General Quick fork() explanation - c

suppose if we have something like this:
printf("A");
fork();
printf("B");
Is the output going to be
1) ABAB
2) ABB
Can you please explain?

The right answer is that it depends on the buffering mode of stdout, which the other answers seem to be ignoring.
When you fork with unflushed buffers and then continue using stdio in both processes (instead of the usual quick execve or _exit in the child process), the stuff that was in the buffer at the time of the fork can be printed twice, once by each process.

It is undefined and ABB, ABAB and AABB is possible. First (ABB) can happen on unbuffered output only; with buffered output both processes will have the A in their output buffer. By calling fflush(3) before the fork(2) you can enforce this behavior.
Order of last chars depends on order of execution; most likely you will get ABAB as in this short example the program won't be interrupted by the scheduler.

The output should read as "ABB"
fork copies the entire program into a new memory space and continues from the fork. Since both processes are running the same code, I would save the process id that is returned from fork, so the rest of your program knows what to do.

the output is simply abb. Fork create one new child process so after execution of fork() two process will run and generally child process first get the chance to executes. See before fork only one process so the out put before fork only once A. Then fork executes and hence two different process now running each will print B and hence output is ABB.

Related

when a child process is created using fork() system call, where the child process starts execution? [duplicate]

This question already has answers here:
printf anomaly after "fork()"
(3 answers)
Closed 8 years ago.
fork() creates a new process and the child process starts to execute from the current state of the parent process.
This is the thing I know about fork() in Linux.
So, accordingly the following code:
int main() {
printf("Hi");
fork();
return 0;
}
needs to print "Hi" only once as per the above.
But on executing the above in Linux, compiled with gcc, it prints "Hi" twice.
Can someone explain to me what is happening actually on using fork() and if I have understood the working of fork() properly?
(Incorporating some explanation from a comment by user #Jack)
When you print something to the "Standard Output" stdout (computer monitor usually, although you can redirect it to a file), it gets stored in temporary buffer initially.
Both sides of the fork inherit the unflushed buffer, so when each side of the fork hits the return statement and ends, it gets flushed twice.
Before you fork, you should fflush(stdout); which will flush the buffer so that the child doesn't inherit it.
stdout to the screen (as opposed to when you're redirecting it to a file) is actually buffered by line ends, so if you'd done printf("Hi\n"); you wouldn't have had this problem because it would have flushed the buffer itself.
printf("Hi"); doesn't actually immediately print the word "Hi" to your screen. What it does do is fill the stdout buffer with the word "Hi", which will then be shown once the buffer is 'flushed'. In this case, stdout is pointing to your monitor (assumedly). In that case, the buffer will be flushed when it is full, when you force it to flush, or (most commonly) when you print out a newline ("\n") character. Since the buffer is still full when fork() is called, both parent and child process inherit it and therefore they both will print out "Hi" when they flush the buffer. If you call fflush(stout); before calling fork it should work:
int main() {
printf("Hi");
fflush(stdout);
fork();
return 0;
}
Alternatively, as I said, if you include a newline in your printf it should work as well:
int main() {
printf("Hi\n");
fork();
return 0;
}
In general, it's very unsafe to have open handles / objects in use by libraries on either side of fork().
This includes the C standard library.
fork() makes two processes out of one, and no library can detect it happening. Therefore, if both processes continue to run with the same file descriptors / sockets etc, they now have differing states but share the same file handles (technically they have copies, but the same underlying files). This makes bad things happen.
Examples of cases where fork() causes this problem
stdio e.g. tty input/output, pipes, disc files
Sockets used by e.g. a database client library
Sockets in use by a server process - which can get strange effects when a child to service one socket happens to inherit a file handle for anohter - getting this kind of programming right is tricky, see Apache's source code for examples.
How to fix this in the general case:
Either
a) Immediately after fork(), call exec(), possibly on the same binary (with necessary parameters to achieve whatever work you intended to do). This is very easy.
b) after forking, don't use any existing open handles or library objects which depend on them (opening new ones is ok); finish your work as quickly as possible, then call _exit() (not exit() ). Do not return from the subroutine that calls fork, as that risks calling C++ destructors etc which may do bad things to the parent process's file descriptors. This is moderately easy.
c) After forking, somehow clear up all the objects and make them all in a sane state before having the child continue. e.g. close underlying file descriptors without flushing data which are in a buffer which is duplicated in the parent. This is tricky.
c) is approximately what Apache does.
printf() does buffering. Have you tried printing to stderr?
Technical answer:
when using fork() you need to make sure that exit() is not called twice (falling off of main is the same as calling exit()). The child (or rarely the parent) needs to call _exit instead. Also, don't use stdio in the child. That's just asking for trouble.
Some libraries have a fflushall() you can call before fork() that makes stdio in the child safe. In this particular case it would also make exit() safe but that is not true in the general case.

