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How do I write a program that tells when my other program ends?
The only way to do a waitpid() or waitid() on a program that isn't spawned by yourself is to become its parent by ptrace'ing it.
Here is an example of how to use ptrace on a posix operating system to temporarily become another processes parent, and then wait until that program exits. As a side effect you can also get the exit code, and the signal that caused that program to exit.:
#include <sys/ptrace.h>
#include <errno.h>
#include <stdio.h>
#include <signal.h>
#include <unistd.h>
#include <sys/wait.h>
int main(int argc, char** argv) {
int pid = atoi(argv[1]);
int status;
siginfo_t si;
switch (ptrace(PTRACE_ATTACH, pid, NULL)) {
case 0:
break;
case -ESRCH:
case -EPERM:
return 0;
default:
fprintf(stderr, "Failed to attach child\n");
return 1;
}
if (pid != wait(&status)) {
fprintf(stderr, "wrong wait signal\n");
return 1;
}
if (!WIFSTOPPED(status) || (WSTOPSIG(status) != SIGSTOP)) {
/* The pid might not be running */
if (!kill(pid, 0)) {
fprintf(stderr, "SIGSTOP didn't stop child\n");
return 1;
} else {
return 0;
}
}
if (ptrace(PTRACE_CONT, pid, 0, 0)) {
fprintf(stderr, "Failed to restart child\n");
return 1;
}
while (1) {
if (waitid(P_PID, pid, &si, WSTOPPED | WEXITED)) {
// an error occurred.
if (errno == ECHILD)
return 0;
return 1;
}
errno = 0;
if (si.si_code & (CLD_STOPPED | CLD_TRAPPED)) {
/* If the child gets stopped, we have to PTRACE_CONT it
* this will happen when the child has a child that exits.
**/
if (ptrace(PTRACE_CONT, pid, 1, si.si_status)) {
if (errno == ENOSYS) {
/* Wow, we're stuffed. Stop and return */
return 0;
}
}
continue;
}
if (si.si_code & (CLD_EXITED | CLD_KILLED | CLD_DUMPED)) {
return si.si_status;
}
// Fall through to exiting.
return 1;
}
}
On Windows, a technique I've used is to create a global named object (such as a mutex with CreateMutex), and then have the monitoring program open that same named mutex and wait for it (with WaitForSingleObject). As soon as the first program exits, the second program obtains the mutex and knows that the first program exited.
On Unix, a usual way to solve this is to have the first program write its pid (getpid()) to a file. A second program can monitor this pid (using kill(pid, 0)) to see whether the first program is gone yet. This method is subject to race conditions and there are undoubtedly better ways to solve it.
If you want to spawn another process, and then do nothing while it runs, then most higher-level languages already have built-ins for doing this. In Perl, for example, there's both system and backticks for running processes and waiting for them to finish, and modules such as IPC::System::Simple for making it easier to figure how the program terminated, and whether you're happy or sad about that having happened. Using a language feature that handles everything for you is way easier than trying to do it yourself.
If you're on a Unix-flavoured system, then the termination of a process that you've forked will generate a SIGCHLD signal. This means your program can do other things your child process is running.
Catching the SIGCHLD signal varies depending upon your language. In Perl, you set a signal handler like so:
use POSIX qw(:sys_wait_h);
sub child_handler {
while ((my $child = waitpid(-1, WNOHANG)) > 0) {
# We've caught a process dying, its PID is now in $child.
# The exit value and other information is in $?
}
$SIG{CHLD} \&child_handler; # SysV systems clear handlers when called,
# so we need to re-instate it.
}
# This establishes our handler.
$SIG{CHLD} = \&child_handler;
There's almost certainly modules on the CPAN that do a better job than the sample code above. You can use waitpid with a specific process ID (rather than -1 for all), and without WNOHANG if you want to have your program sleep until the other process has completed.
Be aware that while you're inside a signal handler, all sorts of weird things can happen. Another signal may come in (hence we use a while loop, to catch all dead processes), and depending upon your language, you may be part-way through another operation!
If you're using Perl on Windows, then you can use the Win32::Process module to spawn a process, and call ->Wait on the resulting object to wait for it to die. I'm not familiar with all the guts of Win32::Process, but you should be able to wait for a length of 0 (or 1 for a single millisecond) to check to see if a process is dead yet.
In other languages and environments, your mileage may vary. Please make sure that when your other process dies you check to see how it dies. Having a sub-process die because a user killed it usually requires a different response than it exiting because it successfully finished its task.
