`fork()` sons are executing in reverse order

Question

I have a code similar to this:

for (i = 0; i < 3; i++)
{
    pid = fork();

    if (pid == 0)
    {
        son_function();
    }

    if (pid < 0)
    {
        exit(1);
    }
}

void son_function(void)
{
    printf("my pid=%d\n", getpid());
    printf("%d: alpha\n", getpid());
    printf("%d: beta\n", getpid());
    printf("%d: charlie\n", getpid());
    exit(0);
}

For some reason I can't understand, the order of execution of son_function() is in reverse order. What I mean is that son_function() is printing the PID numbers from the largest to the smallest.

Another thing that freaks me is that the prints for every son will be in the one after the other, there's no way that two prints from two different processes will print to the screen at the same time.

Sample can be seen here: http://ideone.com/uBYyRX

`getpid()` returns the process ID of the calling process. You should not perhaps expect any correlation between the _variable_ `pid` and the function `getpid()`. — ryyker, Dec 06 '14 at 01:33
@ryyke, I'm not looking for correlation, I'm looking for some sense in the order of execution. Why would the son with the highest `PID` print first? — Quaker, Dec 06 '14 at 01:35
Use an array of _pid_, i.e. `pid_t pid[3];`. Then track them by looking at the unique element (pid[0],...). — ryyker, Dec 06 '14 at 01:40
You're going to have to elaborate your last paragraph. IO streams in C are buffered, with implicit flushes sent per newline. Each of your processes has its *own* stream buffer. After fixing the plethora of syntax and logic errors to make your post actually compilable, a sample output [can be seen here](http://pastebin.com/UGWkXncU). Notice the first two pids. — WhozCraig, Dec 06 '14 at 01:41
You have a race condition. All the child processes are running independently, and the order in which they reach the `printf` instructions is unspecified. If you want the child processes to execute the `printf` in a specific order, you need to use synchronization. — Raymond Chen, Dec 06 '14 at 01:47
@WhozCraig, I really had some typos while paraphrasing the real code, what I wanted to demonstrate in the questions can be seen (including output) here: http://ideone.com/uBYyRX — Quaker, Dec 06 '14 at 01:49
@RaymondChen I understand what you say, and it makes a little sense but shouldn't the first child execute and reach `printf` while the loop in father process is still working? Shouldn't the first child reach `printf` at least before the last child? — Quaker, Dec 06 '14 at 01:52
@Quaker I honestly don't know how that even compiles, and my faith in ideone just went down a notch. None of `fork()`, `getpid()` and `exit()`are standard-declared in `stdio.h`. Regardless, Raymond is correct. — WhozCraig, Dec 06 '14 at 01:54
You are forking so fast that none of the child processes get a chance to run at all until the forking is complete. They then end up racing with each other simultaneously, nobody getting a head start over anybody else. — Raymond Chen, Dec 06 '14 at 02:02

KC-NH · Accepted Answer · 2014-12-08T20:12:02.687

Multiple processes can output to the console at the same time, at least under Windows and Linux.

The reason you may see all of one process before all of the other is due to the way your particular OS schedules threads. A better way to see this behavior is to change son_function to look something like the code below, where each child sleeps for a different amount of time. The reason who lines get interleaved (as noted before) is because printf buffers lines of output.

void son_function()
{
    srandom(getpid());
    int sleepTime = random() % 4; // random sleep between 0 and 3 seconds
    printf("pid [%d] sleep time is %d\n", getpid(), sleepTime);

    printf("my pid = %d\n", getpid());
    sleep(sleepTime);
    printf("alpha = %d\n", getpid());
    sleep(sleepTime);
    printf("beta = %d\n", getpid());
    sleep(sleepTime);
    printf("charlie = %d\n", getpid());
    sleep(sleepTime);
}

Then why aren't there collisions in the printing to the console? Why i.e. child-1 is printing in its' correct order without any other child (i.e. child-2) printing in between his prints? — Quaker, Dec 06 '14 at 09:34

score 0 · Answer 2 · answered Dec 06 '14 at 04:45

To execute the child process in the correct order just add the wait function inside the loop before calling the fork function.

so it will wait till the child terminates and the child process will be terminated in the correct order.

      for (i = 0; i < 3; i++)

            wait();
            pid = fork();

execute the code after adding the wait function.

`fork()` sons are executing in reverse order

2 Answers2