How fork() and scanf() work together?

I tried to see what happens if I read something from keyboard while I have multiple processes with fork() (in my case there are two children and a parent) and I discovered the following problem: I need to tell the parent to wait for children's processes, otherwise the program behaves strangely.
I did a research and I found that the problem is with the parent, he needs to wait for the child's process to end because if the parent's process ends first somehow he closes the STDIN, am I right? But also I found that every process has a copy of STDIN so my question is:
Why it works this way and why only the parent has the problem with STDIN and the children not, I mean why if the child's process ends first doesn't affect STDIN but if the parent's process ends first it does affect STDIN?
Here are my tests:
I ran the program without wait() and after I typed a number the program stopped, but then I pressed enter two more times and the other two messages from printf() appeared.
When I ran the program with wait() everything worked fine, every process called scanf() separately and read a different number.
Well, a lot of stuff is going on here. I will try to explain it step by step.
When you start your terminal, the terminal creates a special file having path /dev/pts/<some number>. Then it starts your shell (which is bash in this case) and links the STDIN, STDOUT and STDERR of the bash process to this special file. This file is called a special file because it doesn't actually exist on your hard disk. Instead, whatever you write to this file, it goes directly to the terminal and the terminal renders it on the screen. (Similarly, whenever you try to read from this file, the read blocks until someone types something at the terminal).
Now when you launch your program by typing ./main, bash calls the fork function in order create a new process. The child process execs your executable file, while the parent process waits for the child to terminate. Your program then calls fork twice and we have three processes trying to read their STDINs, ie the same file /dev/pts/something. (Remember that calling fork and exec duplicates and preserves the file descriptors respectively).
The three processes are in race condition. When you enter something at the terminal, one of the three processes will receive it (99 out of 100 times it would be the parent process since the children have to do more work before reaching scanf statement).
So, parent process prints the number and exits first. The bash process that was waiting for the parent to finish, resumes and puts the STDIN into a so called "non-canonical" mode, and calls read in order to read the next command. Now again, three processes (Child1, Child2 and bash) are trying to read STDIN.
Since the children are trying to read STDIN for a longer time, the next time you enter something it will be received by one of the children, rather than bash. So you think of typing, say, 23. But oops! Just after you press the 2 key, you get Your number is: 2. You didn't even press the Enter key! That happened because of this so called "non-canonical" mode. I won't be going into what and why is that. But for now, to make things easier, use can run your program on sh instead of bash, since sh doesn't put STDIN into non-canonical mode. That will make the picture clear.
TL;DR
No, parent process closing its STDIN doesn't mean that its children or other process won't be able to use it.
The strange behavior you are seeing is because when the parent exits, bash puts the pty (pseudo terminal) into non-canonical mode. If you use sh instead, you won't see that behavior. Read up on pseudo terminals, and line discipline if you want to have a clear understading.
The shell process will resume as soon as the parent exits.
If you use wait to ensure that parents exits last, you won't have any problem, since the shell won't be able to run along with your program.
Normally, bash makes sure that no two foreground processes read from STDIN simultaneously, so you don't see this strange behavior. It does this by either piping STDOUT of one program to another, or by making one process a background process.
Trivia: When a background process tries to read from its STDIN, it is sent a signal SIGTTIN, which stops the process. Though, that's not really relevant to this scenario.
There are several issues that can happen when multiple processes try to do I/O to the same TTY. Without code, we can't tell which may be happening.
Trying to do I/O from a background process group may deliver a signal: SIGTTIN for input (usually enabled), or SIGTTOU for output (usually disabled)
Buffering: if you do any I/O before the fork, any data that has been buffered will be there for both processes. Under some conditions, using fflush may help, but it's better to avoid buffering entirely. Remember that, unlike output buffering, it is impossible to buffer input on a line-by-line basis (although you can only buffer what is available, so it may appear to be line-buffered at first).
Race conditions: if more than one process is trying to read the same pipe-like file, it is undefined which one will "win" and actually get the input each time it is available.

