I am writing a code wherein I am making my threads wait till I do a pthread_cond_broadcast. I have three threads in this code. Threads line2_thread and line3_thread, are running in the order of their priority like they are supposed to run. However, my third thread doesn't even enter its thread function (line4_thread). Can anyone tell me why is my main() not being able to call my line4_thread ?
pthread_cond_t sstart;
pthread_mutex_t sstart_mutex;
void *l3_thread(void *arg){
pthread_mutex_lock(&sstart_mutex);
pthread_cond_wait(&sstart, &sstart_mutex);
pthread_mutex_unlock(&sstart_mutex);
/*do something*/
pthread_exit(NULL);
}
void *l2_thread(void *arg){
pthread_mutex_lock(&sstart_mutex);
pthread_cond_wait(&sstart, &sstart_mutex);
pthread_mutex_unlock(&sstart_mutex);
/*do something*/
pthread_exit(NULL);
}
void *l4_thread(void *arg){
pthread_mutex_lock(&sstart_mutex);
pthread_cond_wait(&sstart, &sstart_mutex);
pthread_mutex_unlock(&sstart_mutex);
/*do something*/
pthread_exit(NULL);
}
int main(){
pthread_cond_init(&sstart, NULL);
//thread creation
pthread_cond_broadcast(&sstart);
pthread_cond_destroy(&sstart);
pthread_mutex_destroy(&sstart_mutex);
return 0;
}
I think you have a few problems here. With apologies (I'm on my phone so typing a long answer is hard) I'm just going to focus on a couple things, since it's not 100% clear to me what you're actually trying to do.
When all your threads start they all try to acquire the mutex, and only one succeeds. Probably l3 but I don't think that's guaranteed here. It then calls the pthread_cond_wait and unlocks the mutex allowing one of the other threads to reach its pthread_cond_wait. But in the meantime. You've allowed your main thread to call pthread_cond_broadcast, and you've taken no steps to synchronize this with the other threads. It may happen before the others get unblocked from waiting for the mutex, and before their wait call, so they could miss the signal and block forever.
Further, I think it's a bit sketchy to immediately call pthread_cond_destroy. Like I said there's no synchronization between your main thread and your worker threads, so it's possible you could call pthread_cond_broadcast followed by pthread_cond_destroy, so some of your threads might be calling pthread_cond_wait on an invalid condition variable and deadlock.
Check the return values of pthread_cond_wait. If I'm right, it might return EINVAL in some cases. But I haven't tested this so there might be a flaw in my reasoning.
Related
I just started learning about threads, mutexes and condition variables and I have this code:
#include <pthread.h>
#include <stdio.h>
#include <stdbool.h>
#include <stdlib.h>
static pthread_cond_t cond = PTHREAD_COND_INITIALIZER;
static pthread_mutex_t mutex;
static pthread_mutexattr_t attr;
volatile int x = 0;
void *thread_one_function (void *dummy) {
printf("In func one.");
while (true) {
pthread_mutex_lock (&mutex);
x = rand ();
pthread_cond_signal (&cond);
pthread_mutex_unlock (&mutex);
}
}
void *thread_two_function (void *dummy) {
printf("In func two.");
pthread_mutex_lock (&mutex);
while (x != 10) {
pthread_cond_wait (&cond, &mutex);
}
printf ("%d\n", x);
pthread_mutex_unlock (&mutex);
printf ("%d\n", x);
}
int main (void){
pthread_mutexattr_init (&attr);
pthread_mutexattr_settype (&attr, PTHREAD_MUTEX_RECURSIVE);
pthread_mutex_init (&mutex, &attr);
pthread_t one, two;
pthread_create(&one,NULL,thread_one_function,NULL);
pthread_create(&two,NULL,thread_two_function,NULL);
//pthread_join(one,NULL); //with this program will not end.
//pthread_join(two,NULL);
return 0;
}
I compile it as gcc prog.c -lpthread -o a.exe
And I am getting no output. Not even that my threads get into those two functions...
What am I doing wrong ? My code is created as a combination from multiple documentations.
Thanks for any help.
The most likely reason for you getting no output is that the main thread returns from the initial call to main() immediately after starting the threads. That terminates the program, including the newly-created threads.
It does not help that neither do the threads' initial messages end with newlines nor do the threads flush the standard output after the printf calls. The standard output is line-buffered when connected to a terminal, so actual delivery of those messages to the terminal will be deferred in your example.
