I am trying to understand how to use fetch and add (atomic operation) in a lock implementation.
I came across this article in Wikipedia, I found it duplicated in at least one other place. The implementation does not make sense and looks to me to have a bug or more in it. Of course I could be missing a subtle point and not really understanding what is being described.
From https://en.wikipedia.org/wiki/Fetch-and-add
<< atomic >>
function FetchAndAdd(address location, int inc) {
int value := *location
*location := value + inc
return value
}
record locktype {
int ticketnumber
int turn
}
procedure LockInit( locktype* lock ) {
lock.ticketnumber := 0
lock.turn := 0
}
procedure Lock( locktype* lock ) {
int myturn := FetchAndIncrement( &lock.ticketnumber ) //must be atomic, since many threads might ask for a lock at the same time
while lock.turn ≠ myturn
skip // spin until lock is acquired
}
procedure UnLock( locktype* lock ) {
FetchAndIncrement( &lock.turn ) //this need not be atomic, since only the possessor of the lock will execute this
}
According to the article they first do LockInit. FetchAndIncrement calls FetchAndAdd with inc set to 1.
If this does not contain a bug I do not understand how it could possibly work.
The first thread to access it will get it:
lock.ticketnumber = 1
lock.turn = 0.
Let's say 5 more accesses to the lock happen before it is released.
lock.ticketnumber = 6
lock.turn = 0
First thread releases the lock.
lock.ticketnumber = 6
lock.turn = 1
Next thread comes in and the status would be
lock.ticketnumber = 7
lock.turn = 1
And the returned value: myturn = 6 (lock.ticketnumber before the faa).
In this case the:
while lock.turn ≠ myturn
can never be true.
Is there a bug in this illustration or am I missing something?
If there is a bug in this implementation what would fix it?
Thanx
Julian
Dang it, I see it now. I found it referring to a general description of the algorithm and then I looked at it more closely.
When a thread calls Lock it spins waiting on the value it got back, for some reason I was thinking it kept calling that function.
When it spins it waits until another thread increments turn and eventually becomes the number of myturn.
Sorry for wasting your time.
https://en.wikipedia.org/wiki/Ticket_lock
Related
I found that there's a macro called PTHRED_MUTEX_ADAPTIVE_NP which is somehow given as a value to a mutex so that the mutex does an adaptive spinning, meaning that it spins in the magnitude of an immediate wakeup through the kernel would last. But how do I utilize this configuration-macro to a thread ?
And as I've developed an improved shared readers-writer lock (it needs only one atomic operation at best in contrast to the three operations given in the Wikipedia-solution) with relative writer-priority (further readers are stalled when there's a writer and the readers before are allowed to proceed) which could also make use of adaptive spinning: how is the number of spinning-cycles calculated ?
I found that there's a macro called PTHRED_MUTEX_ADAPTIVE_NP
Some pthreads implementations provide a macro PTHREAD_MUTEX_ADAPTIVE_NP (note spelling) that is one of the possible values of the kind_np mutex attribute, but neither that attribute nor the macro are standard. It looks like at least BSD and AIX have them, or at least did at one time, but this is not something you should be using in new code.
But how do I utilize this configuration-macro to a thread ?
You don't. Even if you are using a pthreads implementation that supports it, this is the value of a mutex attribute, not a thread attribute. You obtain a mutex with that attribute value by explicitly requesting it when you initialize the mutex. It would look something like this:
pthread_mutexattr_t attr;
pthread_mutex_t mutex;
int rval;
// Return-value checks omitted for brevity and clarity
rval = pthread_mutexattr_init(&attr);
rval = pthread_mutexattr_setkind_np(&attr, PTHREAD_MUTEX_ADAPTIVE_NP);
rval = pthread_mutex_init(&mutex, &attr);
There are other mutex attributes that you can set in analogous ways, which is one of the reasons I wrote this answer. Although you should not be using the kind_np attribute, you can follow this general model for other mutex attributes. There are also thread attributes, which work similarly.
