In our project, we are using semaphore to synchronize between parent and child process. perform the semaphore operation and delete the semaphore for every time synchronize.
After a couple of weeks later semop operation failed with "invalid Argument"
sembuf.sem_num = 0;
sembuf.sem_op = 1;
sembuf.sem_flg = SEM_UNDO;
ret = semop( semid, &sembuf, 1 ); --> invalid arguments couple of weeks later.
semid is proper, I am not seeing any semaphore leak due to the particular process.
In man page i could see the possible reason for invalid argument could see below:
EINVAL The SemaphoreID parameter is not a valid semaphore identifier.
EINVAL The number of individual semaphores for which the calling process requests a SEM_UNDO flag would exceed the limit.
I am suspecting the point 2.
It is multithreaded process. So Howmany semaphore can be used at a time for a procees ?
What is the limit of "SEM_UNDO" ?
1) how to make sure that EINVAL is due to "SEM_UNDO" flag exceed limit. ?.
2) how to avoid the exceed limit ?
3) Is there any kernel variable needs to be configured for this issue (in AIX most of the kernals are dynamic)?
Thanks in advance for helping this
As I've already written in comment, another thread/process of yours has deleted the semaphore you were trying to use. You are not supposed to create and destroy the semaphore at every usage.
Off: I admit SYSV-semaphores have a little quirk, i.e. they are created with value=0 which means locked (or reserved, acquired), instead of value=1 (meaning free (or available)). This can be fixed with a simple trick, example program: http://lzsiga.users.sourceforge.net/sysv_sem_test.c
Related
I need to use task_scheduler_init to limit the number of threads to a number under of cores, however TBB ignores the number & always uses the number of cores (8 in this case).
This doesn't look like normal behavior to me. Please note that it is not a possibility for me to use a different version of TBB.
Snippet:
task_scheduler_init scheduler(nb_thread);
tbb::parallel_for(
tbb::blocked_range<size_t>(0, size),
[&](const tbb::blocked_range<size_t>& subrange) {
int tid = syscall(SYS_gettid);
dragon_draw_raw(
subrange.begin(),
subrange.end(),
dragon,
dragon_width,
dragon_height,
limits,
id
);
});
If the task_scheduler_init instance was not the first call to TBB on this thread, it explains why you see such a behavior. [Almost] any call to TBB initiates auto-initialization and the number of worker threads is fixed at this time, any further call to task_scheduler_init means basically nothing except additional reference to the internal task scheduler object.
If you are in control of the entire application and can make the call to task_scheduler_init object constructor the first call to TBB, it will fix your problem. But if you are writing a component or a library, look at task_arena feature, it allows to control concurrency limits regardless of the current settings.
Start with x = 0. Note there are no memory barriers in any of the code below.
volatile int x = 0
Thread 1:
while (x == 0) {}
print "Saw non-zer0"
while (x != 0) {}
print "Saw zero again!"
Thread 2:
x = 1
Is it ever possible to see the second message, "Saw zero again!", on any (real) CPU? What about on x86_64?
Similarly, in this code:
volatile int x = 0.
Thread 1:
while (x == 0) {}
x = 2
Thread 2:
x = 1
Is the final value of x guaranteed to be 2, or could the CPU caches update main memory in some arbitrary order, so that although x = 1 gets into a CPU's cache where thread 1 can see it, then thread 1 gets moved to a different cpu where it writes x = 2 to that cpu's cache, and the x = 2 gets written back to main memory before x = 1.
Yes, it's entirely possible. The compiler could, for example, have just written x to memory but still have the value in a register. One while loop could check memory while the other checks the register.
It doesn't happen due to CPU caches because cache coherency hardware logic makes the caches invisible on all CPUs you are likely to actually use.
Theoretically, the write race you talk about could happen due to posted write buffering and read prefetching. Miraculous tricks were used to make this impossible on x86 CPUs to avoid breaking legacy code. But you shouldn't expect future processors to do this.
Leaving aside for a second tricks done by the compiler (even ones allowed by language standards), I believe you're asking how the micro-architecture could behave in such scenario. Keep in mind that the code would most likely expand into a busy wait loop of cmp [x] + jz or something similar, which hides a load inside it. This means that [x] is likely to live in the cache of the core running thread 1.
At some point, thread 2 would come and perform the store. If it resides on a different core, the line would first be invalidated completely from the first core. If these are 2 threads running on the same physical core - the store would immediately affect all chronologically younger loads.
