#include <stdlib.h>
#include <sys/types.h>
#include <unistd.h>
int main ()
{
int i = 3;
int pid;
while(i > 0) {
pid = fork();
if(pid > 0) {
exit(0);
} else {
i--;
}
}
}
How many process are created on that program??? I am still confused with the fork() system calls, can anybody help me to explain this?
What does fork() do?
fork() is an interesting call. You can think of it as cloning the state of your program into two exact copies -- the only difference between them will be the return value of fork(). The process that did the fork() receives the process id (pid) of the new process, while the new process receives 0.
With that in mind:
How many processes are created?
Each time you fork, you create a new process and exit the parent. You do this three times, therefore -- three processes are created by forking. This doesn't include the one that you started by starting the process, of course. :)
during start of you program, system creates 1 process (+1)
i=3, program creates new process (+1), parent exits, child continue
i=2, program creates new process (+1), parent exits, child continue
i=1, program creates new process (+1), parent exits, child continue
i=0, program exit
so, totally - 4 processes, 3 created by program, 1 by system
Related
The task is as follows
Write a program that would run another process in memory and leave it running in an infinite loop. When the program is restarted, it must remove the previously started process from memory (you can use kill).
#include <stdio.h>
#include <unistd.h>
#include <signal.h>
#include <stdlib.h>
#include <spawn.h>
#include <sys/wait.h>
int main(void){
int pid = getpid(); // we find out the PID of the current process and store it in a variable
FILE *file = fopen("example.txt", "r"); // getting information from a file about a child process
int filePid = 0;
fscanf(file, "%d", filePid);
fclose(file);
switch (filePid){
case -1:{ // if there is no child process, then run it and write the PID to a file
filePid = fork();
file = fopen("example.txt", "w");
fprintf(file, "%d", filePid);
fclose(file);
break;
}
case 0:{ // if this process is a child, then we go into an infinite loop
for(;;){
sleep(7); // waiting for seven seconds so that the system is not heavily loaded
}
break;
}
default:{ // if this program is started again with a child process, then we send a signal to the child process
kill(filePid, SIGKILL); // we send a signal to the child process so that it ends, and after that we write the information to the file
file = fopen("example.txt", "w"); // we write information to the file that the child process is missing
fprintf(file, "%d", -1);
fclose(file);
}
}
return EXIT_SUCCESS;
}
/Yes, I have to do it through the qnx operating system./
the errors are as follows..I'm a little confused with getpid, because I haven't used the pid variable anywhere.
and another mistake.
I will be grateful for your help.since I'm a little confused...
UPD:
I can't get the value 0
UPD: how could it execute both cases, i mean "if" and "else" blocks at the same time?
how could it execute both cases, i mean "if" and "else" blocks at the same time?
You have to have a clear understanding of how fork works, when you use fork two identical processes are created parent and child and they run simultaneously, when you say filePid = fork() and if the operation is successful then the parent process will hold the process id of the child process and child process will hold 0. So here in parent process filePid == child process ID and in child process filePid == 0.
See man fork
On success, the PID of the child process is returned in the parent,
and 0 is returned in the child. On failure, -1 is returned in the
parent, no child process is created, and errno is set appropriately.
So whatever you put in the if (filePid == 0) block will be executed by the child process and whatever you put in the else block will be executed by the parent process simultaneously.
This will be helpful if you want to know more about fork.
fork() system call creates a child process and return its pid so once child is created you have two process running same instruction after fork() one is checking pid is greater which means its parent which gets child pid and in child context OS will set the pid value to 0 so it will enter condition block intended for child
I am not sure how many processes going to be created in this program? I suppose there is going to be created three child process in a while loop. Then, each child process is going to fork two times which means there are going to be two child process plus the one created by that child process is also going to fork and there is going to be 4 child process for each state. In total, there are 12 processes. Is that true?
