在后台C linux中运行新的子进程

Question

I am trying to run a new process in background so it will be possible to continue working with parent process.

I used fork then execl. I tried to add to the execl command the argument & but it doesn't work:

execl("newproc","newproc","arg1","&",NULL);

Is there any solution?

Answer 1

The child will run in the background since you used fork . The child will keep running in parallel with the parent (if exec succeeded). If you care about whether the child process succeeded or not (and your code should) you should eventually call waitpid to collect its exit status. Otherwise you should call fork twice and have the intermediate process exit without wait ing for the child, so that init adopts the grandchild process.

As @mah said, the & is unnecessary. But another change is needed to that line; execl is a variadic function, and function prototypes therefore don't take care of converting arguments to the correct type. Therefore the final argument should be passed as the correct type - just change it to (char*)NULL .

You mention that your code didn't work. While that could just be because of the spurious & , it may also be because of the first argument. The execl function does not search $PATH for the named program, so unless newproc is actually in the current directory, this execl() invocation will return. When execl returns, that always indicates that there is a problem. The simplest way to solve this is to use execlp() instead of execl() . The alternative approach is to specify an absolute path as the first argument. You can even specify a relative path as the first argument, but this is rarely useful.

Answer 2

& is not a command argument, its a flag that the shell uses to know to run the command in the background. In this case, you're performing the work of the shell... remove the & . Since you state you've called fork() , as long as you're only performing execl() in the child process after fork returns, you're already running in the background.

Answer 3

The problem is that & is not a command line option to programs. Instead, it is merely special shell syntax which puts a command in the background. The distinguishing feature of background programs is that they are not connected to a terminal, and the terminal is not waiting for the process to complete. The proper function is daemon() . Do a man daemon to read up on how it is used.