The main thread should join the other two before it terminates. Your code comments indicate that you had other issues when you did that, but going without joining those threads is not a correct solution if in fact you want the additional threads to run to completion. The problem is not joining the threads, but rather that the threads are not terminating.
There's a pretty clear reason why the thread_one_function thread does not terminate, which I am sure you will recognize if you look for it. But what about the thread_two_function thread? The reasons for that taking a long time to terminate are at least twofold:
the more obvious one is the one hinted in #dbush's comment on the question. Consider the range of the rand() function: how many calls do you think it might take before it returns the one, specific result that thread_two_function is looking for?
moreover, how often is thread_two_function even going to get a chance to check? Consider: thread_one_function runs a tight loop in which it acquires the mutex at the top and releases it at the bottom. Pthreads mutexes do not implement any kind of fairness policy, so when thread_one_function loops around and tries to reacquire the mutex immediately after releasing it, it has a very high probability of succeeding immediately, even though thread_two_function may be trying to acquire the mutex too.
It would make more sense for your threads to take turns. That is, after generating a new number, thread_one_function would wait for thread_two_function to check it before generating another. There are several ways you could implement that. Some involve using the same condition variable in thread_one_function that you do in thread_two_function. (Details left as an exercise.) I would suggest also providing a means by which thread_two_function can tell thread_one_function that it doesn't need to generate any more numbers, and should instead terminate.
Finally, do be aware that volatile has no particular place here. It is not useful or appropriate for synchronization, whereas the mutex alone is entirely sufficient for that. It's not exactly wrong to declare x volatile, but it's extraneous.
I read that main() is single thread itself, so when i create 2 threads in my program like this;
#include<stdio.h>
#include<pthread.h>
#include<windows.h>
void* counting(void * arg){
int i = 0;
for(i; i < 50; i++){
printf("counting ... \n");
Sleep(100);
}
}
void* waiting(void * arg){
int i = 0;
for(i; i < 50; i++){
printf("waiting ... \n");
Sleep(100);
}
}
int main(){
pthread_t thread1;
pthread_t thread2;
pthread_create(&thread1, NULL, counting, NULL);
pthread_create(&thread2, NULL, waiting, NULL);
pthread_join(thread1, NULL);
pthread_join(thread2, NULL);
int i = 0;
for(i; i < 50; i++){
printf("maining ... \n");
Sleep(1000);
}
}
Is main really a thread in that case?
in that case if in main in sleep for some time, shouldn't the main give the CPU to other threads?
Is main a threads itself here? I am confused a bit here.
Is there a specific order to main thread execution?
pthread_join(thread1, NULL);
pthread_join(thread2, NULL);
You asked the thread to wait until thread1 terminates and then wait until thread2 terminates, so that's what it does.
I read that main() is single thread itself
No, you have misunderstood. Every C program has a function named main(). C language semantics of the program start with the initial entry into that function. In that sense, and especially when you supply the parentheses, main(), is a function, not a thread.
However, every process also has a main thread which has a few properties that distinguish it from other threads. That is initially the only thread, so it is that thread that performs the initial entry into the main() function. But it is also that thread that runs all C functions called by main(), and by those functions, etc., so it is not, in general, specific to running only the code appearing directly in the body of main(), if that's what you mean by "main() is a single thread itself".
, so when i create 2 threads in my program like this; [...] Is main really a thread in that case?
There is really a main thread in that case, separate from the two additional threads that it starts.
in that case if in main in sleep for some time, shouldn't the main give the CPU to other threads?
If the main thread slept while either of the other two were alive, then yes, one would expect one or both of the others to get (more) CPU time. And in a sense, that's exactly what happens: the main thread calls pthread_join() on each of the other threads in turn, which causes it to wait (some would say "sleep") until those threads terminate before it proceeds. While it's waiting, it does not contend with the other threads for CPU time, as that's pretty much what "waiting" means. But by the time the main thread reaches the Sleep() call in your program, the other threads have already terminated and been joined, because that's what pthread_join() does. They no longer exist, so naturally they don't run during the Sleep().
Is main a threads itself here?
There is a main thread, yes, and it is the only one in your particular process that executes any of the code in function main(). Nothing gets executed except in some thread or other.
I am confused a bit here. Is there a specific order to main thread execution?
