Is this the correct way to sync threads without mutex.
This code should be running for a long time
#include <boost/thread.hpp>
#include <boost/thread/mutex.hpp>
#include <boost/memory_order.hpp>
#include <atomic>
std::atomic<long> x =0;
std::atomic<long> y =0;
boost::mutex m1;
// Thread increments
void Thread_Func()
{
for(;;)
{
// boost::mutex::scoped_lock lx(m1);
++x;
++y;
}
}
// Checker Thread
void Thread_Func_X()
{
for(;;)
{
// boost::mutex::scoped_lock lx(m1);
if(y > x)
{
// should never hit until int overflows
std::cout << y << "\\" << x << std::endl;
break;
}
}
}
//Test Application
int main(int argc, char* argv[])
{
boost::thread_group threads;
threads.create_thread(Thread_Func);
threads.create_thread(Thread_Func_X);
threads.join_all();
return 0;
}
Without knowing exactly what you're trying to do, it is hard to say it is the "correct" way. That's valid code, it's a bit janky though.
There is no guarantee that the "Checker" thread will ever see the condition y > x. It's theoretically possible that it will never break. In practice, it will trigger at some point but x might not be LONG_MIN and y LONG_MAX. In other words, it's not guaranteed to trigger just as the overflow happens.
Related
I am trying to see the race condition happens in the comsumer-producser problem,
so I made multiple producers and mulitple consumers.
From what I know that I need to provide mutex with semaphore:
Mutex for the race conditions, because muliple producers can access the buffer at the same time. then the data might be corrupted.
And semaphore to provide signaling between the producers and the consumers
The problem here that the sync is happening correctly while I am not using the Mutex (i am using the Semaphore only). is my understanding correct or is there anything wrong to do in the code below:
#include <pthread.h>
#include <stdio.h>
#include <semaphore.h>
#include <stdlib.h>
#include <unistd.h>
int buffer;
int loops = 0;
sem_t empty;
sem_t full;
sem_t mutex; //Adding MUTEX
void put(int value) {
buffer = value;
}
int get() {
int b = buffer;
return b;
}
void *producer(void *arg) {
int i;
for (i = 0; i < loops; i++) {
sem_wait(&empty);
//sem_wait(&mutex);
put(i);
//printf("Data Set from %s, Data=%d\n", (char*) arg, i);
//sem_post(&mutex);
sem_post(&full);
}
}
void *consumer(void *arg) {
int i;
for (i = 0; i < loops; i++) {
sem_wait(&full);
//sem_wait(&mutex);
int b = get();
//printf("Data recieved from %s, %d\n", (char*) arg, b);
printf("%d\n", b);
//sem_post(&mutex);
sem_post(&empty);
}
}
int main(int argc, char *argv[])
{
if(argc < 2 ){
printf("Needs 2nd arg for loop count variable.\n");
return 1;
}
loops = atoi(argv[1]);
sem_init(&empty, 0, 1);
sem_init(&full, 0, 0);
sem_init(&mutex, 0, 1);
pthread_t pThreads[3];
pthread_t cThreads[3];
pthread_create(&cThreads[0], 0, consumer, (void*)"Consumer1");
pthread_create(&cThreads[1], 0, consumer, (void*)"Consumer2");
pthread_create(&cThreads[2], 0, consumer, (void*)"Consumer3");
//Passing the name of the thread as paramter, Ignore attr
pthread_create(&pThreads[0], 0, producer, (void*)"Producer1");
pthread_create(&pThreads[1], 0, producer, (void*)"Producer2");
pthread_create(&pThreads[2], 0, producer, (void*)"Producer3");
pthread_join(pThreads[0], NULL);
pthread_join(pThreads[1], NULL);
pthread_join(pThreads[2], NULL);
pthread_join(cThreads[0], NULL);
pthread_join(cThreads[1], NULL);
pthread_join(cThreads[2], NULL);
return 0;
}
I believe I have the problem figured out. Here's what is happening
When initializing your semaphores you set empty's number of threads to 1 and full's to 0
sem_init(&empty, 0, 1);
sem_init(&full, 0, 0);
sem_init(&mutex, 0, 1);
This means that there is only one "space" for the thread to get into the critical region. In other words, what your program is doing is
produce (empty is now 0, full has 1)
consume (full is now 0, empty has 0)
produce (empty is now 0, full has 1)
...