#include <unistd.h>
#include <sys/types.h>
#include <stdio.h>
#include <stdlib.h>
int main(){
pid_t childpid;
int i=0;
while((childpid>0) && (i<3))
{
childpid=fork();
i++;
}
if(childpid==0){
fork();
fork();
}
exit(0);
}
The total number of processes is 13
while((childpid>0) && (i<3))
will only execute for the parent since the fork() system call returns the pid of the child to the parent process and a 0 to the child process.
Thus the while loop will generate 3 child processes with the childpid variable value 0, and the main parent process which now has the childpid variable value equal to that of the last forked child.
Total processes till now 4
For the condition
if(childpid==0){
fork();
fork();
}
each child process from before spawns 1 child in the first fork call. The second fork will spawn another two new processes from the two proceeses of the first fork. So for each child process we will have spawned processes a total of 3 new processes.
Thus for th 3 child processes from the while loop, we will generate 9 new processes.
So total number of processes would be 4 + 9 = 13
Consider the code:
#include <stdio.h>
#include <errno.h>
#include <sys/types.h>
#include <unistd.h>
/* main --- do the work */
int main(int argc, char **argv)
{
pid_t child;
if ((child = fork()) < 0) {
fprintf(stderr, "%s: fork of child failed: %s\n",
argv[0], strerror(errno));
exit(1);
} else if (child == 0) {
// do something in child
}
} else {
// do something in parent
}
}
My question is from where does in the code the child process starts executing, i.e. which line is executed first??
If it executes the whole code, it will also create its own child process and thing will go on happening continuously which does not happen for sure!!!
If it starts after the fork() command, how does it goes in if statement at first??
It starts the execution of the child in the return of the fork function. Not in the start of the code. The fork returns the pid of the child in the parent process, and return 0 in the child process.
When you execute a fork() the thread is duplicated into memory.
So what effectively happens is that you will have two threads that executes the snippet you posted but their fork() return values will be different.
For the child thread fork() will return 0, so the other branch of the if won't be executed, same thing happens for the father thread.
When fork() is called the operating system assigns a new address space to the new thread that is going to spawn, then starts it, they will both share the same code segment but since the return value will be different they'll execute different parts of the code (if correctly split, like in your example)
The child starts by executing the next instruction (not line) after fork. So in your case it is the assignment of the fork's return value to the child variable.
Well, if i understand your question correctly, i can say to you that your code will run as a process already.When you run a code,it is already a process , so that this process goes if statement anyway. After fork(), you will have another process(child process).
In Unix, a process can create another process, that's why that happens.
Code execution in a child process starts from the next instruction following the fork() system call.
fork() system call just creates a seperate address space for the child process therefore it is a cloned copy of the parent process and the child process has all the memory elements of it's parent's process.
Thus, after spawning a child process through fork(), both processes (the parent process and the child process) resumes the execution right from the next instruction following the fork() system call.
I'm new to c language and Linux. I have a problem related to fork(),getpid()and exec()function.
I wrote a c program using fork() call the code of my program is following"
code:
#include <stdio.h>
#include <sys/types.h>
#include <unistd.h>
#include <stdlib.h>
void fun()
{
printf("\n this is trial for child process");
}
int main (int argc, char const *argv[])
{
int i,status,pid,t;
if(pid=fork()<0)
{
printf("\nfailed to create the process\n");
}
if(pid=fork()==0)
{
printf("\n the child process is created");
fun();
exit(1);
}
while(wait(&status)!=pid);
return 0;
}
The out put of this program is following:
the child process is created
this is trial for child process
the child process is created
this is trial for child process
Now my questions are as follows:
Why the output of program showing same thing twice? The output supposed to be "child process is created this is trial for child process"
Why the output is not according to code ?
Can we have a program which has 4 processes and all the processes perform different task for example one process print "my name". One process print "my age", the other process print "my address ?
How to make multiple process in main function ?
How to control the execution of multiple process ?
what does the exec() function do? Can anyone please explain me the working of exec(), fork(), getpid() with a source code?
Please help this novice fellow.
Your code calls fork() multiple times:
if(pid=fork()<0) /* calls fork() */
{
...
}
if(pid=fork()==0) /* also calls fork() */
{
...
}
Each successful fork() creates a new child process. To make matters worse, the second fork() is called by both the parent and the first child.