As already described, the main thread is initially the only thread. Many programs never have more than that one. Threads other than the main one are created only by the main thread or by another thread that has already been created. Of course, threads cannot run before they are created, nor, by definition, after they have terminated. Threads execute independently of each other, generally without any predefined order, except as is explicitly established via synchronization objects such as mutexes, via for-purpose functions such as pthread_join(), or via cooperative operations on various I/O objects such as pipes.
main() is not a thread but a function, so here's a clear "no" to your initial claim. However, if you read a few definitions of what is a thread, you will find that it is something that can be scheduled, i.e. an ongoing execution of code. Further, a running program will not be able to actually do anything without "ongoing execution of code" without e.g. main() as first entrypoint. So, definitely, every code executed by a program is executed by a thread, without exceptions.
BTW: You can retrieve the thread ID of the current thread. Try running that from main(). It will work and give you a value that distinguishes this call from calls from other threads.
I am trying to better understand how to use pthread_cond_wait() and how it works.
I am just looking for a bit of clarification to an answer I saw on this site.
The answer is the last reply on this page
understanding of pthread_cond_wait() and pthread_cond_signal()
I am wondering how this would look with three threads. Imagine Thread 1 wants to tell Thread 2 and Thread 3 to wake up
For example
pthread_mutex_t mutex;
pthread_cond_t condition;
Thread 1:
pthread_mutex_lock(&mutex);
/*Initialize things*/
pthread_mutex_unlock(&mutex);
pthread_cond_signal(&condition); //wake up thread 2 & 3
/*Do other things*/
Thread 2:
pthread_mutex_lock(&mutex); //mutex lock
while(!condition){
pthread_cond_wait(&condition, &mutex); //wait for the condition
}
pthread_mutex_unlock(&mutex);
/*Do work*/
Thread 3:
pthread_mutex_lock(&mutex); //mutex lock
while(!condition){
pthread_cond_wait(&condition, &mutex); //wait for the condition
}
pthread_mutex_unlock(&mutex);
/*Do work*/
I am wondering if such a setup is valid. Say that Thread 2 and 3 relied on some intialization options that thread 1 needs to process.
First: If you wish thread #1 to wake up thread #2 and #3, it should use pthread_cond_broadcast.
Second: The setup is valid (with broadcast). Thread #2 and #3 are scheduled for wakeup and they will try to reacquire the mutex as part of waking up. One of them will, the other will have to wait for the mutex to be unlocked again. So thread #2 and #3 access the critical section sequentically (to re-evaluate the condition).
If I understand correctly, you want thr#2 and thr#3 ("workers") to block until thr#1 ("boss") has performed some initialization.
Your approach is almost workable, but you need to broadcast rather than signal, and are missing a predicate variable separate from your condition variable. (In the question you reference, the predicate and condition variables were very similarly named.) For example:
pthread_mutex_t mtx;
pthread_cond_t cv;
int initialized = 0; // our predicate of interest, signaled via cv
...
// boss thread
initialize_things();
pthread_mutex_lock(&mtx);
initialized = 1;
pthread_cond_broadcast(&cv);
pthread_mutex_unlock(&mtx);
...
// worker threads
pthread_mutex_lock(&mtx);
while (! initialized) {
pthread_cond_wait(&cv, &mtx);
}
pthread_mutex_unlock(&mtx);
do_things();
That's common enough that you might want to combine the mutex/cv/flag into a single abstraction. (For inspiration, see Python's Event object.) POSIX barriers which are another way of synchronizing threads: every thread waits until all threads have "arrived." pthread_once is another way, as it runs a function once and only once no matter how many threads call it.
pthread_cond_signal wakes up one (random) thread waiting on the cond variable. If you want to wake up all threads waiting on this cond variable use pthread_cond_broadcast.
Depending what you are doing on the critical session there might be also another solution apart from the ones in the previous answers.
Suppose thread1 is executing first (i.e. it is the creator thread) and suppose thread2 and thread3 do not perform any write in to shared resource in the critical session. In this case with pthread_cond_wait you are forcing one thread to wait the other when actually there is no need.
You can use read-write mutex of type pthread_rwlock_t. Basically the thread1 performs a write-lock so the other threads will be blocked when trying to acquire a read-lock.
