So i have some code with a double checked locking solution for reading data files in multi threaded (with openmp) application, which looks something like:
logical, dimension(10,10) :: is_data_loaded
is_data_loaded=.false.
! Other code
subroutine load(i,j)
integer,intent(in) :: i,j ! Indexes into array is_data_loaded
if(is_data_loaded(i,j)) return
!$OMP CRITICAL(load data)
if(.not.is_data_loaded(i,j)) then
call load_single_file(i,j)
is_data_loaded(i,j) = .true.
endif
!$OMP END CRITICAL(load_data)
end subroutine
Where I'm worried that if two threads get to the critical region at the same time (with the same i,j index) the second gets blocked by the first one entering the region but once the first finishes the second thread may start executing the critical block before seeing the updated is_data_loaded flag and thus we get into a problem with two threads updating the same data.
So firstly is this an issue with opemp critical blocks? I'm unsure of the semantics and whether the standard says something like "everything must be consistent across threads before the next thread runs in a critical block" or not. And if it is a problem, would just wrapping the read/writes to is_data_loaded in an omp atomic statement be sufficient?
I think the code is wrong, as the threads might indeed not see the updates of is_data_loaded after another thread has set it from the critical region. While the critical region will ensure that the corresponding memory flushes occur, the thread executing the if(is_data_loaded(i,j)) return might not see the update, as this statement might still see outdated data.
I think adding !$omp flush before the if(is_data_loaded(i,j)) return is needed to ensure that all data has been flushed and is_data_loaded(i,j) is loaded with the most recent data.
Related
The following implementation from Wikipedia:
volatile unsigned int produceCount = 0, consumeCount = 0;
TokenType buffer[BUFFER_SIZE];
void producer(void) {
while (1) {
while (produceCount - consumeCount == BUFFER_SIZE)
sched_yield(); // buffer is full
buffer[produceCount % BUFFER_SIZE] = produceToken();
// a memory_barrier should go here, see the explanation above
++produceCount;
}
}
void consumer(void) {
while (1) {
while (produceCount - consumeCount == 0)
sched_yield(); // buffer is empty
consumeToken(buffer[consumeCount % BUFFER_SIZE]);
// a memory_barrier should go here, the explanation above still applies
++consumeCount;
}
}
says that a memory barrier must be used between the line that accesses the buffer and the line that updates the Count variable.
This is done to prevent the CPU from reordering the instructions above the fence along-with that below it. The Count variable shouldn't be incremented before it is used to index into the buffer.
If a fence is not used, won't this kind of reordering violate the correctness of code? The CPU shouldn't perform increment of Count before it is used to index into buffer. Does the CPU not take care of data dependency while instruction reordering?
Thanks
If a fence is not used, won't this kind of reordering violate the correctness of code? The CPU shouldn't perform increment of Count before it is used to index into buffer. Does the CPU not take care of data dependency while instruction reordering?
Good question.
In c++, unless some form of memory barrier is used (atomic, mutex, etc), the compiler assumes that the code is single-threaded. In which case, the as-if rule says that the compiler may emit whatever code it likes, provided that the overall observable effect is 'as if' your code was executed sequentially.
As mentioned in the comments, volatile does not necessarily alter this, being merely an implementation-defined hint that the variable may change between accesses (this is not the same as being modified by another thread).
So if you write multi-threaded code without memory barriers, you get no guarantees that changes to a variable in one thread will even be observed by another thread, because as far as the compiler is concerned that other thread should not be touching the same memory, ever.
What you will actually observe is undefined behaviour.
It seems, that your question is "can incrementing Count and assigment to buffer be reordered without changing code behavior?".
Consider following code tansformation:
int count1 = produceCount++;
buffer[count1 % BUFFER_SIZE] = produceToken();
Notice that code behaves exactly as original one: one read from volatile variable, one write to volatile, read happens before write, state of program is the same. However, other threads will see different picture regarding order of produceCount increment and buffer modifications.
Both compiler and CPU can do that transformation without memory fences, so you need to force those two operations to be in correct order.
If a fence is not used, won't this kind of reordering violate the correctness of code?
Nope. Can you construct any portable code that can tell the difference?
The CPU shouldn't perform increment of Count before it is used to index into buffer. Does the CPU not take care of data dependency while instruction reordering?
Why shouldn't it? What would the payoff be for the costs incurred? Things like write combining and speculative fetching are huge optimizations and disabling them is a non-starter.
If you're thinking that volatile alone should do it, that's simply not true. The volatile keyword has no defined thread synchronization semantics in C or C++. It might happen to work on some platforms and it might happen not to work on others. In Java, volatile does have defined thread synchronization semantics, but they don't include providing ordering for accesses to non-volatiles.
However, memory barriers do have well-defined thread synchronization semantics. We need to make sure that no thread can see that data is available before it sees that data. And we need to make sure that a thread that marks data as able to be overwritten is not seen before the thread is finished with that data.
my code on OpenMP gets very slow when I add the (*pRandomTrial)++; after generating random number. To g_iRandomTrials[32] I store number of rand() calls from each thread. Each thread writes different index of this array, there are no race conditions, results are OK, but this very easy counter makes the program almost 10 times slower then without counter. Is there some keyword I can use in this case? I tried some setups with firstprivate(g_iRandomTrials), but I was never successfull. When I create int counter in Simulate() function and use the pointer only twice on start and on the end of function, code will run probably much faster, but this seems as somewhat ugly solution, as it doesn't do anything about the problem...
int g_iRandomTrials[32];
...
#pragma omp parallel
{
do
{
...
