fork(), pipe() and exec() process creation and communication - c

I have to write program that create process using pipe().
My first task is to write a parent process that generates four child processes using the fork() function.
Once the fork() is successful, replace the child process with another process rover1, rover2, rover3, and rover4, though all of them have the same code.
The function of the processes is as follows.
Each child process is initially given its own number. It receives a new number from the parent. Using the following formula it creates its own new number as follows and forwards it to the parent:
mynumber = (3 * mynumber + 4 * numberreceived)/7
This process continues until the parent sends the message that the system is stable. The parent also has its initial number. It receives numbers of all the children and computes its new number as follows:
mynumber = (3 * mynumber + numbers sent by all the children)/7
The parent will send this number to all its children. This process will continue until the parent finds that its number is not changing anymore. At that time it will tell the children the system has become stable.
This is what I did but my professor said I have to use exec() to execute the child and replace child process with another child process. I am not sure how to use exec(). Could you please help me with this.
I am attaching only first child generation.
// I included stdio.h, unistd.h stdlib.h and errno.h
int main(void)
{
// Values returned from the four fork() calls
pid_t rover1, rover2, rover3, rover4;
int parentnumber, mynumber1, mynumber2, mynumber3, mynumber4;
int childownnumber1 = 0, status = 1, childownnumber2 = 0,
childownnumber3 = 0, childownnumber4 = 0, numberreceived = 0;
printf("Enter parent number: ");
printf("%d", parentnumber);
printf("Enter each children number");
printf("%d %d %d %d", mynumber1, mynumber2, mynumber3, mynumber4);
// Create pipes for communication between child and parent
int p1[2], p2[2];
// Attempt to open pipe
if (pipe(p1) == -1) {
perror("pipe call error");
exit(1);
}
// Attempt to open pipe
if (pipe(p2) == -1) {
perror("pipe call error");
exit(1);
}
// Parent process generates 4 child processes
rover1 = fork();
// if fork() returns 0, we're in the child process;
// call exec() for each child to replace itself with another process
if (rover1 == 0) {
for(; numberreceived != 1; ) {
close(p1[1]); // Close write end of pipe
close(p2[0]); // Close read end of second pipe
// Read parent's number from pipe
read(p1[0], &numberreceived, sizeof(int));
if (numberreceived == 1) {
// System stable, end child process
close(p1[0]);
close(p2[1]);
_exit(0); // End child process
}
mynumber1 = (int)((3*mynumber1 + 4*numberreceived)/7.0);
printf("\nrover1 number: ");
printf("%i", mynumber1);
// Write to pipe
write(p2[1], &mynumber1, sizeof(int));
}
}
/* Error:
* If fork() returns a negative number, an error happened;
* output error message
*/
if (rover1 < 0) {
fprintf(stderr,
"can't fork, child process 1 not created, error %d\n",
errno);
exit(EXIT_FAILURE);
}
}

The exec family of functions is used to replace the current process with a new process. Note the use of the word replace. Once exec is called, the current process is gone and the new process starts. If you want to create a separate process, you must first fork, and then exec the new binary within the child process.
Using the exec functions is similar to executing a program from the command line. The program to execute as well as the arguments passed to the program are provided in the call to the exec function.
For example, the following execcommand* is the equivalent to the subsequent shell command:
execl("/bin/ls", "/bin/ls", "-r", "-t", "-l", (char *) 0);
/bin/ls -r -t -l
* Note that "arg0" is the command/file name to execute
Since this is homework, it is important to have a good understanding of this process. You could start by reading documentation on pipe, fork, and exec combined with a few tutorials to gain a better understanding each step.
The following links should help to get you started:
IBM developerWorks: Delve into UNIX process creation
YoLinux Tutorial: Fork, Exec and Process control
Pipe, Fork, Exec and Related Topics

If you are supposed to use exec, then you should split your program into two binaries.
Basically, the code that now gets executed by the child should be in the second binary and should be invoked with exec.
Before calling one of the exec family of functions, you'll also need to redirect the pipe descriptors to the new process' standard input/output using dup2. This way the code in the second binary that gets exec'd won't be aware of the pipe and will just read/write to the standard input/output.
It's also worth noting that some of the data you are using now in the child process is inherited from the parent through the fork. When using exec the child won't share the data nor the code of the parent, so maybe you can consider transmitting the needed data through the pipe as well.