Now, the most likely thing to happen on a modern out-of-order machine is that all the loads in the pipeline at this point would be different iterations of the same first loop (since any branch predictor facing so many repetitive "taken" resolution is likely to assume the branch will continue being taken, until proven wrong), so what would happen is that the first load to encounter the new value modified by the other thread will cause the matching branch to simply flush the entire pipe from all younger operations, without the 2nd loop ever having a chance to execute.
However, it's possible that for some reason you did get to the 2nd loop (let's say the predictor issue a not-taken prediction just at the right moment when the loop condition check saw the new value) - in this case, the question boils down to this scenario:
Time -->
----------------------------------------------------------------
thread 1
cmp [x],0 execute
je ... execute (not taken)
...
cmp [x],0 execute
jne ... execute (not taken)
Can_We_Get_Here:
...
thread2
store [x],1 execute
In other words, given that most modern CPUs may execute instructions out of order, can a younger load be evaluated before an older one to the same address, allowing the store (from another thread) to change the value so it may be observed inconsistently by the loads.
My guess is that the above timeline is quite possible given the nature of out-of-order execution engines today, as they simply arbitrate and perform whatever operation is ready. However, on most x86 implementations there are safeguards to protect against such a scenario, since the memory ordering rules strictly say -
8.2.3.2 Neither Loads Nor Stores Are Reordered with Like Operations
Such mechanisms may detect this scenario and flush the machine to prevent the stale/wrong values becoming visible. So The answer is - no, it should not be possible, unless of course the software or the compiler change the nature of the code to prevent the hardware from noticing the relation. Then again, memory ordering rules are sometimes flaky, and i'm not sure all x86 manufacturers adhere to the exact same wording, but this is a pretty fundamental example of consistency, so i'd be very surprised if one of them missed it.
The answer seems to be, "this is exactly the job of the CPU cache coherency." x86 processors implement the MESI protocol, which guarantee that the second thread can't see the new value then the old.
I'm a little confused by the Linux API sem_unlink(), mainly when or why to call it. I've used semaphores in Windows for many years. In Windows once you close the last handle of a named semaphore the system removes the underlying kernel object. But it appears in Linux you, the developer, needs to remove the kernel object by calling sem_unlink(). If you don't the kernel object persists in the /dev/shm folder.
The problem I'm running into, if process A calls sem_unlink() while process B has the semaphore locked, it immediately destroys the semaphore and now process B is no longer "protected" by the semaphore when/if process C comes along. What's more, the man page is confusing at best:
"The semaphore name is removed immediately. The semaphore is destroyed once all other processes that have the semaphore open close it."
How can it destroy the object immediately if it has to wait for other processes to close the semaphore?
Clearly I don't understand the proper use of semaphore objects on Linux. Thanks for any help. Below is some sample code I'm using to test this.
int main(void)
{
sem_t *pSemaphore = sem_open("/MyName", O_CREAT, S_IRUSR | S_IWUSR, 1);
if(pSemaphore != SEM_FAILED)
{
if(sem_wait(pSemaphore) == 0)
{
// Perform "protected" operations here
sem_post(pSemaphore);
}
sem_close(pSemaphore);
sem_unlink("/MyName");
}
return 0;
}
Response to your questions:
In comparison to the semaphore behavior for windows you
describe, POSIX semaphores are Kernel persistent. Meaning that the
semaphore retains it's value even if no process has the semaphore
opened. (the semaphore's reference count would be 0)
If process A calls sem_unlink() while process B has the semaphore
locked. This means the semaphore's reference count is not 0 and will
not be destructed.
Basic operation of sem_close vs sem_unlink, I think will help overall understanding:
sem_close: close's a semaphore, this also done when a process exits. the semaphore still remains in the system.
sem_unlink: will be removed from the system only when the reference count reaches 0 (that is after all processes that have it open, call sem_close or are exited).
References:
Book - Unix Networking Programming-Interprocess Communication by W.Richard Stevens, vol 2, ch10
The sem_unlink() function removes the semaphore identified by name and marks
the semaphore to be destroyed once all processes cease using it (this may mean
immediately, if all processes that had the semaphore open have already closed it).
This question is about the same program I previously asked about. To recap, I have a program with a loop structure like this:
for (int i1 = 0; i1 < N; i1++)
for (int i2 = 0; i2 < N; i2++)
for (int i3 = 0; i3 < N; i3++)
for (int i4 = 0; i4 < N; i4++)
histogram[bin_index(i1, i2, i3, i4)] += 1;
bin_index is a completely deterministic function of its arguments which, for purposes of this question, does not use or change any shared state - in other words, it is manifestly reentrant.
