Leave a binary semaphore locked - semaphore

If a process is used to initial a variable and another process uses that variable
The processes are synchronized like this :
Binary_semaphore s=0;
p1() {
int x =1 ;
semaphore_signal_B (s);
}
p2() {
semaphore_wait_B (s);
int y =x+1 ;
semaphore_signal_B (s); //<---- this line
}
Is the last line in p2 necessary ? knowing that p1 ,p2 and s only gonna be used once .

I think it's wrong to signal the semaphore because it's not doing you any good and it's always better the return the variables to there initial state, which is zero for this case if I'm not wrong.

Related

Do I need a QMutex for variable that is accessed by single statement?

In this document, a QMutex is used to protect "number" from being modified by multiple threads at same time.
I have a code in which a thread is instructed to do different work according to a flag set by another thread.
//In thread1
if(flag)
dowork1;
else
dowork2;
//In thread2
void setflag(bool f)
{
flag=f;
}
I want to know if a QMutex is needed to protect flag, i.e.,
//In thread1
mutex.lock();
if(flag)
{
mutex.unlock();
dowork1;
}
else
{
mutex.unlock();
dowork2;
}
//In thread2
void setflag(bool f)
{
mutex.lock();
flag=f;
mutex.unlock();
}
The code is different from the document in that flag is accessed(read/written) by single statement in both threads, and only one thread modifies the value of flag.
PS:
I always see the example in multi-thread programming tutorials that one thread does "count++", the other thread does "count--", and the tutorials say you should use a Mutex to protect the variable "count". I cannot get the point of using a mutex. Does it mean the execution of single statement "count++" or "count--" can be interrupted in the middle and produce unexpected result? What unexpected results can be gotten?
Does it mean the execution of single statement "count++" or "count--"
can be interrupted in the middle and produce unexpected result? What
unexpected results can be gotten?
Just answering to this part: Yes, the execution can be interrupted in the middle of a statement.
Let's imagine a simple case:
class A {
void foo(){
++a;
}
int a = 0;
};
The single statement ++a is translated in assembly to
mov eax, DWORD PTR [rdi]
add eax, 1
mov DWORD PTR [rdi], eax
which can be seen as
eax = a;
eax += 1;
a = eax;
If foo() is called on the same instance of A in 2 different threads (be it on a single core, or multiple cores) you cannot predict what will be the result of the program.
It can behave nicely:
thread 1 > eax = a // eax in thread 1 is equal to 0
thread 1 > eax += 1 // eax in thread 1 is equal to 1
thread 1 > a = eax // a is set to 1
thread 2 > eax = a // eax in thread 2 is equal to 1
thread 2 > eax += 1 // eax in thread 2 is equal to 2
thread 2 > a = eax // a is set to 2
or not:
thread 1 > eax = a // eax in thread 1 is equal to 0
thread 2 > eax = a // eax in thread 2 is equal to 0
thread 2 > eax += 1 // eax in thread 2 is equal to 1
thread 2 > a = eax // a is set to 1
thread 1 > eax += 1 // eax in thread 1 is equal to 1
thread 1 > a = eax // a is set to 1
In a well defined program, N calls to foo() should result in a == N.
But calling foo() on the same instance of A from multiple threads creates undefined behavior. There is no way to know the value of a after N calls to foo().
It will depend on how you compiled your program, what optimization flags were used, which compiler was used, what was the load of your CPU, the number of core of your CPU,...
NB
class A {
public:
bool check() const { return a == b; }
int get_a() const { return a; }
int get_b() const { return b; }
void foo(){
++a;
++b;
}
private:
int a = 0;
int b = 0;
};
Now we have a class that, for an external observer, keeps a and b equal at all time.
The optimizer could optimize this class into:
class A {
public:
bool check() const { return true; }
int get_a() const { return a; }
int get_b() const { return b; }
void foo(){
++a;
++b;
}
private:
int a = 0;
int b = 0;
};
because it does not change the observable behavior of the program.
However if you invoke undefined behavior by calling foo() on the same instance of A from multiple threads, you could end up if a = 3, b = 2 and check() still returning true. Your code has lost its meaning, the program is not doing what it is supposed to and can be doing about anything.