Simulate();
...
}
}
void Simulate(void)
{
...
int id=omp_get_thread_num();
int*pRandomTrial=g_iRandomTrials+id;
...
while (used[index])
{
index=rand()%50;
(*pRandomTrial)++;
}
}
The reason for the slow down is called false sharing. The answer is padding.
In computer science, false sharing is a performance-degrading usage
pattern that can arise in systems with distributed, coherent caches at
the size of the smallest resource block managed by the caching
mechanism. When a system participant attempts to periodically access
data that will never be altered by another party, but that data shares
a cache block with data that is altered, the caching protocol may
force the first participant to reload the whole unit despite a lack of
logical necessity. The caching system is unaware of activity within
this block and forces the first participant to bear the caching system
overhead required by true shared access of a resource.
https://en.wikipedia.org/wiki/False_sharing
CPUs lock their memory in something called cachelines. These tend to be 64-bytes in length. When one core accesses a variable, it locks the entire cacheline and fetches it from memory. Other cores can no longer access it until the lock is released.
The answer is to pad and align your randomTrials in such a way that no value is within 64-bytes of another. Keep in mind that 64-byte value is most common, but there are architectures were this differs.
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 have this POSIX thread:
void subthread(void)
{
while(!quit_thread) {
// do something
...
// don't waste cpu cycles
if(!quit_thread) usleep(500);
}
// free resources
...
// tell main thread we're done
quit_thread = FALSE;
}
Now I want to terminate subthread() from my main thread. I've tried the following:
quit_thread = TRUE;
// wait until subthread() has cleaned its resources
while(quit_thread);
But it does not work! The while() clause does never exit although my subthread clearly sets quit_thread to FALSE after having freed its resources!
If I modify my shutdown code like this:
quit_thread = TRUE;
// wait until subthread() has cleaned its resources
while(quit_thread) usleep(10);
Then everything is working fine! Could someone explain to me why the first solution does not work and why the version with usleep(10) suddenly works? I know that this is not a pretty solution. I could use semaphores/signals for this but I'd like to learn something about multithreading, so I'd like to know why my first solution doesn't work.
Thanks!
Without a memory fence, there is no guarantee that values written in one thread will appear in another. Most of the pthread primitives introduce a barrier, as do several system calls such as usleep. Using a mutex around both the read and write introduces a barrier, and more generally prevents multi-byte values being visible in partially written state.
You also need to separate the idea of asking a thread to stop executing, and reporting that it has stopped, and appear to be using the same variable for both.
What's most likely to be happening is that your compiler is not aware that quit_thread can be changed by another thread (because C doesn't know about threads, at least at the time this question was asked). Because of that, it's optimising the while loop to an infinite loop.
In other words, it looks at this code:
quit_thread = TRUE;
while(quit_thread);
and thinks to itself, "Hah, nothing in that loop can ever change quit_thread to FALSE, so the coder obviously just meant to write while (TRUE);".
When you add the call to usleep, the compiler has another think about it and assumes that the function call may change the global, so it plays it safe and doesn't optimise it.
Normally you would mark the variable as volatile to stop the compiler from optimising it but, in this case, you should use the facilities provided by pthreads and join to the thread after setting the flag to true (and don't have the sub-thread reset it, do that in the main thread after the join if it's necessary). The reason for that is that a join is likely to be more efficient than a continuous loop waiting for a variable change since the thread doing the join will most likely not be executed until the join needs to be done.
In your spinning solution, the joining thread will most likely continue to run and suck up CPU grunt.
In other words, do something like:
Main thread Child thread
------------------- -------------------
fStop = false
start Child Initialise
Do some other stuff while not fStop:
fStop = true Do what you have to do
Finish up and exit
join to Child
Do yet more stuff
And, as an aside, you should technically protect shared variables with mutexes but this is one of the few cases where it's okay, one-way communication where half-changed values of a variable don't matter (false/not-false).
The reason you normally mutex-protect a variable is to stop one thread seeing it in a half-changed state. Let's say you have a two-byte integer for a count of some objects, and it's set to 0x00ff (255).
Let's further say that thread A tries to increment that count but it's not an atomic operation. It changes the top byte to 0x01 but, before it gets a chance to change the bottom byte to 0x00, thread B swoops in and reads it as 0x01ff.
Now that's not going to be very good if thread B want to do something with the last element counted by that value. It should be looking at 0x0100 but will instead try to look at 0x01ff, the effect of which will be wrong, if not catastrophic.
If the count variable were protected by a mutex, thread B wouldn't be looking at it until thread A had finished updating it, hence no problem would occur.
The reason that doesn't matter with one-way booleans is because any half state will also be considered as true or false so, if thread A was halfway between turning 0x0000 into 0x0001 (just the top byte), thread B would still see that as 0x0000 (false) and keep going (until thread A finishes its update next time around).
And if thread A was turning the boolean into 0xffff, the half state of 0xff00 would still be considered true by thread B so it would do its thing before thread A had finished updating the boolean.
Neither of those two possibilities is bad simply because, in both, thread A is in the process of changing the boolean and it will finish eventually. Whether thread B detects it a tiny bit earlier or a tiny bit later doesn't really matter.
The while(quite_thread); is using the value quit_thread was set to on the line before it. Calling a function (usleep) induces the compiler to reload the value on each test.
In any case, this is the wrong way to wait for a thread to complete. Use pthread_join instead.
You're "learning" multhithreading the wrong way. The right way is to learn to use mutexes and condition variables; any other solution will fail under some circumstances.
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.