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Re: semaphores and handle leaks
Hello again.
Below is C++ code for a fairly simple program which exhibits the
apparent handle leaks I described in my previous posting. I linked
this with the standard STL rather than STLPort and it makes no
difference. This is compiled to an EXE, not a DLL like my real
application.
Again, please feel free to point out if I'm doing somethign wrong.
Thanks
-Morgan McLeod
Software Engineer
National Radio Astronomy Observatory
Charlottesville, Va
#include <stdio.h>
#include <windows.h>
#include <pthread.h>
#include <semaphore.h>
#include <list>
struct listElem {
int num;
sem_t *synchLock;
listElem(int _num, sem_t *_synchLock)
: num(_num),
synchLock(_synchLock)
{}
~listElem()
{}
};
typedef std::list<listElem> semList_t;
semList_t list1;
semList_t list2;
// mutexes to protect the lists:
pthread_mutex_t mutex1;
pthread_mutex_t mutex2;
// flags to tell the threads to stop:
bool shutdownNow;
bool shutdownDone1;
bool shutdownDone2;
// thread 1 processes list1:
void *thread1(void *arg) {
while (true) {
if (shutdownNow) {
shutdownDone1 = true;
pthread_exit(NULL);
}
pthread_mutex_lock(&mutex1);
if (list1.empty())
pthread_mutex_unlock(&mutex1);
else {
// remove the front element from the list:
listElem E = list1.front();
list1.pop_front();
pthread_mutex_unlock(&mutex1);
// save the original semaphore:
sem_t *sem1 = E.synchLock;
// create and initialize a new semaphore.
// substitute it for the original:
sem_t sem2;
E.synchLock = &sem2;
sem_init(E.synchLock, 0, 0);
// put the item in list2 for processing by thread2:
pthread_mutex_lock(&mutex2);
list2.push_back(E);
pthread_mutex_unlock(&mutex2);
// Wait on, then destroy the substitute semaphore:
sem_wait(E.synchLock);
sem_destroy(E.synchLock);
// put back and post on the original semaphore:
E.synchLock = sem1;
sem_post(E.synchLock);
printf("thread1: %d done\n", E.num);
}
Sleep(10);
}
}
// thread2 processes list2:
void *thread2(void *arg) {
while (true) {
if (shutdownNow) {
shutdownDone2 = true;
pthread_exit(NULL);
}
pthread_mutex_lock(&mutex2);
if (list2.empty())
pthread_mutex_unlock(&mutex2);
else {
listElem E = list2.front();
list2.pop_front();
pthread_mutex_unlock(&mutex2);
sem_post(E.synchLock);
printf("thread2: %d done\n", E.num);
}
Sleep(10);
}
}
const int COUNT = 1000;
int main(int, char*[]) {
// Initialize flags:
shutdownNow = shutdownDone1 = shutdownDone2 = false;
// Pause to look at Task Manager. Handles = 8:
Sleep(5000);
pthread_mutex_init(&mutex1, NULL);
pthread_mutex_init(&mutex2, NULL);
sem_t synchLocks[COUNT];
for (int index = 0; index < COUNT; ++index) {
sem_init(&synchLocks[index], 0, 0);
listElem E(index, &synchLocks[index]);
list1.push_back(E);
}
// Handles = 2019. Starts to leak...
pthread_t T1;
pthread_create(&T1, NULL, thread1, NULL);
pthread_t T2;
pthread_create(&T2, NULL, thread2, NULL);
while (!list1.empty() || !list2.empty())
Sleep(10);
// Pause to look at Task Manager. Handles = 2261 (varies):
Sleep(5000);
shutdownNow = true;
while (!shutdownDone1 && !shutdownDone2)
Sleep(10);
for (int index = 0; index < COUNT; ++index)
sem_destroy(&synchLocks[index]);
pthread_mutex_destroy(&mutex1);
pthread_mutex_destroy(&mutex2);
// Pause to look at Task Manager. Handles = 264 (varies):
Sleep(5000);
printf("done\n");
return 0;
}