The functions for this lock are quite self-explanatory:
//They return: 0 if OK, error number on failure
int pthread_rwlock_init(pthread_rwlock_t *restrict rwlock,
const pthread_rwlockattr_t *restrict attr);
int pthread_rwlock_destroy(pthread_rwlock_t *rwlock);
int pthread_rwlock_rdlock(pthread_rwlock_t *rwlock);
int pthread_rwlock_wrlock(pthread_rwlock_t *rwlock);
int pthread_rwlock_unlock(pthread_rwlock_t *rwlock);
int pthread_rwlock_tryrdlock(pthread_rwlock_t *rwlock);
int pthread_rwlock_trywrlock(pthread_rwlock_t *rwlock);
When thread1 has finished its initialization it unlocks. The other threads will perform a read-lock and since more read-locks can co-exist they can execute simultaneously. Again: this is valid if you do not perform any write in thread2&3 on the shared resources.
My problem is that I cannot reuse cancelled pthread. Sample code:
#include <pthread.h>
pthread_t alg;
pthread_t stop_alg;
int thread_available;
void *stopAlgorithm() {
while (1) {
sleep(6);
if (thread_available == 1) {
pthread_cancel(alg);
printf("Now it's dead!\n");
thread_available = 0;
}
}
}
void *algorithm() {
while (1) {
printf("I'm here\n");
}
}
int main() {
thread_available = 0;
pthread_create(&stop_alg, NULL, stopAlgorithm, 0);
while (1) {
sleep(1);
if (thread_available == 0) {
sleep(2);
printf("Starting algorithm\n");
pthread_create(&alg, NULL, algorithm, 0);
thread_available = 1;
}
}
}
This sample should create two threads - one will be created at the program beginning and will try to cancel second as soon it starts, second should be rerunned as soon at it was cancelled and say "I'm here". But when algorithm thread cancelled once it doesn't start once again, it says "Starting algorithm" and does nothing, no "I'm here" messages any more. Could you please tell me the way to start cancelled(immediately stopped) thread once again?
UPD: So, thanks to your help I understood what is the problem. When I rerun algorithm thread it throws error 11:"The system lacked the necessary resources to create another thread, or the system-imposed limit on the total number of threads in a process PTHREAD_THREADS_MAX would be exceeded.". Actually I have 5 threads, but only one is cancelled, others stop by pthread_exit. So after algorithm stopped and program went to standby mode I checked status of all threads with pthread_join - all thread show 0(cancelled shows PTHREAD_CANCELED), as far as I can understand this means, that all threads stopped successfully. But one more try to run algorithm throws error 11 again. So I've checked memory usage. In standby mode before algorithm - 10428, during the algorithm, when all threads used - 2026m, in standby mode after algorithm stopped - 2019m. So even if threads stopped they still use memory, pthread_detach didn't help with this. Are there any other ways to clean-up after threads?
Also, sometimes on pthread_cancel my program crashes with "libgcc_s.so.1 must be installed for pthread_cancel to work"
Several points:
First, this is not safe:
int thread_available;
void *stopAlgorithm() {
while (1) {
sleep(6);
if (thread_available == 1) {
pthread_cancel(alg);
printf("Now it's dead!\n");
thread_available = 0;
}
}
}
It's not safe for at least reasons. Firstly, you've not marked thread_available as volatile. This means that the compiler can optimise stopAlgorithm to read the variable once, and never reread it. Secondly, you haven't ensured access to it is atomic, or protected it by a mutex. Either declare it:
volatile sig_atomic_t thread_available;
(or similar), or better, protect it by a mutex.
But for the general case of triggering one thread from another, you are better using a condition variable (and a mutex), using pthread_condwait or pthread_condtimedwait in the listening thread, and pthread_condbroadcast in the triggering thread.
Next, what's the point of the stopAlgorithm thread? All it does is cancel the algorithm thread after an unpredictable amount of time between 0 and 6 seconds? Why not just sent the pthread_cancel from the main thread?
Next, do you care where your algorithm is when it is cancelled? If not, just pthread_cancel it. If so (and anyway, I think it's far nicer), regularly check a flag (either atomic and volatile as above, or protected by a mutex) and pthread_exit if it's set. If your algorithm does big chunks every second or so, then check it then. If it does lots of tiny things, check it (say) every 1,000 operations so taking the mutex doesn't introduce a performance penalty.
Lastly, if you cancel a thread (or if it pthread_exits), the way you start it again is simply to call pthread_create again. It's then a new thread running the same code.
I'm writing a code in which I have two threads running in parallel.
1st is the main thread which started the 2nd thread.
2nd thread is just a simple thread executing empty while loop.