I found the code in the glibc:
That's the "adaptive" mutex locking code of pthread_mutex_lock
in the glibc 2.31:
else if (__builtin_expect (PTHREAD_MUTEX_TYPE (mutex)
== PTHREAD_MUTEX_ADAPTIVE_NP, 1))
{
if (! __is_smp)
goto simple;
if (LLL_MUTEX_TRYLOCK (mutex) != 0)
{
int cnt = 0;
int max_cnt = MIN (max_adaptive_count (),
mutex->__data.__spins * 2 + 10);
do
{
if (cnt++ >= max_cnt)
{
LLL_MUTEX_LOCK (mutex);
break;
}
atomic_spin_nop ();
}
while (LLL_MUTEX_TRYLOCK (mutex) != 0);
mutex->__data.__spins += (cnt - mutex->__data.__spins) / 8;
}
assert (mutex->__data.__owner == 0);
}
So the spin count is doubled up to a maximum plus 10 first (system configurable or 1000 if thre's no configuration) and after the locking the difference between the actual spins and the predefined spins divided by 8 is added to the next spin-count.
I'm new to atomic techniques and try to implement a safe thread version for the follow code:
// say m_cnt is unsigned
void Counter::dec_counter()
{
if(0==m_cnt)
return;
--m_cnt;
if(0 == m_cnt)
{
// Do seomthing
}
}
Every thread that calls dec_counter must decrement it by one and "Do something" should be done only one time - at when the counter is decremented to 0.
After fighting with it, I did the follow code that does it well (I think), but I wonder if this is the way to do it, or is there a better way. Thanks.
// m_cnt is std::atomic<unsigned>
void Counter::dec_counter()
{
// loop until decrement done
unsigned uiExpectedValue;
unsigned uiNewValue;
do
{
uiExpectedValue = m_cnt.load();
// if other thread already decremented it to 0, then do nothing.
if (0 == uiExpectedValue)
return;
uiNewValue = uiExpectedValue - 1;
// at the short time from doing
// uiExpectedValue = m_cnt.load();
// it is possible that another thread had decremented m_cnt, and it won't be equal here to uiExpectedValue,
// thus the loop, to be sure we do a decrement
} while (!m_cnt.compare_exchange_weak(uiExpectedValue, uiNewValue));
// if we are here, that means we did decrement . so if it was to 0, then do something
if (0 == uiNewValue)
{
// do something
}
}
The thing with atomic is that only that one statement is atomic.
If you write
std::atomic<int> i {20}
...
if (!--i)
...
Then just 1 thread will enter the if.
However, if you split up the change and the test, then other threads can get into the gap, and you may get strange results:
std::atomic<int> i {20}
...
--i;
// other thread(s) can modify i just here
if (!i)
...
Of course you can split the condition test for the decrement by using a local variable:
std::atomic<int> i {20}
...
int j=--i;
// other thread(s) can modify i just here
if (!j)
...
All the simple math operations are generally efficiently supported for small atomics in c++
For more complex types and expressions, you need to use the read/modify/write member methods.
These allow you to read the current value, calculate the new value, and then call compare_exchange_strong or compare_exchange_weak say "if the value has not changed, then store my new value, otherwise give me the new current value" a a single atomic operation. You can stick this in a loop and keep recalculating the new value until you are lucky enough that your thread is the only writer. If there are not too many threads trying too often to change the value this is reasonably efficient as well.
I am new to go and I am trying to understand the way channels in goroutines work. To my understanding, the keyword range could be used to iterate over a the values of the channel up until the channel is closed or the buffer runs out; hence, a for range c will repeatedly loops until the buffer runs out.
I have the following simple function that adds value to a channel:
func main() {
c := make(chan int)
go printchannel(c)
for i:=0; i<10 ; i++ {
c <- i
}
}
I have two implementations of printchannel and I am not sure why the behaviour is different.
Implementation 1:
func printchannel(c chan int) {
for range c {
fmt.Println(<-c)
}
}
output: 1 3 5 7
Implementation 2:
func printchannel(c chan int) {
for i:=range c {
fmt.Println(i)
}
}
output: 0 1 2 3 4 5 6 7 8
And I was expecting neither of those outputs!
Wanted output: 0 1 2 3 4 5 6 7 8 9
Shouldnt the main function and the printchannel function run on two threads in parallel, one adding values to the channel and the other reading the values up until the channel is closed? I might be missing some fundamental go/thread concept here and pointers to that would be helpful.
Feedback on this (and my understanding to channels manipulation in goroutines) is greatly appreciated!
Implementation 1. You're reading from the channel twice - range c and <-c are both reading from the channel.
