I'm trying to create a child process in another process. I am writing both the programs in C language. First I write a dummy process which will be the child process. What it is doing is only to write a string on the screen. It works well on its own. Then I write another program which will be the parent process. However, I can't make it happen. I'm trying to use fork and execl functions together, but I fail. I also want the child process does not terminate until the parent process terminates.
How should I write the parent process?
Thanks.
Here is the code for the child process:
#include <stdio.h>
int main(void) {
while(1) {
printf("*");
sleep(1);
}
}
And here is the parent process:
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
int main(void) {
if (fork()) {
while(1) {
printf("-\n");
sleep(5);
}
} else {
execl("./", "dummy", (char *)0);
}
}
The fork() system call may return three different statuses: failure (<0), parent process (>0) or child process (==0). You must test the return value properly.
int pid = fork();
if (pid < 0) {
/* handle error */
perror("fork");
exit(1);
} else if (pid > 0) {
/* parent code */
} else {
/* child code */
}
Your execl() system call is wrong. The first argument is the path to the program you want to execute, "./" is not valid, it should be something like "./dummy" at least. The next argument is by convention the command name (argv[0] in the executed program), which may be a repetition of the first argument. So:
execl("./dummy", "dummy", NULL);
Also, note that the printf("*") statement in the child program will probably buffer and you won't see anything on the terminal. You must either add a "\n" to the end or call fflush(stdout) to flush the standard output.
Basic use of fork in C
int PID = fork();
if( PID < 0 ) {
//fail
return PID;
}
else if( !PID ) {
//child process
return exec( prog, args );
}
else {
//parent process
return 0;
}
There is no way to force the child process to "not terminate" when it's done (you'll still be able in the parent to wait for it to get info on how it terminated, but that's about it). Apart from that, any of the many examples of fork/exec on the web, such as this one, should work -- why don't you try it and see if it performs as you wish (in which case you'll just need to change whatever you were doing differently in your own attempt). If it doesn't work as desired (except for the impossibility per the first sentence in this A;-), please edit your to add copious detail about how the code behaves differently than you expect it to.
Related
[SOLVED]
So for this practice exercise I was assigned we are practicing process control. I'm simply wondering why it seems that my child process will never execute (when child_pid == 0). Any help on this issue would be greatly appreciated!
Also, safe_fork() is a function our instructor wrote in order to avoid students opening too many processes.
#include "safe-fork.h"
#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>
#include <sys/types.h>
int main(void) {
pid_t child_pid;
int pipefd[2];
int pipe_result;
int count;
pipe_result = pipe(pipefd);
if (pipe_result == -1) {
perror("pipe failure");
exit(1);
}
child_pid = safe_fork();
/*Parent Process*/
if (child_pid > 0) {
dup2(pipefd[0], STDIN_FILENO);
scanf("%d", &count);
if (count > 200) {
printf("Long enough!\n");
exit(0);
}
else {
printf("Too short!\n");
exit(1);
}
}
/* Child Process */
else {
dup2(pipefd[1], STDOUT_FILENO);
execl("/usr/bin/wc", "-w");
}
return 0;
}
Here is a suggestion, check for child_pid == 0 explicitly cause it could also be the case that your safe_fork() function failed, also check for those errors i.e child_pid < 0.
Add debug printf statements to check control flow if you already haven't and finally check if the execl call failed.
Also please have a look at this tutorial for creating pipes in C and using them to communicate between a parent and a child process. As I feel there are some issues with how you are handling those file descriptors. A possible reason that you can't see any output is that wc might still waiting for some input.
And another thing you might need to wait for the child process to complete/do something useful or actually be run before checking for its output. Check the other issues first before you worry about this.
Hope this Helps :)
Yes, But no. First you should carefully understand the meaning of fork().
fork(): creates a child process and it returns into TWO PROCESSES, with different return value in each. In parent, it always returns the number that OS has assigned for the newly created child's process id. At the same it returns in newly created child also, as child process will start executing the code copied from parent just after the fork() statement. In child fork always returns zero. So normally after a fork() system call, people check for returned value by fork to differentiate child process and parent process.
particularly in this case after a fork(), there will be two processes with same code and same program states(except both will have different value assigned in child_pid variable). In parent its a non zero value (which is corresponding to child's pid) and in child, its the value zero as the fork returns zero in child.
Now both processes will execute
if (child_pid > 0)
which is true only for parent so parent process will flow into 'if block' and is false for child so it will directly jump to else part and will execute the else block.
What if child_process id is zero?
OS will always assign a non zero value for a child so that fork() will work. And if you have somehow got child process id as zero then parent will fail to enter in 'if block' and will execute else block which is the same as child.
