The documentation of boost::interprocess::file_lock has some cautions on synchronization limitations:
* It's unspecified if a file_lock synchronizes two threads from the same process.
* It's unspecified if a process can use two file_lock objects pointing to the same file.
But then gives the guidance:
* For each file, use a single file_lock object per process.
* Use the same thread to lock and unlock a file.
I then created the following example:
https://wandbox.org/permlink/dO2rRDCHVWBE3pGrhttps://wandbox.org/permlink/3CNfkA3RcMkwbd2v
#include <fstream>
#include <iostream>
#include <cstdlib>
#include
#include
#include
#include <mutex>
#include <thread>
#include <chrono>
int main()
{
std::ofstream("f.txt");
boost::interprocess::file_lock lock("f.txt");
std::mutex mtx;
const size_t max = 100;
size_t t1Count = 0;
std::thread
t1
(
[&lock,&t1Count,&mtx,max]()
{
while(t1Count < max)
{
//std::lock_guardstd::mutex mtxLock(mtx);
boost::interprocess::scoped_lockboost::interprocess::file_lock guard(lock);
std::cout << "Number 1" << std::endl;
t1Count++;
std::this_thread::sleep_for(std::chrono::milliseconds(1));
}
}
);
size_t t2Count = 0;
std::thread
t2
(
[&lock,&t2Count,&mtx,max]()
{
while(t2Count < max)
{
//std::lock_guardstd::mutex mtxLock(mtx);
boost::interprocess::scoped_lockboost::interprocess::file_lock guard(lock);
std::cout << "Number 2" << std::endl;
t2Count++;
std::this_thread::sleep_for(std::chrono::milliseconds(1));
}
}
);
t1.join();
t2.join();
return 0;
}
It only works properly if the std::mutex is used in addition to the file_lock. That was originally my thought when I read "It's unspecified if a file_lock synchronizes two threads from the same process.", but it seemed like I was following the guidance "Use the same thread to lock and unlock a file.". If I haven't messed anything up, I would probably update the documentation to say something like:
"If utilitizing the file_lock from multiple threads within the same process, synchronize access to the file_lock with an intraprocess synchronization mechanism such as a mutex."
Otherwise, is it a bug?
Thanks,
Rob
Rob Conde