I first wrote this program to use a single thread. Then I converted it to use multiple threads, such that thread n runs all iterations of the outer loop where i1 % nthreads == n. So the function that runs in each thread looks like
for (int i1 = n; i1 < N; i1 += nthreads)
for (int i2 = 0; i2 < N; i2++)
for (int i3 = 0; i3 < N; i3++)
for (int i4 = 0; i4 < N; i4++)
thread_local_histogram[bin_index(i1, i2, i3, i4)] += 1;
and all the thread_local_histograms are added up in the main thread at the end.
Here's the strange thing: when I run the program with just 1 thread for some particular size of the calculation, it takes about 6 seconds. When I run it with 2 or 3 threads, doing exactly the same calculation, it takes about 9 seconds. Why is that? I would expect that using 2 threads would be faster than 1 thread since I have a dual-core CPU. The program does not use any mutexes or other synchronization primitives so two threads should be able to run in parallel.
For reference: typical output from time (this is on Linux) for one thread:
real 0m5.968s
user 0m5.856s
sys 0m0.064s
and two threads:
real 0m9.128s
user 0m10.129s
sys 0m6.576s
The code is at http://static.ellipsix.net/ext-tmp/distintegral.ccs
P.S. I know there are libraries designed for exactly this kind of thing that probably could have better performance, but that's what my last question was about so I don't need to hear those suggestions again. (Plus I wanted to use pthreads as a learning experience.)
To avoid further comments on this: When I wrote my reply, the questioner hasn't posted a link to his source yet, so I could not tailor my reply to his specific issues. I was only answering the general question what "can" cause such an issue, I never said that this will necessarily apply to his case. When he posted a link to his source, I wrote another reply, that is exactly only focusing on his very issue (which is caused by the use of the random() function as I explained in my other reply). However, since the question of this post is still "What can make a program run slower when using more threads?" and not "What makes my very specific application run slower?", I've seen no need to change my rather general reply either (general question -> general response, specific question -> specific response).
1) Cache Poisoning
All threads access the same array, which is a block of memory. Each core has its own cache to speed up memory access. Since they don't just read from the array but also change the content, the content is changed actually in the cache only, not in real memory (at least not immediately). The problem is that the other thread on the other core may have overlapping parts of memory cached. If now core 1 changes the value in the cache, it must tell core 2 that this value has just changed. It does so by invalidating the cache content on core 2 and core 2 needs to re-read the data from memory, which slows processing down. Cache poisoning can only happen on multi-core or multi-CPU machines. If you just have one CPU with one core this is no problem. So to find out if that is your issue or not, just disable one core (most OSes will allow you to do that) and repeat the test. If it is now almost equally fast, that was your problem.
2) Preventing Memory Bursts
Memory is read fastest if read sequentially in bursts, just like when files are read from HD. Addressing a certain point in memory is actually awfully slow (just like the "seek time" on a HD), even if your PC has the best memory on the market. However, once this point has been addressed, sequential reads are fast. The first addressing goes by sending a row index and a column index and always having waiting times in between before the first data can be accessed. Once this data is there, the CPU starts bursting. While the data is still on the way it sends already the request for the next burst. As long as it is keeping up the burst (by always sending "Next line please" requests), the RAM will continue to pump out data as fast as it can (and this is actually quite fast!). Bursting only works if data is read sequentially and only if the memory addresses grow upwards (AFAIK you cannot burst from high to low addresses). If now two threads run at the same time and both keep reading/writing memory, however both from completely different memory addresses, each time thread 2 needs to read/write data, it must interrupt a possible burst of thread 1 and the other way round. This issue gets worse if you have even more threads and this issue is also an issue on a system that has only one single-core CPU.
BTW running more threads than you have cores will never make your process any faster (as you mentioned 3 threads), it will rather slow it down (thread context switches have side effects that reduce processing throughput) - that is unlike you run more threads because some threads are sleeping or blocking on certain events and thus cannot actively process any data. In that case it may make sense to run more threads than you have cores.
Everything I said so far in my other reply holds still true on general, as your question was what "can"... however now that I've seen your actual code, my first bet would be that your usage of the random() function slows everything down. Why?
See, random keeps a global variable in memory that stores the last random value calculated there. Each time you call random() (and you are calling it twice within a single function) it reads the value of this global variable, performs a calculation (that is not so fast; random() alone is a slow function) and writes the result back there before returning it. This global variable is not per thread, it is shared among all threads. So what I wrote regarding cache poisoning applies here all the time (even if you avoided it for the array by having separated arrays per thread; this was very clever of you!). This value is constantly invalidated in the cache of either core and must be re-fetched from memory. However if you only have a single thread, nothing like that happens, this variable never leaves cache after it has been initially read, since it's permanently accessed again and again and again.