Waiting process to return

Consider:
int main()
{
if (fork() == 0){
printf("a");
}
else{
printf("b");
waitpid(-1, NULL, 0);
}
printf("c");
exit(0);
}
(from Computer Systems, Bryant - O'Hallaron).
We are asked for all the possible output sequences.
I answered: acbc, abcc, bacc.
However, I am missing one output compared to the solution (bcac). I thought this output was not possible because the parent process waits for its child to return before printing c (waitpid). Is this not true? Why? And, in that case, what is the difference between the code above and the same without the waitpid line?
I don't see any way bcac is possible. At first I expected some trickery based on the stdio buffers being flushed in an unexpected order. But even then:
The child won't output c until after it has output a. Therefore the first c in bcac must have come from the parent.
The parent won't output c until after the waitpid completes. But that can't happen until after the child is finished, including the final stdio flush that happens during exit(). Therefore the first c is always from the child.
Proof by contradiction has been achieved... the output can't be bcac.
Well, there is one thing you could do to mess up the order. You could exec the program inside a process that already has a child which is about to exit. If the pre-existing child exits before the new child prints a, then the main process will detect that exit with waitpid, and go ahead and print its stuff and possibly exit before the child prints anything.
This is something to watch out for in setuid programs: don't assume that because your program only created one child process that it only has one child process. If you're in an advanced defensive-code learning context this answer makes sense. In a unix-newbie context it doesn't seem relevant, and it's probably better to just say bcac is impossible, even though it's technically not true.
It's tricky, but the call to waitpid can be interrupted (returns -1 and errno is EINTR). In this case the parent can output c before the child outputs anything and bcac is possible.
To prevent bcac from occurring, either a signal mask needs to be set, or, better, the waitpid return value is checked and if it was interrupted gets called again.

fork starts executing form where?

To my previous question about segmentation fault ,I got very useful answers.Thanks for those who have responded.
#include<stdio.h>
main()
{
printf("hello");
int pid = fork();
wait(NULL);
}
output: hellohello.
In this the child process starts executing form the beginning.
If Iam not wrong , then how the program works if I put the sem_open before fork()
(ref answers to :prev questions)
I need a clear explanation about segmentation fault which happens occasionally and not always. And why not always... If there is any error in coding then it should occur always right...?
fork creates a clone of your process. Conceptually speaking, all state of the parent also ends up in the child. This includes:
CPU registers (including the instruction pointer, which defines where in the code your program is)
Memory (as an optimization your kernel will most likely mark all pages as copy-on-write, but semantically speaking it should be the same as copying all memory.)
File descriptors
Therefore... Your program will not "start running" from anywhere... All the state that you had when you called fork will propagate to the child. The child will return from fork just as the parent will.
As for what you can do after a fork... I'm not sure about what POSIX says, but I wouldn't rely on semaphores doing the right thing after a fork. You might need an inter-process semaphore (see man sem_open, or the pshared parameter of sem_init). In my experience cross-process semaphores aren't really well supported on free Unix type OS's... (Example: Some BSDs always fail with ENOSYS if you ever try to create one.)
#GregS mentions the duplicated "hello" strings after a fork. He is correct to say that stdio (i.e. FILE*) will buffer in user-space memory, and that a fork leads to the string being buffered in two processes. You might want to call fflush(stdout); fflush(stderr); and flush any other important FILE* handles before a fork.
No, it starts from the fork(), which returns 0 in the child or the child's process ID in the parent.
You see "hello" twice because the standard output is buffered, and has not actually been written at the point of the fork. Both parent and child then actually write the buffered output. If you fflush(stdout); after the printf(), you should see it only once.

What is the purpose of fork()?