I need to create a program that takes input from the user and executes it just like it does in the terminal. I am using the execvp() function for this purpose. The requirement of the program is to keep taking input from the user unless the quit call is encountered. The problem here is that the current program is replaced after the execvp() call. So, using a goto is not an option either. I found this Fork–exec article but it doesn't tell how to create an indefinite number of processes. Here is my code:
#include <unistd.h>
#include <stdio.h>
#include <string.h>
void main() {
char *args[4];
char inputCommand[100];
fgets (inputCommand, 100, stdin);
printf ("Splitting string \"%s\" into tokens:\n",inputCommand);
/* Perfrom string tokenization here */
execvp(args[0], args);
}
fork() can be called an indefinite number of times; as long as the return value indicates that you're the parent process, you can continue to operate as usual and call it again.
Thus, you can have a loop within which you fork, call an execvp() if you're the child process, and continue to the next iteration if you're the parent.
Execvp replaces the current process image with the command you run. So it cancels your C program. To produce the desired effect, you should fork before execvp. It would look something like this:
int status = 0;
pid_t pid = fork();
if(pid > 0) {
waitpid(pid, &status, 0);
// is parent
} else if(pid == 0) {
execvp(*args, args);
// is child
} else {
// fork didn't work
}
I am currently trying to run a fork function in C where in the child section of the code,
I am trying to execute a command using exacvp, but before the execution I am trying a printf function which never executes. I ran this in debug and I have noticed that the pid is never assigned 0. I did try a simple fork example on a separate project and it worked smoothly. Does anyone have an idea why the child section never executes?
int startProcesses(int background) {
int i = 0;
while(*(lineArray+i) != NULL) {
int pid;
int status;
char *processName;
pid = fork();
if (pid == 0) {
printf("I am child");
// Child Process
processName = strtok(lineArray[i], " ");
execvp(processName, lineArray[i]);
i++;
continue;
} else if (!background) {
// Parent Process
waitpid(pid, &status, 0);
i++;
if(WEXITSTATUS(status)) {
printf(CANNOT_RUN_ERROR);
return 1;
}
} else {
i++;
continue;
}
}
return 0;
}
stdio files are flushed on program's exit, but execvp directly replaces the process with another image, bypassing the flush-at-exit mechanism and leaving the I am child message in memory never to be sent to the screen. Add an explicit fflush(stdout) before execvp, or end your string with \n so that it is automatically flushed when running on a TTY.
Note that execvp never exits, and if it does, it is because it has failed to execute the new process. At that point, the only thing the child can do is to report an error and call _exit(127) (or similar exit status). If the child continues, an incorrectly configured command name will cause it to execute the rest of the parent's loop in parallel with the parent. This process will continue for other descendants, effectively creating a fork bomb that can grind your system to a halt.
EDIT:
I am trying to write a simple smoketest, where all options and reasonable parameters are tested.
I used popen() to execute the program that should be tested.
Using this approach does not work, because if the process dies with a signal (SIGINT, SIGSEGV...) the pipe from popen() does not tell me what happend.
Writing a signal handler did not help since popen creates a new process that receives the signals but not my smoketest.
Thanks to the answers i used pipe(), fork() and execv() to create my own popen()-version.
When the program now segfaults there is the problem that the pipe is useless (a read caused weird behavior -> blocked the process until i send a sigkill to the parent!)
To avoid this i tried different things and my solution is the following (it is simple but it took me a while to figure it out). so here is my example-code:
static int child_dead = 0;
void sigaction_sigchld(int signal) { /* Child died */
child_dead = 1;
}
int main(int argc, char *argv[], char *env[])
{
char *crashing_program = "/program_path/and_name";
int ret;
int byte;
pid = fork();
if(pid == 0) /* Child */
{
execve(crashing_program, argv, env);
/* if execve returns that it mus have failed! */
fprintf(stderr, "Exec failed\n");
_exit(-1);
} else /* Parent */
{
if(!child_dead)
{
byte = read(pipe_out[1], line, BUFFSIZE);
if(!byte){
perror("Smoketest:Line:xxx");
} else
{
fprintf(stdout, line);
}
}
wait(&child_status);
/*
check if the child died with SIGSEGV etc
*/
}
This seems to work fine as long as i only have one child at a time which is sufficient for me though. I anyone has a better idea or any tipps for me i would be glad to update this entry.
Last but not least: Of course using this method it is probably impossible to do any cleanup.
Cheers.
See the documentation for waitpid(2). There are a bunch of macros you can use to test how the child process was terminated. In particular, you can use WIFSIGNALED() and WTERMSIG() to test if the child process was terminated by a signal, and if so, which signal:
int status = pclose(...);
if (WIFSIGNALED(status))
{
// It was terminated by a signal
if (WTERMSIG(status) == SIGSEGV)
{
// It was terminated by a segfault
}
}
Edit: As stated in the comments, you'd rather make use of fork and exec, then use waitpid(2) to correctly update 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