From here you can imagine more complex cases, like if A manages network connections, you can end up sending the data for client #10 to client #6. If your program is running in a factory, you can end up activating the wrong tool.
If you want the definition of undefined behavior you can look here : https://en.cppreference.com/w/cpp/language/ub
and in the C++ standard
For a better understanding of UB you can look for CppCon talks on the topic.
For any standard object (including bool) that is accessed from multiple threads, where at least one of the threads may modify the object's state, you need to protect access to that object using a mutex, otherwise you will invoke undefined behavior.
As a practical matter, for a bool that undefined behavior probably won't come in the form of a crash, but more likely in the form of thread B sometimes not "seeing" changes made to the bool by thread A, due to caching and/or optimization issues (e.g. the optimizer "knows" that the bool can't change during a function call, so it doesn't bother checking it more than once)
If you don't want to guard your accesses with a mutex, the other option is to change flag from a bool to a std::atomic<bool>; the std::atomic<bool> type has exactly the semantics you are looking for, i.e. it can be read and/or written from any thread without invoking undefined behavior.
Look here for an explanation: Do I have to use atomic<bool> for "exit" bool variable?
To synchronize access to flag you can make it a std::atomic<bool>.
Or you can use a QReadWriteLock together with a QReadLocker and a QWriteLocker. Compared to using a QMutex this gives you the advantage that you do not need to care about the call to QMutex::unlock() if you use exceptions or early return statements.
Alternatively you can use a QMutexLocker if the QReadWriteLock does not match your use case.
QReadWriteLock lock;
...
//In thread1
{
QReadLocker readLocker(&lock);
if(flag)
dowork1;
else
dowork2;
}
...
//In thread2
void setflag(bool f)
{
QWriteLocker writeLocker(&lock);
flag=f;
}
Keeping your program expressing its intent (ie. accessing shared vars under locks) is a big win for program maintenance and clarity. You need to have some pretty good reasons to abandon that clarity for obscure approaches like the atomics and devising consistent race conditions.
Good reasons include you have measured your program spending too much time toggling the mutex. In any decent implementation, the difference between a non-contested mutex and an atomic is minute -- the mutex lock and unlock typical employ an optimistic compare-and-swap, returning quickly. If your vendor doesn't provide a decent implementation, you might bring that up with them.
In your example, dowork1 and dowork2 are invoked with the mutex locked; so the mutex isn't just protecting flag, but also serializing these functions. If that is just an artifact of how you posed the question, then race conditions (variants of atomics travesty) are less scary.
In your PS (dup of comment above):
Yes, count++ is best thought of as:
mov $_count, %r1
ld (%r1), %r0
add $1, %r0, %r2
st %r2,(%r1)
Even machines with natural atomic inc (x86,68k,370,dinosaurs) instructions might not be used consistently by the compiler.
So, if two threads do count--; and count++; at close to the same time, the result could be -1, 0, 1. (ignoring the language weenies that say your house might burn down).
barriers:
if CPU0 executes:
store $1 to b
store $2 to c
and CPU1 executes:
load barrier -- discard speculatively read values.
load b to r0
load c to r1
Then CPU1 could read r0,r1 as: (0,0), (1,0), (1,2), (0,2).
This is because the observable order of the memory writes is weak; the processor may make them visible in an arbitrary fashion.
So, we change CPU0 to execute:
store $1 to b
store barrier -- stop storing until all previous stores are visible
store $2 to c
Then, if CPU1 saw that r1 (c) was 2, then r0 (b) has to be 1. The store barrier enforces that.
For me, its seems to be more handy to use a mutex here.
In general not using mutex when sharing references could lead to
problems.
The only downside of using mutex here seems to be, that you will slightly decrease the performance, because your threads have to wait for each other.
