How to use Fork() to create only 2 child processes

问题:

I'm starting to learn some C and while studying the fork, wait functions I got to a unexpected output. At least for me.

Is there any way to create only 2 child processes from the parent?

Here my code:

#include <sys/types.h>
#include <stdio.h>
#include <unistd.h>
#include <sys/wait.h>

int main ()
{
    /* Create the pipe */
    int fd [2];
    pipe(fd);

    pid_t pid;
    pid_t pidb;


    pid = fork ();
    pidb = fork ();

    if (pid < 0)
    {
        printf ("Fork Failed\n");
        return -1;
    }
    else if (pid == 0)
    {
        //printf("I'm the child\n");
    }
    else 
    {
        //printf("I'm the parent\n");
    }

    printf("I'm pid %d\n",getpid());

    return 0;
}

And Here is my output:

I'm pid 6763
I'm pid 6765
I'm pid 6764
I'm pid 6766

Please, ignore the pipe part, I haven't gotten that far yet. I'm just trying to create only 2 child processes so I expect 3 "I'm pid ..." outputs only 1 for the parent which I will make wait and 2 child processes that will communicate through a pipe.

Let me know if you see where my error is.

回答1:

pid = fork (); #1
pidb = fork (); #2

Let us assume the parent process id is 100, the first fork creates another process 101. Now both 100 & 101 continue execution after #1, so they execute second fork. pid 100 reaches #2 creating another process 102. pid 101 reaches #2 creating another process 103. So we end up with 4 processes.

What you should do is something like this.

if(fork()) # parent
    if(fork()) #parent
    else # child2
else #child1

回答2:

After you create process , you should check the return value. if you don't , the seconde fork() will be executed by both the parent process and the child process, so you have four processes.

if you want to create 2 child processes , just :

if (pid = fork()) {
    if (pid = fork()) {
        ;
    } 
} 

You can create n child processes like this:

for (i = 0; i < n; ++i) {
    pid = fork();
    if (pid > 0) {   /* I am the parent, create more children */
        continue;
    } else if (pid == 0) { /* I am a child, get to work */
        break;
    } else {
        printf("fork error\n");
        exit(1);
    }
}

回答3:

When a fork statement is executed by the parent, a child process is created as you'd expect. You could say that the child process also executes the fork statement but returns a 0, the parent, however, returns the pid. All code after the fork statement is executed by both, the parent and the child.

In your case what was happening was that the first fork statement created a child process. So presently there's one parent, P1, and one child, C1.

Now both P1 and C1 encounter the second fork statement. The parent creates another child (c2) as you'd expect, but even the child, c1 creates a child process (c3). So in effect you have P1, C1, C2 and C3, which is why you got 4 print statement outputs.

A good way to think about this is using trees, with each node representing a process, and the root node is the topmost parent.

回答4:

you can check the value as if ( pid < 0 ) process creation unsuccessful this tells if the child process creation was unsuccessful.. fork returns the process id of the child process if getpid() is used from parent process..

回答5:

You can create a child process within a child process. This way you can have 2 copies of the original parent process.

int main (void) {
    pid_t pid, pid2;
    int status;

    pid = fork();

    if (pid == 0) { //child process
        pid2 = fork();
        int status2;

        if (pid2 == 0) { //child of child process
            printf("friends!\n");
        }
        else {
            printf("my ");
            fflush(stdout);
            wait(&status2);
        }
    }
    else { //parent process
        printf("Hello ");
        fflush(stdout);
        wait(&status);
    }

    return 0;
}

This prints the following:

Hello my friends!