Now I want to pause / suspend the execution of 2nd thread by 1st thread who created it.
And after some time I want to resume the execution of 2nd thread (by issuing some command or function) from where it was paused / suspended.
This question is not about how to use mutexes, but how to suspend a thread.
In Unix specification there is a thread function called pthread_suspend, and another called pthread_resume_np, but for some reason the people who make Linux, FreeBSD, NetBSD and so on have not implemented these functions.
So to understand it, the functions simply are not there. There are workarounds but unfortunately it is just not the same as calling SuspendThread on windows. You have to do all kinds of non-portable stuff to make a thread stop and start using signals.
Stopping and resuming threads is vital for debuggers and garbage collectors. For example, I have seen a version of Wine which is not able to properly implement the "SuspendThread" function. Thus any windows program using it will not work properly.
I thought that it was possible to do it properly using signals based on the fact that JVM uses this technique of signals for the Garbage collector, but I have also just seen some articles online where people are noticing deadlocks and so on with the JVM, sometimes unreproducable.
So to come around to answer the question, you cannot properly suspend and resume threads with Unix unless you have a nice Unix that implements pthread_suspend_np. Otherwise you are stuck with signals.
The big problem with Signals is when you have about five different libraries all linked in to the same program and all trying to use the same signals at the same time. For this reason I believe that you cannot actually use something like ValGrind and for example, the Boehm GC in one program. At least without major coding at the very lowest levels of userspace.
Another answer to this question could be. Do what Linuz Torvalds does to NVidia, flip the finger at him and get him to implement the two most critical parts missing from Linux. First, pthread_suspend, and second, a dirty bit on memory pages so that proper garbage collectors can be implemented. Start a large petition online and keep flipping that finger. Maybe by the time Windows 20 comes out, they will realise that Suspending and resuming threads, and having dirty bits is actually one of the fundamental reasons Windows and Mac are better than Linux, or any Unix that does not implement pthread_suspend and also a dirty bit on virtual pages, like VirtualAlloc does in Windows.
I do not live in hope. Actually for me I spent a number of years planning my future around building stuff for Linux but have abandoned hope as a reliable thing all seems to hinge on the availability of a dirty bit for virtual memory, and for suspending threads cleanly.
As far as I know you can't really just pause some other thread using pthreads. You have to have something in your 2nd thread that checks for times it should be paused using something like a condition variable. This is the standard way to do this sort of thing.
I tried suspending and resuming thread using signals, here is my solution. Please compile and link with -pthread.
Signal SIGUSR1 suspends the thread by calling pause() and SIGUSR2 resumes the thread.
From the man page of pause:
pause() causes the calling process (or thread) to sleep until a
signal is delivered that either terminates the process or causes the
invocation of a
signal-catching function.
#include <stdio.h>
#include <unistd.h>
#include <pthread.h>
#include <signal.h>
// Since I have only 2 threads so using two variables,
// array of bools will be more useful for `n` number of threads.
static int is_th1_ready = 0;
static int is_th2_ready = 0;
static void cb_sig(int signal)
{
switch(signal) {
case SIGUSR1:
pause();
break;
case SIGUSR2:
break;
}
}
static void *thread_job(void *t_id)
{
int i = 0;
struct sigaction act;
pthread_detach(pthread_self());
sigemptyset(&act.sa_mask);
act.sa_flags = 0;
act.sa_handler = cb_sig;
if (sigaction(SIGUSR1, &act, NULL) == -1)
printf("unable to handle siguser1\n");
if (sigaction(SIGUSR2, &act, NULL) == -1)
printf("unable to handle siguser2\n");
if (t_id == (void *)1)
is_th1_ready = 1;
if (t_id == (void *)2)
is_th2_ready = 1;
while (1) {
printf("thread id: %p, counter: %d\n", t_id, i++);
sleep(1);
}
return NULL;
}
int main()
{
int terminate = 0;
int user_input;
pthread_t thread1, thread2;
pthread_create(&thread1, NULL, thread_job, (void *)1);
// Spawned thread2 just to make sure it isn't suspended/paused
// when thread1 received SIGUSR1/SIGUSR2 signal
pthread_create(&thread2, NULL, thread_job, (void *)2);
while (!is_th1_ready && !is_th2_ready);
while (!terminate) {
// to test, I am sensing signals depending on input from STDIN
printf("0: pause thread1, 1: resume thread1, -1: exit\n");
scanf("%d", &user_input);
switch(user_input) {
case -1:
printf("terminating\n");
terminate = 1;
break;
case 0:
printf("raising SIGUSR1 to thread1\n");
pthread_kill(thread1, SIGUSR1);
break;
case 1:
printf("raising SIGUSR2 to thread1\n");
pthread_kill(thread1, SIGUSR2);
break;
}
}
pthread_kill(thread1, SIGKILL);
pthread_kill(thread2, SIGKILL);
return 0;
}
There is no pthread_suspend(), pthread_resume() kind of APIs in POSIX.
