _exit(0); is used, the command prompt immediately returns on the shell.

This is the trick.

回答19:

The simplest form of backgrounding is:

if (fork() != 0) exit(0);

In Unix, if you want to background an disassociate from the tty completely, you would do:

1. Close all descriptors which may access a tty (usually 0, 1, and 2).
2. if (fork() != 0) exit(0);
3. setpgroup(0,getpid()); /* Might be necessary to prevent a SIGHUP on shell exit. */
4. signal(SIGHUP,SIG_IGN); /* just in case, same as using nohup to launch program. */
5. fd=open("/dev/tty",O_RDWR);
6. ioctl(fd,TIOCNOTTY,0); /* Disassociates from the terminal */
7. close(fd);
8. if (fork() != 0) exit(0); /* just for good measure */

That should fully daemonize your program.

回答20:

If you need a script to have the PID of the program, you can still get it after a fork.

When you fork, save the PID of the child in the parent process. When you exit the parent process, either output the PID to STD{OUT,ERR} or simply have a return pid; statement at the end of main(). A calling script can then get the pid of the program, although it requires a certain knowledge of how the program works.