It's as if you had a token (or, if you like, a mutex), and you pass that token between consumers and producers. That is actually what the consumer-producer problem is all about, only that in most cases we are worried about having several consumers and producers working at the same time (which means you have more than one token). Here, because you have only one token, you basically have what one mutex would do.
Hope it helped :)
I am trying to make a child class of LeafSystem whose output is sinusoidal and its derivative.
I wrote the code and try to plot it but signal logger doesn't log correctly.
#include "drake/systems/framework/leaf_system.h"
#include "drake/systems/analysis/simulator.h"
#include "drake/systems/framework/diagram.h"
#include "drake/systems/framework/diagram_builder.h"
#include "drake/systems/primitives/signal_logger.h"
#include "drake/common/proto/call_python.h"
class Sinusoid : public drake::systems::LeafSystem<double>
{
public:
Sinusoid (double tstart, double freq, double amp, double offset) :
m_freq(freq), m_amp(amp), m_offset(offset), m_tstart(tstart) {
this->DeclareVectorOutputPort(
drake::systems::BasicVector<double>(2), &Sinusoid::output);
}
private:
void output(const drake::systems::Context<double>& c, drake::systems::BasicVector<double>* output) const {
double t(c.get_time());
double tknot(t - m_tstart);
if (t > m_tstart) {
output->SetAtIndex(0, std::sin(tknot*m_freq + m_offset)*m_amp);
output->SetAtIndex(1, std::cos(tknot*m_freq + m_offset)*m_amp*m_freq);
} else {
output->SetAtIndex(0, 0.0);
output->SetAtIndex(1, 0.0);
}
}
double m_freq{0.0}, m_amp{0.0}, m_offset{0.0}, m_tstart{0.0};
};
int main(int argc, char *argv[])
{
// Add System and Connect
drake::systems::DiagramBuilder<double> builder;
auto system = builder.AddSystem<Sinusoid>(1.0, 2.*M_PI*1., 3., 0.);
auto logger = LogOutput(system->get_output_port(0), &builder);
auto diagram = builder.Build();
// Construct Simulator
drake::systems::Simulator<double> simulator(*diagram);
// Run simulation
simulator.StepTo(100);
// Plot with Python
auto sample_time = logger->sample_times();
auto sample_data = logger->data();
std::cout << sample_time.size() << std::endl;
for (int i = 0; i < sample_time.size(); ++i) {
std::cout << sample_time(i) << " : " << sample_data(i, 0) << " " << sample_data(i, 1) << std::endl;
}
std::cout << "END" << std::endl;
return 0;
}
The output of the code is
2
0 : 0 0
0 : 0 0
END
Whatever number I used in StepTo function, signal logger only cate 2 data whose sampled times are both 0.
The code looks good. Note that TrajectorySource does this almost exactly (and used SingleOutputVectorSource as a base class, which you might consider, too). The only problem is that you do not have anything telling the simulator that there is a reason to evaluate the output port. The logger block will pull on that for every publish event, but you haven't told the simulator to publish.
The solution is to call
simulator.set_publish_every_timestep(true)
http://drake.mit.edu/doxygen_cxx/classdrake_1_1systems_1_1_simulator.html#aef1dc6aeb821503379ab1dd8c6044562
If you want to further control the timestep of the integrator, you could set the parameters of the integrator (e.g. simalator.get_integerator), then calls like set_fixed_step_mode.
i have a problem with the pthread library in a C-Application for Linux.
In my Application a Thread is started over and over again.
But I allways wait until the Thread is finished before starting it.
At some point the thread doesn't start anymore and I get an out of memory error.
The solution I found is to do a pthread_join after the thread has finished.
Can anyone tell me why the Thread doesn't end correctly?
Here is an Example Code, that causes the same Problem.
If the pthread_join isn't called the Process stops at about 380 calls of the Thread:
#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
#include <pthread.h>
#include <unistd.h>
volatile uint8_t check_p1 = 0;
uint32_t stack_start;
void *thread1(void *ch)
{
static int counter = 0;
int i;
int s[100000];
char stack_end;
srand(time(NULL) + counter);
for (i = 0; i < (sizeof (s)/sizeof(int)); i++) //do something
{
s[i] = rand();
}
counter++;
printf("Thread %i finished. Stacksize: %u\n", counter, ((uint32_t) (stack_start)-(uint32_t) (&stack_end)));
check_p1 = 1; // Mark Thread as finished
return 0;
}
int main(int argc, char *argv[])
{
pthread_t p1;
int counter = 0;
stack_start = (uint32_t)&counter; // save the Address of counter
while (1)
{
counter++;
check_p1 = 0;
printf("Start Thread %i\n", counter);
pthread_create(&p1, NULL, thread1, 0);
while (!check_p1) // wait until thread has finished
{
usleep(100);
}
usleep(1000); // wait a little bit to be really sure that the thread is finished
//pthread_join(p1,0); // crash without pthread_join
}
return 0;
}
The solution I found is to do a pthread_join after the thread has finished.