Mostly condition variables can be used to control the execution of other threads.
The condition variable mechanism allows threads to suspend execution
and relinquish the processor until some condition is true. A condition
variable must always be associated with a mutex to avoid a race
condition created by one thread preparing to wait and another thread
which may signal the condition before the first thread actually waits
on it resulting in a deadlock.
For more info
Pthreads
Linux Tutorial Posix Threads
If you can use processes instead, you can send job control signals (SIGSTOP / SIGCONT) to the second process. If you still want to share the memory between those processes, you can use SysV shared memory (shmop, shmget, shmctl...).
Even though I haven't tried it myself, it might be possible to use the lower-level clone() syscall to spawn threads that don't share signals. With that, you might be able to send SIGSTOP and SIGCONT to the other thread.
For implementing the pause on a thread, you need to make it wait for some event to happen. Waiting on a spin-lock mutex is CPU cycle wasting. IMHO, this method should not be followed as the CPU cycles could have been used up by other processes/threads.
Wait on a non-blocking descriptor (pipe, socket or some other). Example code for using pipes for inter-thread communication can be seen here
Above solution is useful, if your second thread has more information from multiple sources than just the pause and resume signals. A top-level select/poll/epoll can be used on non-blocking descriptors. You can specify the wait time for select/poll/epoll system calls, and only that much micro-seconds worth of CPU cycles will be wasted.
I mention this solution with forward-thinking that your second thread will have more things or events to handle than just getting paused and resumed. Sorry if it is more detailed than what you asked.
Another simpler approach can be to have a shared boolean variable between these threads.
Main thread is the writer of the variable, 0 - signifies stop. 1 - signifies resume
Second thread only reads the value of the variable. To implement '0' state, use usleep for sime micro-seconds then again check the value. Assuming, few micro-seconds delay is acceptable in your design.
To implement '1' - check the value of the variable after doing certain number of operations.
Otherwise, you can also implement a signal for moving from '1' to '0' state.
You can use mutex to do that, pseudo code would be:
While (true) {
/* pause resume */
lock(my_lock); /* if this is locked by thread1, thread2 will wait until thread1 */
/* unlocks it */
unlock(my_lock); /* unlock so that next iteration thread2 could lock */
/* do actual work here */
}
You can suspend a thread simply by signal
pthread_mutex_t mutex;
static void thread_control_handler(int n, siginfo_t* siginfo, void* sigcontext) {
// wait time out
pthread_mutex_lock(&mutex);
pthread_mutex_unlock(&mutex);
}
// suspend a thread for some time
void thread_suspend(int tid, int time) {
struct sigaction act;
struct sigaction oact;
memset(&act, 0, sizeof(act));
act.sa_sigaction = thread_control_handler;
act.sa_flags = SA_RESTART | SA_SIGINFO | SA_ONSTACK;
sigemptyset(&act.sa_mask);
pthread_mutex_init(&mutex, 0);
if (!sigaction(SIGURG, &act, &oact)) {
pthread_mutex_lock(&mutex);
kill(tid, SIGURG);
sleep(time);
pthread_mutex_unlock(&mutex);
}
}
Not sure if you will like my answer or not. But you can achieve it this way.
If it is a separate process instead of a thread, I have a solution (This might even work for thread, maybe someone can share your thoughts) using signals.
There is no system currently in place to pause or resume the execution of the processes. But surely you can build one.
Steps I would do if I want it in my project:
Register a signal handler for the second process.
Inside the signal handler, wait for a semaphore.
Whenever you want to pause the other process, just send in a signal
that you registered the other process with. The program will go into
sleep state.
When you want to resume the process, you can send a different signal
again. Inside that signal handler, you will check if the semaphore is
locked or not. If it is locked, you will release the semaphore. So
the process 2 will continue its execution.
If you can implement this, please do share your feedack, if it worked for you or not. Thanks.