Related

Do *Unix shells call the pipe() function when encountering the "pipe character"? [duplicate]

I am working on a tiny shell(tsh) implemented in C(it's an assignment). One part of assignment belongs to PIPING. I have to pipe a command's output to another command. e.g:ls -l | sort
When I run the shell, every command that I execute on it, is processed by a child process that it spawns. After the child finishes the result is returned. For piping I wanted to implement a harcoded example first to check how it works. I wrote a method, that partially works. The problems is when I run the pipe command, after child process finishes, the whole program quits with it! Obviously I am not handling the child process signal properly(Method code below).
My Question:
How does process management with pipe() works? if i run a command ls -l | sort does it create a child process for ls -l and another process for sort ? From the piping examples that I have seen so far, only one process is created(fork()).
When the second command (sort from our example) is processed, how can i get its process ID?
EDIT: Also while running this code I get the result twice. don't know why it runs twice, there is no loop in there.
Here is my code:
pid_t pipeIt(void){
pid_t pid;
int pipefd[2];
if(pipe(pipefd)){
unix_error("pipe");
return -1;
}
if((pid = fork()) <0){
unix_error("fork");
return -1;
}
if(pid == 0){
close(pipefd[0]);
dup2(pipefd[1],1);
close(pipefd[1]);
if(execl("/bin/ls", "ls", (char *)NULL) < 0){
unix_error("/bin/ls");
return -1;
}// End of if command wasn't successful
}// End of pid == 0
else{
close(pipefd[1]);
dup2(pipefd[0],0);
close(pipefd[0]);
if(execl("/usr/bin/tr", "tr", "e", "f", (char *)NULL) < 0){
unix_error("/usr/bin/tr");
return -1;
}
}
return pid;
}// End of pipeIt
Yes, the shell must fork to exec each subprocess. Remember that when you call one of the execve() family of functions, it replaces the current process image with the exec'ed one. Your shell cannot continue to process further commands if it directly execs a subprocess, because thereafter it no longer exists (except as the subprocess).
To fix it, simply fork() again in the pid == 0 branch, and exec the ls command in that child. Remember to wait() for both (all) child processes if you don't mean the pipeline to be executed asynchronously.
Yes, you do need to call fork at least twice, once for each program in the pipeline. Remember that exec replaces the program image of the current process, so your shell stops existing the moment you start running sort or (tr).

Pipe function in Linux shell write in C

My mini-shell program accepts pipe command, for example, ls -l | wc -l and uses excevp to execute these commands.
My problem is if there is no fork() for execvp, the pipe command works well but the shell terminates afterward. If there is a fork() for execvp, dead loop happens. And I cannot fix it.
code:
void run_pipe(char **args){
int ps[2];
pipe(ps);
pid_t pid = fork();
pid_t child_pid;
int child_status;
if(pid == 0){ // child process
close(1);
close(ps[0]);
dup2(ps[1], 1);
//e.g. cmd[0] = "ls", cmd[1] = "-l"
char ** cmd = split(args[index], " \t");
//if fork here, program cannot continue with infinite loop somewhere
if(fork()==0){
if (execvp(cmd[0],cmd)==-1){
printf("%s: Command not found.\n", args[0]);
}
}
wait(0);
}
else{ // parent process
close(0);
close(ps[1]);
dup2(ps[0],0);
//e.g. cmd[0] = "wc", cmd[1] = "-l"
char ** cmd = split(args[index+1], " \t");
//if fork here, program cannot continue with infinite loop somewhere
if(fork()==0){
if (execvp(cmd[0],cmd)==-1){
printf("%s: Command not found.\n", args[0]);
}
}
wait(0);
waitpid(pid, &child_status, 0);
}
}
I know fork() is needed for excevp in order to not terminate the shell program, but I still cannot fix it. Any help will be appreciated, thank you!
How should I make two children parallel?
pid = fork();
if( pid == 0){
// child
} else{ // parent
pid1 = fork();
if(pid1 == 0){
// second child
} else // parent
}
is this correct?
Yes, execvp() replaces the program in which it is called with a different one. If you want to spawn another program without ending execution of the one that does the spawning (i.e. a shell) then that program must fork() to create a new process, and have the new process perform the execvp().
Your program source exhibits a false parallelism that probably either confuses you or reflects a deeper confusion. You structure the behavior of the first child forked in just the same way as the behavior of the parent process after the fork, but what should be parallel is the behavior of the first child and the behavior of the second child.
One outcome is that your program has too many forks. The initial process should fork exactly twice -- once for each child it wants to spawn -- and neither child should fork because it's already a process dedicated to one of the commands you want to run. In your actual program, however, the first child does fork. That case is probably rescued by the child also wait()ing for the grandchild, but it's messy and poor form.
Another outcome is that when you set up the second child's file descriptors, you manipulate the parent's, prior to forking, instead of manipulating the child's after forking. Those changes will persist in the parent process, which I'm pretty confident is not what you want. This is probably why the shell seems to hang: when run_pipe() returns (the shell's standard input has been changed to the read end of the pipe).
Additionally, the parent process should close both ends of the pipe after the children have both been forked, for more or less the same reason that the children must each close the end they are not using. In the end, there will be exactly one open copy of the file descriptor for each end of the pipe, one in one child and the other in the other. Failing to do this correctly can also cause a hang under some circumstances, as the processes you fork may not terminate.
Here's a summary of what you want the program to do:
The original process sets up the pipe.
The original process forks twice, once for each command.
Each subprocess manipulates its own file descriptors to use the correct end of the pipe as the appropriate standard FD, and closes the other end of the pipe.
Each subprocess uses execvp() (or one of the other functions in that family) to run the requested program
the parent closes its copies of the file descriptors for both ends of the pipe
the parent uses wait() or waitpid() to collect two children.
Note, too, that you should check the return values of all your function calls and provide appropriate handling for errors.