That is the correct solution. You must do that, or you leak thread resources.
Can anyone tell me why the Thread doesn't end correctly?
It does end correctly, but you must join it in order for the thread library to know: "yes, he is really done with this thread; no need to hold resources any longer".
This is exactly the same reason you must use wait (or waitpid, etc.) in this loop:
while (1) {
int status;
pid_t p = fork();
if (p == 0) exit(0); // child
// parent
wait(&status); // without this wait, you will run out of OS resources.
}
I'm wondering why the following code gives unexpected output: a can get 110...!
pthread_t th[nbT];
void * func(void *d)
{
while(a<100)
{
pthread_mutex_lock(&l);
cout <<a <<" in thread "<<pthread_self()<<"\n";
a+=1;
pthread_mutex_unlock(&l);
}
return NULL;
}
int main(int argc, const char* argv[])
{
for(int i=0;i<nbT;i++)
pthread_create(&(th[i]), NULL, func, NULL);
for(int i=0;i<nbT;i++)
pthread_join(th[i],NULL);
}
The problem is that you get the lock (mutex) after checking the condition, so you don't know if it's still true or not once you get the lock. You should just do a simple double-check:
while(a<100)
{
pthread_mutex_lock(&l);
cout <<a <<" in thread "<<pthread_self()<<"\n";
if (a<100) a+=1; // <== Added condition here!
pthread_mutex_unlock(&l);
}
How do you end a long running Lua script?
I have two threads, one runs the main program and the other controls a user supplied Lua script. I need to kill the thread that's running Lua, but first I need the script to exit.
Is there a way to force a script to exit?
I have read that the suggested approach is to return a Lua exception. However, it's not garanteed that the user's script will ever call an api function ( it could be in a tight busy loop). Further, the user could prevent errors from causing his script to exit by using a pcall.
You could use setjmp and longjump, just like the Lua library does internally. That will get you out of pcalls and stuff just fine without need to continuously error, preventing the script from attempting to handle your bogus errors and still getting you out of execution. (I have no idea how well this plays with threads though.)
#include <stdio.h>
#include <setjmp.h>
#include "lua.h"
#include "lualib.h"
#include "lauxlib.h"
jmp_buf place;
void hook(lua_State* L, lua_Debug *ar)
{
static int countdown = 10;
if (countdown > 0)
{
--countdown;
printf("countdown: %d!\n", countdown);
}
else
{
longjmp(place, 1);
}
}
int main(int argc, const char *argv[])
{
lua_State* L = luaL_newstate();
luaL_openlibs(L);
lua_sethook(L, hook, LUA_MASKCOUNT, 100);
if (setjmp(place) == 0)
luaL_dostring(L, "function test() pcall(test) print 'recursing' end pcall(test)");
lua_close(L);
printf("Done!");
return 0;
}
You could set a variable somewhere in your program and call it something like forceQuitLuaScript. Then, you use a hook, described here to run every n instructions. After n instructions, it'll run your hook which just checks if forceQuitLuaScript is set, and if it is do any clean up you need to do and kill the thread.
Edit: Here's a cheap example of how it could work, only this is single threaded. This is just to illustrate how you might handle pcall and such:
#include <stdlib.h>
#include "lauxlib.h"
void hook(lua_State* L, lua_Debug *ar)
{
static int countdown = 10;
if (countdown > 0)
{
--countdown;
printf("countdown: %d!\n", countdown);
}
else
{
// From now on, as soon as a line is executed, error
// keep erroring until you're script reaches the top
lua_sethook(L, hook, LUA_MASKLINE, 0);
luaL_error(L, "");
}
}
int main(int argc, const char *argv[])
{
lua_State* L = luaL_newstate();
luaL_openlibs(L);
lua_sethook(L, hook, LUA_MASKCOUNT, 100);
// Infinitely recurse into pcalls
luaL_dostring(L, "function test() pcall(test) print 'recursing' end pcall(test)");
lua_close(L);
printf("Done!");
return 0;
}
The way to end a script is to raise an error by calling error. However, if the user has called the script via pcall then this error will be caught.