All the best,
Paul
Are you on Windows ? If so, the following should solve the problem - you need to pass the process ID:
bool WaitForProcessExit( DWORD _dwPID )
{
HANDLE hProc = NULL;
bool bReturn = false;
hProc = OpenProcess(SYNCHRONIZE, FALSE, _dwPID);
if(hProc != NULL)
{
if ( WAIT_OBJECT_0 == WaitForSingleObject(hProc, INFINITE) )
{
bReturn = true;
}
}
CloseHandle(hProc) ;
}
return bReturn;
}
Note: This is a blocking function. If you want non-blocking then you'll need to change the INFINITE to a smaller value and call it in a loop (probably keeping the hProc handle open to avoid reopening on a different process of the same PID).
Also, I've not had time to test this piece of source code, but I lifted it from an app of mine which does work.
Most operating systems its generally the same kind of thing....
you record the process ID of the program in question and just monitor it by querying the actives processes periodically
In windows at least, you can trigger off events to do it...
Umm you can't, this is an impossible task given the nature of it.
Let's say you have a program foo that takes as input another program foo-sub.
Foo {
func Stops(foo_sub) { run foo_sub; return 1; }
}
The problem with this all be it rather simplistic design is that quite simply if foo-sub is a program that never ends, foo itself never ends. There is no way to tell from the outside if foo-sub or foo is what is causing the program to stop and what determines if your program simply takes a century to run?
Essentially this is one of the questions that a computer can't answer. For a more complete overview, Wikipedia has an article on this.
This is called the "halting problem" and is not solvable.
See http://en.wikipedia.org/wiki/Halting_problem
If you want analyze one program without execution than it's unsolvable problem.
Related
I have some code that needs to benchmark multiple algorithms. But before they can be benchmarked they need to get prepared. I would like to do this preparation multi-threaded, but the benchmarking needs to be sequential. Normally I would make threads for the preparation, wait for them to finish with join and let the benchmarking be done in the main thread. However the preparation and benchmarking are done in a seperate process after a fork because sometimes the preparation and the benchmarking may take too long. (So there is also a timer process made by a fork which kills the other process after x seconds.) And the preparation and benchmarking have to be done in the same process otherwise the benchmarking does not work. So I was wondering if I make a thread for every algorithm if there is a way to let them run concurrently until a certain point, then let them all wait untill the others reach that point and then let them do the rest of the work sequentially.
Here is the code that would be executed in a thread:
void prepareAndBenchmark(algorithm) {
//The timer thread that stops the worker after x seconds
pid_t timeout_pid = fork();
if (timeout_pid == 0) {
sleep(x);
_exit(0);
}
//The actual work
pid_t worker_pid = fork();
if (worker_pid == 0) {
//Concurrently:
prepare(algorithm)
//Concurrently up until this point
//At this point all the threads should run sequentially one after the other:
double result = benchmark(algorithm)
exit(0);
}
int status;
pid_t exited_pid = wait(&status);
if (exited_pid == worker_pid) {
kill(timeout_pid, SIGKILL);
if(status == 0) {
//I would use pipes to get the result of the benchmark.
} else {
//Something went wrong
}
} else {
//It took too long.
kill(worker_pid, SIGKILL);
}
wait(NULL);
}
I have also read that forking in threads migth give problems, would it be a problem in this code?
I think I could use mutex to have only one thread benchmarking at a time, but I don't want to have a thread benchmarking while others are still preparing.
Hi I have a simple question, however the timing issue is troubling me. Assume this is the code.
#include <stdio.h>
int main() {
int p = fork();
if (p==0) {
printf("ok\n");
sleep(1);
} else {
printf("hey!");
sleep(1);
}
printf("done!");
return 0;
}
My question is, will "done!" always be executed twice when the sleep is 1sec for both parent and child. Because I notice that when I increase the sleep to 10 seconds in the child process (p==0 case), I only see "done!" once.
I think when you increase sleep time parent process exited faster and stdout file descriptor closed. note that child and parent process shared their file descriptors.
if you want you can use _exit() in your parent process so when it exited, child process file descriptors will not be closed. in this way after 10 sec you see "done!" in your terminal. for use of this method you must use printf("done!\n") to flush your buffer manually because _exit() did not flush your buffer.
If you want you can use something like wait() in your parent process to issue wait on your child process.
There are lot of questions on fork() but I am little bit confused in this code.I am analyzing a code in c++ in that I have got this function.
int daemon(int nochdir, int noclose)
{
switch (fork())
{
case 0: break;
case -1: return -1;
default: _exit(0); /* exit the original process */
}
if (setsid() < 0) /* shoudn't fail */
return -1;
/* dyke out this switch if you want to acquire a control tty in */
/* the future -- not normally advisable for daemons */
printf("Starting %s [ OK ]\n",AGENT_NAME);
switch (fork())
{
case 0: break;
case -1: return -1;
default: _exit(0);
}
if (!nochdir)
{
chdir("/");
}
if (!noclose)
{
dup(0);
dup(1);
}
return 0;
}
So the fork will create an exact copy of the code from where the fork() has been called. so,
Is switch executed twice or once?