using execvp to execute commands that I have in an array

I have a commands array and I want to execute each command in this array but I couldn't seem to get it working so I have
childPid = fork();
for(int i =0;i < numOfCommands;i++)
{
if(childPid == 0)
{
execvp(commands[i], argv);
perror("exec failure");
exit(1);
}
else
{
wait(&child_status);
}
}
What this does, is that it only executes the 1st command in my array but doesn't proceed any further, how would I continue ?
And what if i want the order for the commands to executed randomly and the results be intermixed so do I have to use fork then ?
You need to use fork in any case, if you want to execute more than one program. From man exec: (emphasis added)
The exec() family of functions replaces the current process image with a new process image.
…
The exec() functions return only if an error has occurred.
By using fork, you create a new process with the same image, and you can replace the image in the child process by calling exec without affecting the parent process, which is then free to fork and exec as many times as it wants to.
Don't forget to wait for the child processes to terminate. Otherwise, when they die they will become zombies. There is a complete example in the wait manpage, linked above.

How system function in C works

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;
}
}

After fork, do the parent and child process share the file descriptor created by pipe?

int main()
{
int data_processed;
int file_pipes[2];
const char some_data[] = "123";
char buffer[BUFSIZ + 1];
pid_t fork_result;
memset(buffer, '\0', sizeof(buffer));
if (pipe(file_pipes) == 0) {
fork_result = fork();
if (fork_result == -1) {
fprintf(stderr, "Fork failure");
exit(EXIT_FAILURE);
}
// We've made sure the fork worked, so if fork_result equals zero, we're in the child process.
if (fork_result == 0) {
data_processed = read(file_pipes[0], buffer, BUFSIZ);
printf("Read %d bytes: %s\n", data_processed, buffer);
exit(EXIT_SUCCESS);
}
// Otherwise, we must be the parent process.
else {
data_processed = write(file_pipes[1], some_data,
strlen(some_data));
printf("Wrote %d bytes\n", data_processed);
}
}
exit(EXIT_SUCCESS);
}
Based on my understanding, the child process created by fork doesn't share variables with its parent process. Then, why here the parent can write to one file descriptor and child process can get the data by reading from another file descriptor. Is this because they are controled somehow by the pipe function internally?
File descriptors, including pipes, are duplicated on fork -- the child process ends up with the same file descriptor table, including stdin/out/err and the pipes, as the parent had immediately before the fork.
Based on my understanding, the child process created by fork doesn't share variables with its parent process.
This isn't entirely true -- changes to variables are not shared with the parent, but the values that the parent had immediately prior to the fork are all visible to the child afterwards.
In any case, pipes exist within the operating system, not within the process. As such, data written to one end of the pipe becomes visible to any other process holding a FD for the other end. (If more than one process tries to read the data, the first process to try to read() data gets it, and any other processes miss out.)
The variables are not shared e.g. if you write file_pipes[0] = 999 in the child, it will not be reflected in the parent. The file descriptors are shared (FD number x in the child refers to the same thing as FD number x in the parent). This is why (for example) you can redirect the output of a shell script which executes other commands (because they share the same standard output file descriptor).
You're right - ordinary variables aren't shared between the parent and the child.
However, pipes are not variables. They're a pseudo-file specifically designed to connect two independent processes together. When you write to a pipe, you're not changing a variable in the current process - you're sending data off to the operating system and asking it to make that data available to the next process to read from the pipe.
It's just like when you write to a real, on-disk file - except that the data isn't written to disk, it's just made available at the other end of the pipe.

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