It seems like you could terminate the thread externally (from your main thread) since the lua script is user supplied and you can't signal it to exit.
If that isn't an option, you could try the debug API. You could use lua_sethook to enable you to regain control assuming you have a way to gracefully terminate your thread in the hook.
I haven't found a way to cleanly kill a thread that is executing a long running lua script without relying on some intervention from the script itself. Here are some approaches I have taken in the past:
If the script is long running it is most likely in some loop. The script can check the value of some global variable on each iteration. By setting this variable from outside of the script you can then terminate the thread.
You can start the thread by using lua_resume. The script can then exit by using yield().
You could provide your own implementation of pcall that checks for a specific type of error. The script could then call error() with a custom error type that your version of pcall could watch for:
function()
local there_is_an_error = do_something()
if (there_is_an_error) then
error({code = 900, msg = "Custom error"})
end
end
possibly useless, but in the lua I use (luaplayer or PGELua), I exit with
os.exit()
or
pge.exit()
If you're using coroutines to start the threads, you could maybe use coroutine.yield() to stop it.
You might wanna take look at
https://github.com/amilamad/preemptive-task-scheduler-for-lua
project. its preemptive scheduler for lua.
It uses a lua_yeild function inside the hook. So you can suspend your lua thread. It also uses longjmp inside but its is much safer.
session:destroy();
Use this single line code on that where you are want to destroy lua script.
lua_KFunction cont(lua_State* L);
int my_yield_with_res(lua_State* L, int res) {
cout << " my_yield_with_res \n" << endl;
return lua_yieldk(L, 0, lua_yield(L, res), cont(L));/* int lua_yieldk(lua_State * L, int res, lua_KContext ctx, lua_KFunction k);
Приостанавливает выполнение сопрограммы(поток). Когда функция C вызывает lua_yieldk, работающая
сопрограмма приостанавливает свое выполнение и вызывает lua_resume, которая начинает возврат данной сопрограммы.
Параметр res - это число значений из стека, которые будут переданы в качестве результатов в lua_resume.
Когда сопрограмма снова возобновит выполнение, Lua вызовет заданную функцию продолжения k для продолжения выполнения
приостановленной C функции(смотрите §4.7). */
};
int hookFunc(lua_State* L, lua_Debug* ar) {
cout << " hookFunc \n" << endl;
return my_yield_with_res(L, 0);// хук./
};
lua_KFunction cont(lua_State* L) {// функция продолжения.
cout << " hooh off \n" << endl;
lua_sethook(L, (lua_Hook)hookFunc, LUA_MASKCOUNT, 0);// отключить хук foo.
return 0;
};
struct Func_resume {
Func_resume(lua_State* L, const char* funcrun, unsigned int Args) : m_L(L), m_funcrun(funcrun), m_Args(Args) {}
//имена функций, кол-во агрументов.
private:
void func_block(lua_State* L, const char* functionName, unsigned int Count, unsigned int m_Args) {
lua_sethook(m_L, (lua_Hook)hookFunc, LUA_MASKCOUNT, Count); //вызов функции с заданной паузой.
if (m_Args == 0) {
lua_getglobal(L, functionName);// получить имя функции.
lua_resume(L, L, m_Args);
}
if (m_Args != 0) {
int size = m_Args + 1;
lua_getglobal(L, functionName);
for (int i = 1; i < size; i++) {
lua_pushvalue(L, i);
}
lua_resume(L, L, m_Args);
}
};
public:
void Update(float dt) {
unsigned int Count = dt * 100.0;// Время работы потока.
func_block(m_L, m_funcrun, Count, m_Args);
};
~Func_resume() {}
private:
lua_State* m_L;
const char* m_funcrun; // имя функции.
unsigned int m_Count;// число итерации.
unsigned int m_Args;
};
const char* LUA = R"(
function main(y)
--print(" func main arg, a = ".. a.." y = ".. y)
for i = 1, y do
print(" func main count = ".. i)
end
end
)";
int main(int argc, char* argv[]) {
lua_State* L = luaL_newstate();/*Функция создает новое Lua состояние. */
luaL_openlibs(L);
luaL_dostring(L, LUA);
//..pushlua(L, 12);
pushlua(L, 32);
//do {
Func_resume func_resume(L, "main", 2);
func_resume.Update(1.7);
lua_close(L);
// } while (LUA_OK != lua_status(L)); // Пока поток не завершен.
return 0;
};