What kind of errors could happen ?
Like somebody in the comments said its a different situation if
your share fundamental datatype e.g. int, bool, float
or a object references. I added some qt code
example, which emphases 2 possible problems during NOT using mutex. The problem #3 is a fundamental one and pretty well described in details by Benjamin T and his nice answer.
Blockquote
main.cpp
#include <QCoreApplication>
#include <QThread>
#include <QtDebug>
#include <QTimer>
#include "countingthread.h"
int main(int argc, char *argv[])
{
QCoreApplication a(argc, argv);
int amountThread = 3;
int counter = 0;
QString *s = new QString("foo");
QMutex *mutex = new QMutex();
//we construct a lot of thread
QList<CountingThread*> threadList;
//we create all threads
for(int i=0;i<amountThread;i++)
{
CountingThread *t = new CountingThread();
#ifdef TEST_ATOMIC_VAR_SHARE
t->addCounterdRef(&counter);
#endif
#ifdef TEST_OBJECT_VAR_SHARE
t->addStringRef(s);
//we add a mutex, which is shared to read read write
//just used with TEST_OBJECT_SHARE_FIX define uncommented
t->addMutexRef(mutex);
#endif
//t->moveToThread(t);
threadList.append(t);
}
//we start all with low prio, otherwise we produce something like a fork bomb
for(int i=0;i<amountThread;i++)
threadList.at(i)->start(QThread::Priority::LowPriority);
return a.exec();
}
countingthread.h
#ifndef COUNTINGTHREAD_H
#define COUNTINGTHREAD_H
#include <QThread>
#include <QtDebug>
#include <QTimer>
#include <QMutex>
//atomic var is shared
//#define TEST_ATOMIC_VAR_SHARE
//more complex object var is shared
#define TEST_OBJECT_VAR_SHARE
// we add the fix
#define TEST_OBJECT_SHARE_FIX
class CountingThread : public QThread
{
Q_OBJECT
int *m_counter;
QString *m_string;
QMutex *m_locker;
public :
void addCounterdRef(int *r);
void addStringRef(QString *s);
void addMutexRef(QMutex *m);
void run() override;
};
#endif // COUNTINGTHREAD_H
countingthread.cpp
#include "countingthread.h"
void CountingThread::run()
{
//forever
while(1)
{
#ifdef TEST_ATOMIC_VAR_SHARE
//first use of counter
int counterUse1Copy= (*m_counter);
//some other operations, here sleep 10 ms
this->msleep(10);
//we will retry to use a second time
int counterUse2Copy= (*m_counter);
if(counterUse1Copy != counterUse2Copy)
qDebug()<<this->thread()->currentThreadId()<<" problem #1 found, counter not like we expect";
//we increment afterwards our counter
(*m_counter) +=1; //this works for fundamental types, like float, int, ...
#endif
#ifdef TEST_OBJECT_VAR_SHARE
#ifdef TEST_OBJECT_SHARE_FIX
m_locker->lock();
#endif
m_string->replace("#","-");
//this will crash here !!, with problem #2,
//segmentation fault, is not handle by try catch
m_string->append("foomaster");
m_string->append("#");
if(m_string->length()>10000)
qDebug()<<this->thread()->currentThreadId()<<" string is: " << m_string;
#ifdef TEST_OBJECT_SHARE_FIX
m_locker->unlock();
#endif
#endif
}//end forever
}
void CountingThread::addCounterdRef(int *r)
{
m_counter = r;
qDebug()<<this->thread()->currentThreadId()<<" add counter with value: " << *m_counter << " and address : "<< m_counter ;
}
void CountingThread::addStringRef(QString *s)
{
m_string = s;
qDebug()<<this->thread()->currentThreadId()<<" add string with value: " << *m_string << " and address : "<< m_string ;
}
void CountingThread::addMutexRef(QMutex *m)
{
m_locker = m;
}
If you follow up the code you are able to perform 2 tests.
If you uncomment TEST_ATOMIC_VAR_SHARE and comment TEST_OBJECT_VAR_SHARE in countingthread.h
your will see
problem #1 if you use your variable multiple times in your thread, it could be changes in the background from another thread, besides my expectation there was no app crash or weird exception in my build environment during execution using an int counter.
If you uncomment TEST_OBJECT_VAR_SHARE and comment TEST_OBJECT_SHARE_FIX and comment TEST_ATOMIC_VAR_SHARE in countingthread.h
your will see
problem #2 you get a segmentation fault, which is not possible to handle via try catch. This appears because multiple threads are using string functions for editing on the same object.
If you uncomment TEST_OBJECT_SHARE_FIX too you see the right handling via mutex.
problem #3 see answer from Benjamin T
What is Mutex:
I really like the chicken explanation which vallabh suggested.
I also found an good explanation here

