I have read that system function make use of execl, fork and wait functions internally. So, I tried to simulate working of system without using it. But I am not able to achieve the same working.
When we call a program using system function the code below(after) system() function call also executes. So to simulate system function i wrote this code below:
int main()
{
printf("In controller Start: %d\n\n",getpid());
system("./prog1");
printf("Forking New Process %d\n\n",fork());
printf("Process Id: %d\n\n",getpid());
execl("./infinite",0);
printf("In controller End\n\n");
return 0;
}
In the above code after running "infinite" program the last line does not get printed.
i.e. printf("In controller End\n\n");
What to do in order to print the last line and also execute the "infinite" program without using system function.
It would be great if someone can explain the step by step working of system function like which function is called by system first and so on.
Why execution is not continuing to last line like it must have did if we made a simple function call other than execl.
Foot notes:-
infinite: is a binary file created using C code.
The last line doesn't get printed because it is never executed. The execl function never returns if everything went okay, instead it replaces your program with the one in the call.
I highly recommend you read the manual pages for fork and execl.
In short, fork splits the current process into two, and returns differently depending on if it returns to the parent or the child process. In the child process you then does your exec call, while the parent process continues to do what it wants. The parent must however wait on the child process to finish, or the child process will become what is called a "zombie" process.
In your code, both the parent and the child processes calls exec.
this is basis of fork
/*previous code*/
if((cpid=fork())<0){
printf("\n\tFORK ERROR");
exit(1);
}
if(cpid==0){ /*SON*/
/*CODE FOR SON-your `execl("./infinite",0);` goes here*/
}else{ /*FATHER*/
/*CODE FOR FATHER-your `printf("In controller End\n\n");` */
}
dont forget that when making a fork memory and variables are copied to the SON pid
In your example you do the same thing in both the parent and the child process. You have to check the return value of fork, which indicates if you are in the parent or the child, and then exec in the child, while you wait in your main process.
When you call fork(), both the parent and child process continue executing the same code from that point, but the return value of fork() is different for each. Generally you would do some conditional logic based on that return value.
I would imagine that system() does something like this:
int childpid = fork();
if (childpid) {
/* This is the parent */
wait( childpid );
} else {
/* This is the child */
execl( program_name );
}
Since execl() replaces the current executable with a new one, the child will run that executable then end. The parent will wait for the child to complete then continue.
You are not performing any kind of conditional statement based on the return value of fork. If you don't make sure that one process does the exec and one does something else then both will do the same thing.
You usually want to check against 0 and then execute the program you want to run. 0 signals that everything went ok and you are in the child process.
int main()
{
int pid;
printf("In controller Start: %d\n\n",getpid());
system("./prog1");
pid = fork();
printf("Forking New Process %d\n\n",pid);
printf("Process Id: %d\n\n",getpid());
if (pid == 0) { /* Son process : execute the command */
execl("./infinite",0);
} else { /* Original process : keep working */
printf("In controller End\n\n");
return 0;
}
}
Related
My original code is too long to post so let me try to write a simple version
//this commands() function will be called multiple times
int commands()
{
pid_t pid = fork();
if(pid<0)
{
printf("Failed to open child process\n");
return -1;
}
if(pid == 0)
{
//we are in child
child_function(); // calls some function to do some work
//note that this function end up calling execvp() so it never comes back
}
else
{
int status=0;
int_returned_pid = waitpid(-1, &status,0);
if(status !=0)
return 1;
else
return 0;
}
}
so what happens is, when the program calls commands first time, it successfully creates a child, child process finishes and command returns.
But when commands function is called second time, again child is successfully created and child_function() is called. after this, the program just stops and it feels like its waiting for input or someting (it does not crash).
After spending several hours here, I have no idea what is the problem, Please help.
Edit: I am using pipes in my program and basically when commands function is called the first time, I need its children to read someting from stdin, do some stuff and then write to pipe. And then upon calling commands function second time, i need its children to read from that pipe that the 1st call children wrote to and then to write to stdout.
Do I need to close these pipes in a parent process ?
I have been asked this question for homework, and am having trouble figuring it out. If anyone can help me i would really appreciate it.
What Linux library function is like a fork(), but the parent process is terminated?
I'm fairly certain that whoever assigned you this homework is looking for the exec() family of functions, from the POSIX API header <unistd.h>, because there is nothing else that more closely resembles the sort of functionality you describe.
The exec() family of functions executes a new process and replaces the currently running process address space with the newly executed process.
From the man page:
The exec() family of functions replaces the current process image with
a new process image.
It's not exactly the same as "terminating" the parent process, but in practice it results in a similar situation where the parent process address space is erased (replaced) with the address space of the child process.
What Linux library function is like a fork(), but the parent process
is terminated?
The parent process should not terminate because , it must wait for the child processes to finish executing , after which they will be in a state called "zombie state", now it is the responsibility of the parent to clean up the leftovers of the child process. The parent process can terminate without cleaning up the child processes, but then, it is not a proper way to do it, as the exit status of the child processes should be collected and checked by the parent process.
Here is an example, to demonstrate , what i just said...
#include<stdio.h>
#include<unistd.h>
#include<sys/wait.h>
int main()
{
pid_t cpid = 1 ;
int status;
cpid = fork();
// Load a application to the child using execl() , and finish the job
printf("Parent waiting for child to terminate\n");
int wait_stat = waitpid(-1,&status,0); // Parent will hang here till all child processes finish executing..
if (wait_stat < 0)
{
perror("waitpid error");
exit(-1);
}
// WIFEXITED and WEXITSTATUS are macros to get exit status information, of the child process
if (WIFEXITED (status))
{
printf("Child of id %u cleaned up\n",wait_stat);
printf("Exit status of application = %u\n",WEXITSTATUS(status));
}
}
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 have a Makefile that looks something like this:
sometarget:
command_one # calls fork()
command_two
Here's the problem I'm running into when I do make sometarget:
command_one starts and eventually calls fork().
The child process execs something and finishes early, returning control to make before all the processing of command_one is done.
command_two then executes before the parent completes, causing the sequence to fail (as it depends on command_one finishing completely).
I can change command_one (the fork() and exec() have to stay, though), and I'd rather not change the Makefile if possible. Is there a way to prevent the child process from returning (on Linux)? I'm thinking the answer is no, but I've been wrong before...
It sounds like your command_one looks something like this pseudo-code:
main() {
pid_t child = fork(); /* ignore error for sake of example */
if (child) {
/* some work in the parent */
exit;
} else {
/* some work in the child */
}
exit;
}
If you insert a waitpid(2) or wait(2) (or any member of the family) immediately before the parent's exit, it'll make sure both child and parent are finished before make(1) moves onto the next command. It'll look something more like this:
main() {
pid_t child = fork(); /* ignore error for sake of example */
if (child) {
/* some work in the parent */
exit;
} else {
/* some work in the child */
}
waitpid(child, &status, 0); /* NEW LINE */
exit(&status);
}
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