Implementation 2. That's the correct approach. The reason you might not see 9 printed is that two goroutines might run in parallel threads. In that case it might go like this:
main goroutine sends 9 to the channel and blocks until it's read
second goroutine receives 9 from the channel
main goroutine unblocks and exits. That terminates whole program which doesn't give second goroutine a chance to print 9
In case like that you have to synchronize your goroutines. For example, like so
func printchannel(c chan int, wg *sync.WaitGroup) {
for i:=range c {
fmt.Println(i)
}
wg.Done() //notify that we're done here
}
func main() {
c := make(chan int)
wg := sync.WaitGroup{}
wg.Add(1) //increase by one to wait for one goroutine to finish
//very important to do it here and not in the goroutine
//otherwise you get race condition
go printchannel(c, &wg) //very important to pass wg by reference
//sync.WaitGroup is a structure, passing it
//by value would produce incorrect results
for i:=0; i<10 ; i++ {
c <- i
}
close(c) //close the channel to terminate the range loop
wg.Wait() //wait for the goroutine to finish
}
As to goroutines vs threads. You shouldn't confuse them and probably should understand the difference between them. Goroutines are green threads. There're countless blog posts, lectures and stackoverflow answers on that topic.
In implementation 1, range reads into channel once, then again in Println. Hence you're skipping over 2, 4, 6, 8.
In both implementations, once the final i (9) has been sent to goroutine, the program exits. Thus goroutine does not have the time to print out 9. To solve it, use a WaitGroup as has been mentioned in the other answer, or a done channel to avoid semaphore/mutex.
func main() {
c := make(chan int)
done := make(chan bool)
go printchannel(c, done)
for i:=0; i<10 ; i++ {
c <- i
}
close(c)
<- done
}
func printchannel(c chan int, done chan bool) {
for i := range c {
fmt.Println(i)
}
done <- true
}
The reason your first implementation only returns every other number is because you are, in effect "taking" from c twice each time the loop runs: first with range, then again with <-. It just happens that you're not actually binding or using the first value taken off the channel, so all you end up printing is every other one.
An alternative approach to your first implementation would be to not use range at all, e.g.:
func printchannel(c chan int) {
for {
fmt.Println(<-c)
}
}
I could not replicate the behavior of your second implementation, on my machine, but the reason for that is that both of your implementations are racy - they will terminate whenever main ends, regardless of what data may be pending in a channel or however many goroutines may be active.
As a closing note, I'd warn you not to think about goroutines as explicitly being "threads", though they have a similar mental model and interface. In a simple program like this it's not at all unlikely that Go might just do it all using a single OS thread.
Your first loop does not work as you have 2 blocking channel receivers and they do not execute at the same time.
When you call the goroutine the loop starts, and it waits for the first value to be sent to the channel. Effectively think of it as <-c .
When the for loop in the main function runs it sends 0 on the Chan. At this point the range c recieves the value and stops blocking the execution of the loop.
Then it is blocked by the reciever at fmt.println(<-c) . When 1 is sent on the second iteration of the loop in main the recieved at fmt.println(<-c) reads from the channel, allowing fmt.println to execute thus finishing the loop and waiting for a value at the for range c .
Your second implementation of the looping mechanism is the correct one.
The reason it exits before printing to 9 is that after the for loop in main finishes the program goes ahead and completes execution of main.
In Go func main is launched as a goroutine itself while executing. Thus when the for loop in main completes it goes ahead and exits, and as the print is within a parallel goroutine that is closed, it is never executed. There is no time for it to print as there is nothing to block main from completing and exiting the program.
One way to solve this is to use wait groups http://www.golangprograms.com/go-language/concurrency.html
In order to get the expected result you need to have a blocking process running in main that provides enough time or waits for confirmation of the execution of the goroutine before allowing the program to continue.
I want to atomically add 1 to a counter under certain conditions, but I'm not sure if following is correct in a threaded environment:
void UpdateCounterAndLastSessionIfMoreThan60Seconds() const {
auto currentTime = timeProvider->GetCurrentTime();
auto currentLastSession = lastSession.load();
bool shouldIncrement = (currentTime - currentLastSession >= 1 * 60);
if (shouldIncrement) {
auto isUpdate = lastSession.compare_exchange_strong(currentLastSession, currentTime);
if (isUpdate)
changes.fetch_add(1);
}
}
private:
std::shared_ptr<Time> timeProvider;
mutable std::atomic<time_t> lastSession;
mutable std::atomic<uint32_t> changes;
I don't want to increment changes multiple times if 2 threads simultaneously evaluate to shouldIncrement = true and isUpdate = true also (only one should increment changes in that case)
I'm no C++ expert, but it looks to me like you've got a race condition between the evaluation of "isUpdate" and the call to "fetch_add(1)".