If twice then in the switch what if the child executes first? Will it just break or go to the other statements?
What If the parent executes? will the main process be terminated and child will continue?
Edit:
So the switch will also run twice once with parent and once with child. and behaves on the return values.
And the final thing is, the daemon is a predefined function and it has been redefined and used like user created daemon. How it will create the daemon process and what the
`if (!nochdir)
{
chdir("/");
}`
and
if (!noclose)
{
dup(0);
dup(1);
}
I am calling this function like this.
if (daemon(0, 0) < 0)
{
printf("Starting %s [ Failed ]\n",AGENT_NAME);
exit(2);
}
Is switch executed twice or once?
It is said that fork is the function that is called once but returns twice, that is once in each process: once in parent and once in a child.
man :
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
It might return just once (-1): only in parent if child wasn't created. It always returns in the parent ( -1 on error, > 0 on success).
If twice then in the switch what if the child executes first? Will it
just break or go to the other statements?
It is unknown whether child or parent returns first. After fork() all the memory segments are copied into child, but it continues with the correct value 0 returned from the fork(). Parent continues with pid of the child. You use return value of fork in the code to determine whether you are child or parent. Maybe this will get more clear if you write code this way
int daemon( int nochdir, int noclose)
{
pid_t pid; /* to drive logic in the code */
if ( ( pid = Fork()) < 0) /* fork and remember actual value returned to pid */
return -1;
if( pid > 0)
_exit(0); /* exit the original process */
// here pid is 0, i.e. the child
What If the parent executes? will the main process be terminated and
child will continue?
What if the parent exit() is called before any child instructions? Then yes, parent will terminate, child will do on its own. Both the parent and the child processes possess the same code segments, but execute independently of each other (unless you added some synchronization).
http://linux.die.net/man/2/fork
Yes, when the parent executes it will continue in the default: case as the switch will have returned the child process id.
The common convention of saying that fork() is a function which is called once and returns two times is a bit obfuscating as it only returns once in each process space. The question is whether a child was created or not which determines which of the two ways a parent returns. The parent never gets a result of '0' from fork(), only either -1 or >0. The child always (if at all) gets zero.
If the child wasn't created, then fork() never returns in its process space.
Unless there's an error, fork will return twice: once in the parent process and once in the child process. fork creates a copy of the current process, then continues execution in both processes and you can determine by the return value. Note that the copy (child) is not a "perfect" copy: for example, in the child, all threads are terminated except for the one executing fork. The exact behavior is a bit complex.
It's not specified whether the parent or child process continues execution first. This depends on your OS and might even be totally random on your OS. Since they are two separate processes (which happen to run the same code) the order doesn't matter. The parent process will get a return value >0 (or -1 on error) and thus execute the default: label. The child process will get a return value of 0 and thus execute the case 0: label. This return value of fork is how the parent process knows it's a parent and the child process that it is a child (the child can query its own PID using getpid(2) and the PID of its parent using getppid(2)).
Yes, the parent runs into the default: label and executes _exit, thus terminating. The child will continue to run (note that here setsid() is very important; without it, the child would not continue to run if the shell session of your parent exits). This is the usual pattern for creating a daemon: when you run the program, it spawns actual main program (the daemon) through forking and then exits. For example, in the shell, you'll see the program exits quickly, but when you enter ps you can see that there's a process with the same name (your daemon).
My program is supposed to limit the number of child processes to 3.
With the code below, waitpid stalls my parent process so I can't create more child processes after the first one. If I don't use waitpid then I don't know when a child process quits to decrease the number of alive processes.
int numProcs = 0;
while(1==1) {
/*
* inserts code that waits for incoming input
*/
numProcs++;
pid = fork();
if (pid == 0) {
doStuff(); // may exit anytime, based on user input
} else {
if (numProcs > 3) {
wait(&status);
numProcs--;
} else {
waitpid(pid, &status, 0); // PROBLEM!
numProcs--;
}
}
}
I've been searching for this problem the whole day. Can somebody help?
At the risk of being obvious, you basically want to just drop the else clause. The logic you're looking for is something like:
int max_active = 3; // or whatever
int number_active = 0;
bool done = false;
for (; !done; ++number_active) {
// wait for something to do;
GetSomeWork();
// wait for something to finish, if necessary.
for (; number_active >= max_active; --number_active)
wait(&status);
pid = fork();
if (pid < 0)
ReportErrorAndDie();
if (pid == 0)
DoTheWorkAndExit();
}
This actually lets you change the value of max_active without restarting, which is the only justification for the for loop around the wait() call.