using atomic c++11 to implement a thread safe down counter to zero

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.

Threads and the addres space

I have the following programs :
char **ptr; /* global */
int main()
{
int i;
pthread_t tid;
char *msgs[N] = {
"Hello from foo",
"Hello from bar"
};
ptr = msgs;
for (i = 0; i < 2; i++)
Pthread_create(&tid,
NULL,
thread,
(void *)i);
Pthread_exit(NULL);
}
/* thread routine */
void *thread(void *vargp)
{
int myid = (int)vargp;
static int svar = myid;
f1 (myid, svar);
printf("[%d]: %s (svar=%d)\n",
myid, ptr[myid], ++svar);
}
/* subroutine f1 */
void f1(id, id2)
{
If (id == id2){
something();
}else{
somethingelse();
}
}
And I had a few questions about :
1) Where is thread() stored in the “address space”?
2) How many copies of the variable “myid” do we have and where are they
stored?
3) How many copies of the variable “svar” do we have and where is it stored?
4) After thread 1 calls f1() which subroutine (something() or somethingelse())
will be called next?
5) How many copies of ptr (defined external to main()) do we have?
Any help would be greatly appreciated!
Thanks :-)
1) Where is thread() stored in the “address space”?
Wherever the compiler/linker/OS decide to put it. Typically some form of "code segment". There is usually one per module, but there may be more, depending on the compiler/linker/OS configuration.
2) How many copies of the variable “myid” do we have and where are
they stored?
2 copies. You create 2 threads, and each has its own copy of myid on its stack.
3) How many copies of the variable “svar” do we have and where is it stored?
There is 1 copy of svar; it's a static variable. It will be stored in a global data segment. This may be the same data segment as variables declared at global scope, or it may be separate, depending on the compiler/linker/OS configuration.
4) After thread 1 calls f1() which subroutine (something() or somethingelse()) > will be called next?
The first thread to execute the static int svar=myid line will set svar; the second thread through will leave svar unchanged (assuming C++11-style thread-safe statics --- on older versions of MSVC, or gcc with thread-safe statics disabled this is a data race and undefined behaviour)
This may be thread 1 or thread 2. Whichever thread is "first" will then call something, while the other thread will call somethingelse.
5) How many copies of ptr (defined external to main()) do we have?
One. It is a global variable.

Assigning return value of an atomic function

I'm trying to implement a barrier function, such that when a thread calls waitBarrier() it will wait until all other n threads have called the function, after which all will proceed, i.e. a sort of synchronization construct.
I have following code:
int i = 0; // Shared variable. Initialized as 0 at the beginning.
waitBarrier() {
// CAS = Compare-and-swap, the first argument holds "old_val" the second the new
i = CAS(i, i+1);
// Spin until all n threads (number of all threads known prior) have been "here"
while (i != n) {}
}
If this gets accessed by n threads, will this function work? Is the assignment of the return value of an atomic function atomic? Or could race conditions occur?
First of all you would have to specify the address of a register, the value of which you are comparing and swapping with. This can be done with either of the following:
CAS(int* reg, int oldValue, int newValue)
or
reg.CAS(int oldValue, int newValue)
Assuming your line now would be:
i = i.CAS(i, i+1)
Imagine two threads calling waitBarrier() at the same time.
Assuming arguments of an atomic function are being evaluated non-atomically, i.e. both threads will actually call i.CAS(0,1)
Whoever atomic call is being executed first will successfully set the shared variable i to 1.
Since CAS does always return the old value, by having an assignment i = OLD_VALUE_OF_i you're actually resetting the shared variable back to 0.
Not only that but imagine you would omit that assignment completely and just made the CAS call, whoever thread executes CAS second will compare the value of the shared value (which now would be 1) with the initial value of i (at evaluation time of the arguments that was 0) which will fail and therefore the shared variable will only be incremented once!
Taking those two aspects into consideration, your code would have to look as following:
int i = 0;
waitBarrier() {
// Atomically increment the shared value i by 1
do {
int value = i;
} while (CAS(i, value, value + 1));
// Wait until all threads have passed the barrier
while (i != n) {}
}

Getting invalid argument from semop

I've set(semget + init to 1) a semaphore in process A. Forked A, and got B.
Forked B and got C (code of processes B,C is at another .c file, so I am passing the semid as a global integer with extern int semid).
Into the process C code, I try to apply down(semid) and getting "invalid argument" error.
What I am doing in the code for down function is this:
struct sembuf sem_d;
sem_d.sem_num = 0;
sem_d.sem_op = -1;
sem_d.sem_flg = 0;
if ( semop(semid, &sem_d, 1) == -1 )
{
perror("error with down function");
return -1;
}
What am I doing wrong?
I've also have reassured that the semid from when the semaphore is initialized is the same before semop.
Also, in process A,B I am using wait(-1).
I'm not sure if you are allowed to use semget() over forks - it's a different process space after all.
semget() is part of the old System V semaphores anyway.
I would recommend to switch over to the POSIX semaphores - sem_open(), sem_wait() and friends and use named semaphores. Then open the same semaphore names in each process.

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