So I think the answer to your question "Is this thread safe?" is "No, it is not".
It is at least a bit iffy, as following scenario will show:
First thread 1 does these:
auto currentTime = timeProvider->GetCurrentTime();
auto currentLastSession = lastSession.load();
bool shouldIncrement = (currentTime - currentLastSession >= 1 * 60);
Then thread 2 does the same 3 statements, but so that currentTime is more than it just was for Thread 1.
Then thread 1 continues to update lastSession with it's time, which is less than thread 2's time.
Then thread 2 gets its turn, but fails to update lastSession, because thread 1 changed the value already.
So end result is, lastSession is outdated, because thread 2 failed to update it to the latest value. This might not matter in all cases, situation might be fixed very soon after, but it's one ugly corner which might break some assumptions somewhere, if not with current code then after some later changes.
Another thing to note is, lastSession and chnages are not atomically in sync. Other threads occasionally see changed lastSession with changes counter still not incremeted for that change. Again this is something which might not matter, but it's easy to forget something like this and accidentally code something which assumes they are in sync.
I'm not immediately sure if you can make this 100% safe with just atomics. Wrap it in a mutex instead.
I am working on multithread programming and I am stuck on something.
In my program there are two tasks and two types of robots for carrying out the tasks:
Task 1 requires any two types of robot and
task 2 requires 2 robot1 type and 2 robot2 type.
Total number of robot1 and robot2 and pointers to these two types are given for initialization. Threads share these robots and robots are reserved until a thread is done with them.
Actual task is done in doTask1(robot **) function which takes pointer to a robot pointer as parameter so I need to pass the robots that I reserved. I want to provide concurrency. Obviously if I lock everything it will not be concurrent. robot1 is type of Robot **. Since It is used by all threads before one thread calls doTask or finish it other can overwrite robot1 so it changes things. I know it is because robot1 is shared by all threads. Could you explain how can I solve this problem? I don't want to pass any arguments to thread start routine.
rsc is my struct to hold number of robots and pointers that are given in an initialization function.
void *task1(void *arg)
{
int tid;
tid = *((int *) arg);
cout << "TASK 1 with thread id " << tid << endl;
pthread_mutex_lock (&mutexUpdateRob);
while (rsc->totalResources < 2)
{
pthread_cond_wait(&noResource, &mutexUpdateRob);
}
if (rsc->numOfRobotA > 0 && rsc->numOfRobotB > 0)
{
rsc->numOfRobotA --;
rsc->numOfRobotB--;
robot1[0] = &rsc->robotA[counterA];
robot1[1] = &rsc->robotB[counterB];
counterA ++;
counterB ++;
flag1 = true;
rsc->totalResources -= 2;
}
pthread_mutex_unlock (&mutexUpdateRob);
doTask1(robot1);
pthread_mutex_lock (&mutexUpdateRob);
if(flag1)
{
rsc->numOfRobotA ++;
rsc->numOfRobotB++;
rsc->totalResources += 2;
}
if (totalResource >= 2)
{
pthread_cond_signal(&noResource);
}
pthread_mutex_unlock (&mutexUpdateRob);
pthread_exit(NULL);
}
If robots are global resources, threads should not dispose of them. It should be the duty of the main thread exit (or cleanup) function.
Also, there sould be a way for threads to locate unambiguously the robots, and to lock their use.
The robot1 array seems to store the robots, and it seems to be a global array. However:
its access is not protected by a mutex (pthread_mutex_t), it seems now that you've taken care of that.
Also, the code in task1 is always modifying entries 0 and 1 of this array. If two threads or more execute that code, the entries will be overwritten. I don't think that it is what you want. How will that array be used afterwards?
In fact, why does this array need to be global?
The bottom line is this: as long as this array is shared by threads, they will have problems working concurrently. Think about it this way:
You have two companies using robots to work, but they're using the same truck (robot1) to move the robots around. How are these two companies supposed to function properly, and efficiently with only one truck?