In many programs and man pages of Linux, I have seen code using fork(). Why do we need to use fork() and what is its purpose?
fork() is how you create new processes in Unix. When you call fork, you're creating a copy of your own process that has its own address space. This allows multiple tasks to run independently of one another as though they each had the full memory of the machine to themselves.
Here are some example usages of fork:
Your shell uses fork to run the programs you invoke from the command line.
Web servers like apache use fork to create multiple server processes, each of which handles requests in its own address space. If one dies or leaks memory, others are unaffected, so it functions as a mechanism for fault tolerance.
Google Chrome uses fork to handle each page within a separate process. This will prevent client-side code on one page from bringing your whole browser down.
fork is used to spawn processes in some parallel programs (like those written using MPI). Note this is different from using threads, which don't have their own address space and exist within a process.
Scripting languages use fork indirectly to start child processes. For example, every time you use a command like subprocess.Popen in Python, you fork a child process and read its output. This enables programs to work together.
Typical usage of fork in a shell might look something like this:
int child_process_id = fork();
if (child_process_id) {
// Fork returns a valid pid in the parent process. Parent executes this.
// wait for the child process to complete
waitpid(child_process_id, ...); // omitted extra args for brevity
// child process finished!
} else {
// Fork returns 0 in the child process. Child executes this.
// new argv array for the child process
const char *argv[] = {"arg1", "arg2", "arg3", NULL};
// now start executing some other program
exec("/path/to/a/program", argv);
}
The shell spawns a child process using exec and waits for it to complete, then continues with its own execution. Note that you don't have to use fork this way. You can always spawn off lots of child processes, as a parallel program might do, and each might run a program concurrently. Basically, any time you're creating new processes in a Unix system, you're using fork(). For the Windows equivalent, take a look at CreateProcess.
If you want more examples and a longer explanation, Wikipedia has a decent summary. And here are some slides here on how processes, threads, and concurrency work in modern operating systems.
fork() is how Unix create new processes. At the point you called fork(), your process is cloned, and two different processes continue the execution from there. One of them, the child, will have fork() return 0. The other, the parent, will have fork() return the PID (process ID) of the child.
For example, if you type the following in a shell, the shell program will call fork(), and then execute the command you passed (telnetd, in this case) in the child, while the parent will display the prompt again, as well as a message indicating the PID of the background process.
$ telnetd &
As for the reason you create new processes, that's how your operating system can do many things at the same time. It's why you can run a program and, while it is running, switch to another window and do something else.
fork() is used to create child process. When a fork() function is called, a new process will be spawned and the fork() function call will return a different value for the child and the parent.
If the return value is 0, you know you're the child process and if the return value is a number (which happens to be the child process id), you know you're the parent. (and if it's a negative number, the fork was failed and no child process was created)
http://www.yolinux.com/TUTORIALS/ForkExecProcesses.html
fork() is basically used to create a child process for the process in which you are calling this function. Whenever you call a fork(), it returns a zero for the child id.
pid=fork()
if pid==0
//this is the child process
else if pid!=0
//this is the parent process
by this you can provide different actions for the parent and the child and make use of multithreading feature.
fork() will create a new child process identical to the parent. So everything you run in the code after that will be run by both processes — very useful if you have for instance a server, and you want to handle multiple requests.
System call fork() is used to create processes. It takes no arguments and returns a process ID. The purpose of fork() is to create a new process, which becomes the child process of the caller. After a new child process is created, both processes will execute the next instruction following the fork() system call. Therefore, we have to distinguish the parent from the child. This can be done by testing the returned value of fork():
If fork() returns a negative value, the creation of a child process was unsuccessful.
fork() returns a zero to the newly created child process.
fork() returns a positive value, the process ID of the child process, to the parent. The returned process ID is of type pid_t defined in sys/types.h. Normally, the process ID is an integer. Moreover, a process can use function getpid() to retrieve the process ID assigned to this process.
Therefore, after the system call to fork(), a simple test can tell which process is the child. Please note that Unix will make an exact copy of the parent's address space and give it to the child. Therefore, the parent and child processes have separate address spaces.
Let us understand it with an example to make the above points clear. This example does not distinguish parent and the child processes.
#include <stdio.h>
#include <string.h>
#include <sys/types.h>
#define MAX_COUNT 200
#define BUF_SIZE 100
void main(void)
{
pid_t pid;
int i;
char buf[BUF_SIZE];
fork();
pid = getpid();
for (i = 1; i <= MAX_COUNT; i++) {
sprintf(buf, "This line is from pid %d, value = %d\n", pid, i);
write(1, buf, strlen(buf));
}
}
Suppose the above program executes up to the point of the call to fork().
If the call to fork() is executed successfully, Unix will make two identical copies of address spaces, one for the parent and the other for the child.
Both processes will start their execution at the next statement following the fork() call. In this case, both processes will start their execution at the assignment
pid = .....;
Both processes start their execution right after the system call fork(). Since both processes have identical but separate address spaces, those variables initialized before the fork() call have the same values in both address spaces. Since every process has its own address space, any modifications will be independent of the others. In other words, if the parent changes the value of its variable, the modification will only affect the variable in the parent process's address space. Other address spaces created by fork() calls will not be affected even though they have identical variable names.