The obvious complaint is that number_active in my version doesn't actually tell you how many processes are active, which is true. It tells you how many processes you haven't wait()'ed for, which means that you might keep some zombies (but the number is limited). If you're constantly running at or close to the maximum number of tasks, this doesn't matter, and unless your maximum is huge, it doesn't matter anyway, since the only Design Requirement was that you don't use more than the maximum number of tasks, and consequently you only have to know that the number active is not more than the maximum.
If this really bothers you and you want to clean the tasks up, you can put:
for (; waitpid(-1, &status, WNOHANG) > 0; --number_active) {}
before the other for loop, which will reap the zombies before checking if you need to block. (I can't remember if waitpid(-1, &status WNOHANG) returns an error if there are no processes at all, but in any event there's no point continuing the loop on an error.)
You have two problems with your code, but the second one is masked by the first.
Your immediate problem is that waitpid(pid, &status, 0); will block until the process with the specified pid terminates. You want to add the WNOHANG option as a third parameter to the waitpid() call. That will ensure that the call dosn't block.
This will add a new problem: You will have to check for yourself is any child process have terminated. You can do that using the WIFEXITED macro:
} else {
waitpid (-1, &status, WNOHANG);
if (WIFEXITED(status)) {
numProcs--;
}
}
The second problem is that your original code only waits for the latest pid to be created. You should instead wait for -1, which is all child processes.
I am kind of newbie on C++, and working on a simple program on Linux which is supposed to invoke another program in the same directory and get the output of the invoked program without showing output of the invoked program on console. This is the code snippet that I am working on:
pid_t pid;
cout<<"General sentance:"<<endl<<sentence<<endl;
cout<<"==============================="<<endl;
//int i=system("./Satzoo");
if(pid=fork()<0)
cout<<"Process could not be created..."<<endl;
else
{
cout<<pid<<endl;
execv("./Satzoo",NULL);
}
cout<<"General sentance:"<<endl<<sentence<<endl;
cout<<"==============================="<<endl;
One of the problem I encounter is that I am able to print the first two lines on console but I cant print the last two lines. I think the program stops working when I invoke the Satzoo program.
Another thing is that this code invokes Satzoo program twice, I dont know why? I can see the output on screen twice. On the other hand if I use system() instead of execv(), then the Satzoo works only once.
I haven't figured out how to read the output of Satzoo in my program.
Any help is appreciated.
Thanks
You aren't distinguisng between the child and the parent process after the call to fork(). So both the child and the parent run execv() and thus their respective process images are replaced.
You want something more like:
pid_t pid;
printf("before fork\n");
if((pid = fork()) < 0)
{
printf("an error occurred while forking\n");
}
else if(pid == 0)
{
/* this is the child */
printf("the child's pid is: %d\n", getpid());
execv("./Satzoo",NULL);
printf("if this line is printed then execv failed\n");
}
else
{
/* this is the parent */
printf("parent continues execution\n");
}
The fork() function clones the current process and returns different values in each process. In the "parent" process, it returns the pid of the child. In the child process, it returns zero. So you would normally invoke it using a model like this:
if (fork() > 0) {
cout << "in parent" << endl;
} else {
cout << "in child" << endl;
exit(0);
}
I have omitted error handling in the above.
In your example, both of the above code paths (both parent and child) fall into the else clause of your call to fork(), causing both of them to execv("./Satzoo"). That is why your program runs twice, and why you never reach the statements beyond that.
Instead of using fork() and doing everything manually (properly managing process execution is a fair amount of work), you may be interested in using the popen() function instead:
FILE *in = popen("./Satzoo", "r");
// use "in" like a normal stdio FILE to read the output of Satzoo
pclose(in);
From the fork() manpage:
RETURN VALUE
Upon successful completion, fork() shall return 0 to the child process and shall return the process ID of the child process to the parent process. Both processes shall continue to execute from the fork() function. Otherwise, -1 shall be returned to the parent process, no child process shall be created, and errno shall be set to indicate the error.
You check to make sure it succeeds, but not whether the pid indicates we're in the child or the parent. Thus, both the child and the parent do the same thing twice, which means that your program gets executed twice and the ending text is never printed. You need to check the return value of fork() more than just once.
exec - The exec() family of functions replaces the current process image with a new process image.
system - Blocks on execution of the command. Execution of the calling program continues after the system command returns
There are three return value tests you want with fork
0: you are the child
-1: error
other: you are the parent
You ran the other program from both the child and the parent...