CODEWITHSUNDEEP
cunixparent-childchild-process
Hakan Svensson
Hakan Svensson

Reputation: 3

Creating a child process on Unix systems?

I'm trying to create a child process in another process. I am writing both the programs in C language. First I write a dummy process which will be the child process. What it is doing is only to write a string on the screen. It works well on its own. Then I write another program which will be the parent process. However, I can't make it happen. I'm trying to use fork and execl functions together, but I fail. I also want the child process does not terminate until the parent process terminates.

How should I write the parent process?

Thanks.

Here is the code for the child process:

#include <stdio.h>

int main(void) {
  while(1) {
    printf("*");
    sleep(1);
  }
}

And here is the parent process:

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

int main(void) {
  if (fork()) {
    while(1) {
      printf("-\n");
      sleep(5);
    }
  } else {
    execl("./", "dummy", (char *)0);
  }
}

Upvotes: 0

Views: 5383

Answers (3)

zellio
zellio

Reputation: 32484

Basic use of fork in C

int PID = fork();

if( PID < 0 ) {
    //fail
    return PID;
}
else if( !PID ) {
    //child process
    return exec( prog, args );
} 
else {
    //parent process
    return 0;
}

Upvotes: 1

Juliano
Juliano

Reputation: 41387

The fork() system call may return three different statuses: failure (<0), parent process (>0) or child process (==0). You must test the return value properly.

int pid = fork();

if (pid < 0) {
  /* handle error */
  perror("fork");
  exit(1);
} else if (pid > 0) {
  /* parent code */
} else {
  /* child code */
}

Your execl() system call is wrong. The first argument is the path to the program you want to execute, "./" is not valid, it should be something like "./dummy" at least. The next argument is by convention the command name (argv[0] in the executed program), which may be a repetition of the first argument. So:

execl("./dummy", "dummy", NULL);

Also, note that the printf("*") statement in the child program will probably buffer and you won't see anything on the terminal. You must either add a "\n" to the end or call fflush(stdout) to flush the standard output.

Upvotes: 7

Alex Martelli
Alex Martelli

Reputation: 881507

There is no way to force the child process to "not terminate" when it's done (you'll still be able in the parent to wait for it to get info on how it terminated, but that's about it). Apart from that, any of the many examples of fork/exec on the web, such as this one, should work -- why don't you try it and see if it performs as you wish (in which case you'll just need to change whatever you were doing differently in your own attempt). If it doesn't work as desired (except for the impossibility per the first sentence in this A;-), please edit your to add copious detail about how the code behaves differently than you expect it to.

Upvotes: 0

Related Questions