What is the reason of using write rather than printf? It is because printf() is "buffered," meaning printf() will group the output of a process together. While buffering the output for the parent process, the child may also use printf to print out some information, which will also be buffered. As a result, since the output will not be send to screen immediately, you may not get the right order of the expected result. Worse, the output from the two processes may be mixed in strange ways. To overcome this problem, you may consider to use the "unbuffered" write.
If you run this program, you might see the following on the screen:
................
This line is from pid 3456, value 13
This line is from pid 3456, value 14
................
This line is from pid 3456, value 20
This line is from pid 4617, value 100
This line is from pid 4617, value 101
................
This line is from pid 3456, value 21
This line is from pid 3456, value 22
................
Process ID 3456 may be the one assigned to the parent or the child. Due to the fact that these processes are run concurrently, their output lines are intermixed in a rather unpredictable way. Moreover, the order of these lines are determined by the CPU scheduler. Hence, if you run this program again, you may get a totally different result.
You probably don't need to use fork in day-to-day programming if you are writing applications.
Even if you do want your program to start another program to do some task, there are other simpler interfaces which use fork behind the scenes, such as "system" in C and perl.
For example, if you wanted your application to launch another program such as bc to do some calculation for you, you might use 'system' to run it. System does a 'fork' to create a new process, then an 'exec' to turn that process into bc. Once bc completes, system returns control to your program.
You can also run other programs asynchronously, but I can't remember how.
If you are writing servers, shells, viruses or operating systems, you are more likely to want to use fork.
Multiprocessing is central to computing. For example, your IE or Firefox can create a process to download a file for you while you are still browsing the internet. Or, while you are printing out a document in a word processor, you can still look at different pages and still do some editing with it.
Fork creates new processes. Without fork you would have a unix system that could only run init.
Fork() is used to create new processes as every body has written.
Here is my code that creates processes in the form of binary tree.......It will ask to scan the number of levels upto which you want to create processes in binary tree
#include<unistd.h>
#include<fcntl.h>
#include<stdlib.h>
int main()
{
int t1,t2,p,i,n,ab;
p=getpid();
printf("enter the number of levels\n");fflush(stdout);
scanf("%d",&n);
printf("root %d\n",p);fflush(stdout);
for(i=1;i<n;i++)
{
t1=fork();
if(t1!=0)
t2=fork();
if(t1!=0 && t2!=0)
break;
printf("child pid %d parent pid %d\n",getpid(),getppid());fflush(stdout);
}
waitpid(t1,&ab,0);
waitpid(t2,&ab,0);
return 0;
}
OUTPUT
enter the number of levels
3
root 20665
child pid 20670 parent pid 20665
child pid 20669 parent pid 20665
child pid 20672 parent pid 20670
child pid 20671 parent pid 20670
child pid 20674 parent pid 20669
child pid 20673 parent pid 20669
First one needs to understand what is fork () system call. Let me explain
fork() system call creates the exact duplicate of parent process, It makes the duplicate of parent stack, heap, initialized data, uninitialized data and share the code in read-only mode with parent process.
Fork system call copies the memory on the copy-on-write basis, means child makes in virtual memory page when there is requirement of copying.
Now Purpose of fork():
Fork() can be used at the place where there is division of work like a server has to handle multiple clients, So parent has to accept the connection on regular basis, So server does fork for each client to perform read-write.
fork() is used to spawn a child process. Typically it's used in similar sorts of situations as threading, but there are differences. Unlike threads, fork() creates whole seperate processes, which means that the child and the parent while they are direct copies of each other at the point that fork() is called, they are completely seperate, neither can access the other's memory space (without going to the normal troubles you go to access another program's memory).
fork() is still used by some server applications, mostly ones that run as root on a *NIX machine that drop permissions before processing user requests. There are some other usecases still, but mostly people have moved to multithreading now.
The rationale behind fork() versus just having an exec() function to initiate a new process is explained in an answer to a similar question on the unix stack exchange.
Essentially, since fork copies the current process, all of the various possible options for a process are established by default, so the programmer does not have supply them.
In the Windows operating system, by contrast, programmers have to use the CreateProcess function which is MUCH more complicated and requires populating a multifarious structure to define the parameters of the new process.
So, to sum up, the reason for forking (versus exec'ing) is simplicity in creating new processes.
Fork() system call use to create a child process. It is exact duplicate of parent process. Fork copies stack section, heap section, data section, environment variable, command line arguments from parent.
refer: http://man7.org/linux/man-pages/man2/fork.2.html
Fork() was created as a way to create another process with shared a copy of memory state to the parent. It works the way it does because it was the most minimal change possible to get good threading capabilities in time-slicing mainframe systems that previously lacked this capability. Additionally, programs needed remarkably little modification to become multi-process, fork() could simply be added in the appropriate locations, which is rather elegant. Basically, fork() was the path of least resistance.
Originally it actually had to copy the entire parent process' memory space. With the advent of virtual memory, it has been hacked and changed to be more efficient, with copy-on-write mechanisms avoiding the need to actual copy any memory.
However, modern systems now allow the creation of actual threads, which simply share the parent process' actual heap. With modern multi-threading programming paradigms and more advanced languages, it's questionable whether fork() provides any real benefit, since fork() actually prevents processes from communicating through memory directly, and forces them to use slower message passing mechanisms.

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