Can someone explain fork()?

Question

Can someone explain fork() and give examples of its uses?

What I understood by some online sources is that:

1. fork() creates a child process that would run the same as the parent process.
2. 2^n process will be created, n = number of fork().
3. ID of the Child Process will always be 0, while the ID of the Parent Process will be something else - A Positive Integer != 0.

And I have another question, for that first please take a look at my code below:

#include<stdio.h>
#include<sys/types.h>
#include<unistd.h>
int main() {
    int id = fork();
    int i = 45;
    fork();
    fork();
    if(id == 0) {
        printf("\nChild Process id = %d",id);
        printf("\ni = %d ; i = %d",i++,i);  
    }
    else {
        printf("\nParent Process id = %d",id);
        printf("\ni = %d ; i = %d",i,i+2);
    }
    return 0;
}   
/*
O/P:
Parent Process id = 9048
i = 45 ; i = 47
Parent Process id = 9048                                                                                                                                                        
i = 45 ; i = 47
Parent Process id = 9048
i = 45 ; i = 47
Parent Process id = 9048
i = 45 ; i = 47
Child Process id = 0
i = 45 ; i = 46
Child Process id = 0
i = 45 ; i = 46
Child Process id = 0
i = 45 ; i = 46
Child Process id = 0
*/

Why doesn't the last child process display the second printf() i.e.,

printf("\ni = %d ; i = %d",i++,i);

Can you please explain why does the second printf() doesn't get executed while the first printf() gets executed for the 8th time - Child Process.

Answer 1

Correct usage of fork implies testing immediately if the current process is the child (fork returned 0) or the parent (fork returned the pid of the child), and act accordingly. In your case, both parent and child launch a couple or spurious processes, which will not be able to know what they are, since the value returned is ignored. A couple of them will think they are parent processes.

Normally the child process will exec a new executable, if something serious has to be done.

Answer 2

fork() creates child process that would run same as the parent process.

True. fork() creates a child process which is a duplicate of it's parent process.

---

2^n process will be created, n = number of fork().

Depends on the code!

int i = 0;
while (i < n)
{
    fork();
    i++;
}

Here, assuming no fork() failed, then yes, there will be 2^n processes.

---

ID of the Child Process will always be 0, while ID of the Parent Process will be something else - A Positive Integer != 0.

The fork() return value in child process is 0! not id, but return value! the return value of fork() in parent process is the actual pid of the child process.

---

To your code now

1. int main() {
2.    int id = fork(); 
3.    int i = 45;
4.    fork();
5.    fork();
6.    if(id == 0) {
7.        printf("\nChild Process id = %d",id);
8.        printf("\ni = %d ; i = %d",i++,i);  
9.    }
10.   else {
11.       printf("\nParent Process id = %d",id);
12.       printf("\ni = %d ; i = %d",i,i+2);
13.   }
14.   return 0;
15. } 

At line 4, there are 2 processes

parent         |     child_1
    id > 0     |     id == 0
    i == 45    |     i == 45     

Now a fork is invoked

    parent     |     child_1    |    child_OF_1   |   child_2
    id > 0     |     id == 0    |    id == 0      |   id > 0
    i == 45    |     i == 45    |    i == 45      |   i == 45

child_1 and child_OF_1 will have id == 0 and enter the if statement, while child_2 and parent will go to the else statement and print 47

I ignored the 3rd fork() which by then you have 8 processes, of which, 4 will go inside the if statement while the others will enter the else block. This is why you see

i = 45 ; i = 47

4 times!

Answer 3

All the output you expect is there, it's just that you write it out in the background without a terminating linefeed. This means that some of the output may be obscured or overwritten by the shell prompt appearing too soon and in the middle of the program's output.

To see exactly how this is possible, here's strace output for some of the processes in your program:

3215092 write(1, "\n", 1)               = 1
3215092 write(1, "Child Process id = 0\n", 21) = 21
3215092 write(1, "i = 45 ; i = 46", 15) = 15
3215094 write(1, "\n", 1)               = 1
3215094 write(1, "Child Process id = 0\n", 21) = 21
3215094 write(1, "i = 45 ; i = 46", 15) = 15

And here's how Zsh draws its prompt. First a missing linefeed indicator and then the prompt itself. Note in particular that the prompt starts with \r, a carriage return:

3215036 write(10, "\33[1m\33[3m%\33[23m\33[1m\33[0m          "..., 306) = 306
3215036 write(10, "\r\33[0m\33[23m\33[24m\33[Jmyhostname.i"..., 34) = 34

Since you fork processes in the background that write to the terminal, all your forked processes are in a race with Zsh to write their output.

Here is one possible, valid ordering that this may result in:

1. 3215036 write(10, "\33[1m\33[3m%\33[23m\33[1m\33[0m          "..., 306) = 306
2. 3215092 write(1, "\n", 1)               = 1
3. 3215092 write(1, "Child Process id = 0\n", 21) = 21
4. 3215092 write(1, "i = 45 ; i = 46", 15) = 15
5. 3215094 write(1, "\n", 1)               = 1
6. 3215094 write(1, "Child Process id = 0\n", 21) = 21
7. 3215094 write(1, "i = 45 ; i = 46", 15) = 15
8. 3215036 write(10, "\r\33[0m\33[23m\33[24m\33[Jmyhostname.i"..., 34) = 34

Steps 1-7 happen fairly uneventfully, resulting in a screen like this (cursor marked with |):

[...]
i = 45 ; i = 46
Child Process id = 0
i = 45 ; i = 46|

Since you have no terminating linefeeds, the cursor ends up after instead of under the last string.

Now, as you can see in write 8, Zsh starts its default prompt with a leading carriage return. This means that will go to the start of the line:

[...]
i = 45 ; i = 46
Child Process id = 0
|i = 45 ; i = 46

and then proceed to overwrite the line with the prompt:

[...]
i = 45 ; i = 46
Child Process id = 0
myhostname.internal% 

In effect, even though the last line of output was written to screen, the Zsh prompt effectively erased it by overwriting it using the carriage return.

You can use redirection to output to a file to more clearly see what output your program produces:

./foo > myfile

Opening the file in an editor shows your expected result:

Parent Process id = 3135806
i = 45 ; i = 47
Parent Process id = 3135806
i = 45 ; i = 47
Parent Process id = 3135806
i = 45 ; i = 47
Child Process id = 0
i = 45 ; i = 46
Parent Process id = 3135806
i = 45 ; i = 47
Child Process id = 0
i = 45 ; i = 46
Child Process id = 0
i = 45 ; i = 46
Child Process id = 0
i = 45 ; i = 46

You can similarly add your own terminating linefeed and delay to show the output in your terminal:

./foo; sleep 5; echo

This gives the same result as above.

A correct program would output linefeeds after and not before lines, and would wait() for each child process. (Please note that it won't work if you fix only one of these, while leaving the other)