pthread_cond_signal does not wake up the thread that waits - pthreads

Question

I am writing a program which creates a thread that prints 10 numbers. When it prints 5 of them, it waits and it is notifying the main thread and then it continues for the next 5 numbers
This is test.c
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <time.h>
#include <pthread.h>
#include <unistd.h>
int rem = 10;
int count = 5;
pthread_mutex_t mtx;
pthread_cond_t cond1;
pthread_cond_t cond2;
void *f(void *arg)
{
int a;
srand(time(NULL));
while (rem > 0) {
a = rand() % 100;
printf("%d\n",a);
rem--;
count--;
if (count==0) {
printf("time to wake main thread up\n");
pthread_cond_signal(&cond1);
printf("second thread waits\n");
pthread_cond_wait(&cond2, &mtx);
printf("second thread woke up\n");
}
}
pthread_exit(NULL);
}
int main()
{
pthread_mutex_init(&mtx, 0);
pthread_cond_init(&cond1, 0);
pthread_cond_init(&cond2, 0);
pthread_t tids;
pthread_create(&tids, NULL, f, NULL);
while(1) {
if (count != 0) {
printf("main: waiting\n");
pthread_cond_wait(&cond1, &mtx);
printf("5 numbers are printed\n");
printf("main: waking up\n");
pthread_cond_signal(&cond2);
break;
}
pthread_cond_signal(&cond2);
if (rem == 0) break;
}
pthread_join(tids, NULL);
}
The output of the program is:
main: waiting
//5 random numbers
time to wake main thread up
second thread waits
5 numbers are printed
main: waking up
Since I do pthread_cond_signal(&cond2);I thought that the thread will wake up and prints the rest numbers but this is not the case. Any ideas why? Thanks in advance.

Answer 1

Summary
The issues have been summarized in comments, or at least most of them. So as to put an actual answer on record, however:
Pretty much nothing about the program's use of shared variables and synchronization objects is correct. It's behavior is undefined, and the specific manifestation observed is just one of the more likely in a universe of possible behaviors.
Accessing shared variables
If two different threads access (read or write) the same non-atomic object during their runs, and at least one of the accesses is a write, then all accesses must be properly protected by synchronization actions.
There is a variety of these, too large to cover comprehensively in a StackOverflow answer, but among the most common is to use a mutex to guard access. In this approach, a mutex is created in the program and designated for protecting access to one or more shared variables. Each thread that wants to access one of those variables locks the mutex before doing so. At some later point, the thread unlocks the mutex, lest other threads be permanently blocked from locking the mutex themselves.
Example:
pthread_mutex_t mutex; // must be initialized before use
int shared_variable;
// ...
void *thread_one_function(void *data) {
int rval;
// some work ...
rval = pthread_mutex_lock(&mutex);
// check for and handle lock failure ...
shared_variable++;
// ... maybe other work ...
rval = pthread_mutex_unlock(&mutex);
// check for and handle unlock failure ...
// more work ...
}
In your program, the rem and count variables are both shared between threads, and access to them needs to be synchronized. You already have a mutex, and using it to protect accesses to these variables looks like it would be appropriate.
Using condition variables
Condition variables have that name because they are designed to support a specific thread interaction pattern: that one thread wants to proceed past a certain point only if a certain condition, which depends on actions performed by other threads, is satisfied. Such requirements arise fairly frequently. It is possible to implement this via a busy loop, in which the thread repeatedly tests the condition (with proper synchronization) until it is true, but this is wasteful. Condition variables allow such a thread to instead suspend operation until a time when it makes sense to check the condition again.
The correct usage pattern for a condition variable should be viewed as a modification and specialization of the busy loop:
the thread locks a mutex guarding the data on which the condition is to be computed;
the thread tests the condition;
if the condition is satisfied then this procedure ends;
otherwise, the thread waits on a designated condition variable, specifying the (same) mutex;
when the thread resumes after its wait, it loops back to (2).
Example:
pthread_cond_t cv; // must be initialized before use
void *thread_two_function(void *data) {
int rval;
// some work ...
rval = pthread_mutex_lock(&mutex);
// check for and handle lock failure ...
while (shared_variable < 5) {
rval = pthread_cond_wait(&cv, &mutex);
// check for and handle wait failure ...
}
// ... maybe other work ...
rval = pthread_mutex_unlock(&mutex);
// check for and handle unlock failure ...
// more work ...
}
Note that
the procedure terminates (at (3)) with the thread still holding the mutex locked. The thread has an obligation to unlock it, but sometimes it will want to perform other work under protection of that mutex first.
the mutex is automatically released while the thread is waiting on the CV, and reacquired before the thread returns from the wait. This allows other threads the opportunity to access shared variables protected by the mutex.
it is required that the thread calling pthread_cond_wait() have the specified mutex locked. Otherwise, the call provokes undefined behavior.
this pattern relies on threads to signal or broadcast to the CV at appropriate times to notify any then-waiting other threads that they might want to re-evaluate the condition for which they are waiting. That is not modeled in the examples above.
multiple CVs can use the same mutex.
the same CV can be used in multiple places and with different associated conditions. It makes sense to do this when all the conditions involved are affected by the same or related actions by other threads.
condition variables do not store signals. Only threads that are already blocked waiting for the specified CV are affected by a pthread_cond_signal() or pthread_cond_broadcast() call.
Your program
Your program has multiple problems in this area, among them:
Both threads access shared variables rem and count without synchronization, and some of the accesses are writes. The behavior of the whole program is therefore undefined. Among the common manifestations would be that the threads do not observe each other's updates to those variables, though it's also possible that things seem to work as expected. Or anything else.
Both threads call pthread_cond_wait() without holding the mutex locked. The behavior of the whole program is therefore undefined. "Undefined" means "undefined", but it is plausible that the UB would manifest as, for example, one or both threads failing to return from their wait after the CV is signaled.
Neither thread employs the standard pattern for CV usage. There is no clear associated condition for either one, and the threads definitely don't test one. That leaves an implied condition of "this CV has been signaled", but that is unsafe because it cannot be tested before waiting. In particular, it leaves open this possible chain of events:
The main thread blocks waiting on cond1.
The second thread signals cond1.
The main thread runs all the way at least through signaling cond2 before the second thread proceeds to waiting on cond2.
Once (3) occurs, the program cannot avoid deadlock. The main thread breaks from the loop and tries to join the second thread, meanwhile the second thread reaches its pthread_cond_wait() call and blocks awaiting a signal that will never arrive.
That chain of events can happen even if the issues called out in the previous points is corrected, and it could manifest exactly the observable behavior you report.