Further to make things even worse, glibc has a thread-safe version of random() - I just verified that by looking at the source. While this seems to be a good idea in practice, it means that each random() call will cause a mutex to be locked, memory to be accessed, and a mutex to be unlocked. Thus two threads calling random exactly the same moment will cause one thread to be blocked for a couple of CPU cycles. This is implementation specific, though, as AFAIK it is not required that random() is thread safe. Most standard lib functions are not required to be thread-safe, since the C standard is not even aware of the concept of threads in the first place. When they are not calling it the same moment, the mutex will have no influence on speed (as even a single threaded app must lock/unlock the mutex), but then cache poisoning will apply again.
You could pre-build an array with random numbers for every thread, containing as many random number as each thread needs. Create it in the main thread before spawning the threads and add a reference to it to the structure pointer you hand over to every thread. Then get the random numbers from there.
Or just implement your own random number generator if you don't need the "best" random numbers on the planet, that works with per-thread memory for holding its state - that one might be even faster than the system's built-in generator.
If a Linux only solution works for you, you can use random_r. It allows you to pass the state with every call. Just use a unique state object per thread. However this function is a glibc extension, it is most likely not supported by other platforms (neither part of the C standards nor of the POSIX standards AFAIK - this function does not exist on Mac OS X for example, it may neither exist in Solaris or FreeBSD).
Creating an own random number generator is actually not that hard. If you need real random numbers, you shouldn't use random() in the first place. Random only creates pseudo-random numbers (numbers that look random, but are predictable if you know the generator's internal state). Here's the code for one that produces good uint32 random numbers:
static uint32_t getRandom(uint32_t * m_z, uint32_t * m_w)
{
*m_z = 36969 * (*m_z & 65535) + (*m_z >> 16);
*m_w = 18000 * (*m_w & 65535) + (*m_w >> 16);
return (*m_z << 16) + *m_w;
}
It's important to "seed" m_z and m_w in a proper way somehow, otherwise the results are not random at all. The seed value itself should already be random, but here you could use the system random number generator.
uint32_t m_z = random();
uint32_t m_w = random();
uint32_t nextRandom;
for (...) {
nextRandom = getRandom(&m_z, &m_w);
// ...
}
This way every thread only needs to call random() twice and then uses your own generator. BTW, if you need double randoms (that are between 0 and 1), the function above can be easily wrapped for that:
static double getRandomDouble(uint32_t * m_z, uint32_t * m_w)
{
// The magic number below is 1/(2^32 + 2).
// The result is strictly between 0 and 1.
return (getRandom(m_z, m_w) + 1) * 2.328306435454494e-10;
}
Try to make this change in your code and let me know how the benchmark results are :-)
You are seeing cache line bouncing. I'm really surprised that you don't get wrong results, due to race conditions on the histogram buckets.
One possibility is that the time taken to create the threads exceeds the savings gained by using threads. I would think that N is not very large, if the elapsed time is only 6 seconds for a O(n^4) operation.
There's also no guarantee that multiple threads will run on different cores or CPUs. I'm not sure what the default thread affinity is with Linux - it may be that both threads run on a single core which would negate the benefits of a CPU-intensive piece of code such as this.
This article details default thread affinity and how to change your code to ensure threads run on specific cores.
Even though threads don't access the same elements of the array at the same, the whole array may sit in a few memory pages. When one core/processor writes to that page, it has to invalidate its cache for all other processors.
Avoid having many threads working over the same memory space. Allocate separate data for each thread to work upon, then join them together when the calculation finishes.
Off the top of my head:
Context switches
Resource contention
CPU contention (if they aren't getting split to multiple CPUs).
Cache thrashing
David,
Are you sure you run a kernel that supports multiple processors? If only one processor is utilized in your system, spawning additional CPU-intensive threads will slow down your program.
And, are you sure support for threads in your system actually utilizes multiple processors? Does top, for example, show that both cores in your processor utilized when you run your program?
I understand about race conditions and how with multiple threads accessing the same variable, updates made by one can be ignored and overwritten by others, but what if each thread is writing the same value (not different values) to the same variable; can even this cause problems? Could this code:
GlobalVar.property = 11;
(assuming that property will never be assigned anything other than 11), cause problems if multiple threads execute it at the same time?