If you're trying to create a single child process, you should call fork() just once:
pid_t pid; /* note the correct return type of fork() */
...
pid = fork();
if (pid < 0)
{
...
}
else if (pid == 0)
{
...
}
If you want to create multiple child processes, you can have the parent process call fork() in a loop.
As to questions like "what does exec do?", my advice is to learn how to use man and then come back with specific questions if there's something in the manpages that remains unclear.
In this code you are creating Three process not including your main process.
pid=fork()
is itself a statement , which forks a new process even though it is inside an if statement condition. After the first fork() call the remaining codes will be executed twice. so next fork call will be called twice. You have already created a new process.
fork returns zero to itself and its
process id to its parent
That is consider a process A forks B (not from your code)
pid = fork();
printf("pid is : %d",pid);
printf statement executes twice(one for A and one for B). For A it prints(A is parent)
pid is : 512 //some integer value
process id
and B prints
pid is : 0
So in your question
if(pid=fork()==0)
{
printf("\n the child process is created");
fun();
exit(1);
}
this is the second fork which is already executing twice. So each of this execution creates a new child process. For both childs pid value is 0. So your print statement executes, which is what you see in the output. But for both parents a pid value will be there and your if condition fails, so it wont print. These two childs are your second and third processes..So in short you create 3 processes along with the main process
The output is generated twice because you are forking twice:
if(pid=fork()<0) // Fork #1
{
printf("\nfailed to create the process\n");
}
if(pid=fork()==0) // Fork #2
May be it look childish for most of you but I am unable to understand this small piece of code.
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/types.h>
#include <sys/wait.h>
int main(int argc, char** argv) {
int i, pid;
pid = fork();
printf("Forking the pid: %d\n",pid);
for(i =0; i<5; i++)
printf("%d %d\n", i,getpid());
if(pid)
wait(NULL);
return (0);
}
Out put of this program is
Forking the pid: 2223
0 2221
1 2221
2 2221
3 2221
4 2221
Forking the pid: 0
0 2223
1 2223
2 2223
3 2223
4 2223
Press [Enter] to close the terminal ...
In the for loop the printf command is used once. Why "Forking the pid" and after that the pid's are printed twice. How this is working? Can anybody explain me this? Thanks in advance.
Can anybody explain me why we have to use wait here? What I understood from the man pages is wait retuns the control to parent process? Is what I understood is correct?Is it necessary to use wait after forking a process?
Operating system : ubuntu, compiler : gcc, IDE : netbeans
But that' exactly what fork does. You forked the process and everything after the fork is done twice because now you have two processes executing the same printing code. You are basically asking why fork forks. fork forks because is is supposed to fork. That's what it's for.
After fork the parent and the child processes are generally executed in parallel, meaning that the nice sequential output you see in your example is not guaranteed. You might have easily ended up with line-interleaved output from two processes.
wait function in your case is executed from the parent process only. It makes it wait until the child process terminates, and only after that the parent process proceeds to terminate as well. Calling wait in this particular example is not really critical, since the program does nothing after that, it just terminates. But, for example, if you wanted to receive some feedback from the child process into the parent process and do some additional work on that feedback in the parent process, you'd have to use wait to wait for the child process to complete its execution.
The fork() call makes a new process. The rest of the code is then executed from each of the 2 processes. (Man page)
You're printing in both processes. Put your printing loop in an else clause of the if (pid):
pid = fork();
if(pid)
{
printf("Child pid: %d\n",pid);
wait(NULL);
}
else
{
for(i =0; i<5; i++)
printf("%d %d\n", i,getpid());
}
You see, fork returns twice, once in the parent process and once in the child process. It returns 0 in the child and the pid of the created process in the parent.
Because both the parent and child process are outputting their results.
See here: http://en.wikipedia.org/wiki/Fork_(operating_system)#Example_in_C for a good example.
fork creates a new process, and returns in both the old process (the parent) and in the new process (the child).
You can tell which one you are in by looking at the return value from fork. In the parent process it returns the PID of the child process. In the child process it return 0.