The problem comes when you read that state back, and do something about it. Writing is a red herring - it is true that as long as this is a single word most environments guarantee the write will be atomic, but that doesn't mean that a larger piece of code that includes this fragment is thread-safe. Firstly, presumably your global variable contained a different value to begin with - otherwise if you know it's always the same, why is it a variable? Second, presumably you eventually read this value back again?
The issue is that presumably, you are writing to this bit of shared state for a reason - to signal that something has occurred? This is where it falls down: when you have no locking constructs, there is no implied order of memory accesses at all. It's hard to point to what's wrong here because your example doesn't actually contain the use of the variable, so here's a trivialish example in neutral C-like syntax:
int x = 0, y = 0;
//thread A does:
x = 1;
y = 2;
if (y == 2)
print(x);
//thread B does, at the same time:
if (y == 2)
print(x);
Thread A will always print 1, but it's completely valid for thread B to print 0. The order of operations in thread A is only required to be observable from code executing in thread A - thread B is allowed to see any combination of the state. The writes to x and y may not actually happen in order.
This can happen even on single-processor systems, where most people do not expect this kind of reordering - your compiler may reorder it for you. On SMP even if the compiler doesn't reorder things, the memory writes may be reordered between the caches of the separate processors.
If that doesn't seem to answer it for you, include more detail of your example in the question. Without the use of the variable it's impossible to definitively say whether such a usage is safe or not.
It depends on the work actually done by that statement. There can still be some cases where Something Bad happens - for example, if a C++ class has overloaded the = operator, and does anything nontrivial within that statement.
I have accidentally written code that did something like this with POD types (builtin primitive types), and it worked fine -- however, it's definitely not good practice, and I'm not confident that it's dependable.
Why not just lock the memory around this variable when you use it? In fact, if you somehow "know" this is the only write statement that can occur at some point in your code, why not just use the value 11 directly, instead of writing it to a shared variable?
(edit: I guess it's better to use a constant name instead of the magic number 11 directly in the code, btw.)
If you're using this to figure out when at least one thread has reached this statement, you could use a semaphore that starts at 1, and is decremented by the first thread that hits it.
I would expect the result to be undetermined. As in it would vary from compiler to complier, langauge to language and OS to OS etc. So no, it is not safe
WHy would you want to do this though - adding in a line to obtain a mutex lock is only one or two lines of code (in most languages), and would remove any possibility of problem. If this is going to be two expensive then you need to find an alternate way of solving the problem
In General, this is not considered a safe thing to do unless your system provides for atomic operation (operations that are guaranteed to be executed in a single cycle).
The reason is that while the "C" statement looks simple, often there are a number of underlying assembly operations taking place.
Depending on your OS, there are a few things you could do:
Take a mutual exclusion semaphore (mutex) to protect access
in some OS, you can temporarily disable preemption, which guarantees your thread will not swap out.
Some OS provide a writer or reader semaphore which is more performant than a plain old mutex.
Here's my take on the question.
You have two or more threads running that write to a variable...like a status flag or something, where you only want to know if one or more of them was true. Then in another part of the code (after the threads complete) you want to check and see if at least on thread set that status... for example
bool flag = false
threadContainer tc
threadInputs inputs
check(input)
{
...do stuff to input
if(success)
flag = true
}
start multiple threads
foreach(i in inputs)
t = startthread(check, i)
tc.add(t) // Keep track of all the threads started
foreach(t in tc)
t.join( ) // Wait until each thread is done
if(flag)
print "One of the threads were successful"
else
print "None of the threads were successful"
I believe the above code would be OK, assuming you're fine with not knowing which thread set the status to true, and you can wait for all the multi-threaded stuff to finish before reading that flag. I could be wrong though.
If the operation is atomic, you should be able to get by just fine. But I wouldn't do that in practice. It is better just to acquire a lock on the object and write the value.
Assuming that property will never be assigned anything other than 11, then I don't see a reason for assigment in the first place. Just make it a constant then.
Assigment only makes sense when you intend to change the value unless the act of assigment itself has other side effects - like volatile writes have memory visibility side-effects in Java. And if you change state shared between multiple threads, then you need to synchronize or otherwise "handle" the problem of concurrency.
When you assign a value, without proper synchronization, to some state shared between multiple threads, then there's no guarantees for when the other threads will see that change. And no visibility guarantees means that it it possible that the other threads will never see the assignt.
Compilers, JITs, CPU caches. They're all trying to make your code run as fast as possible, and if you don't make any explicit requirements for memory visibility, then they will take advantage of that. If not on your machine, then somebody elses.