Modified version oft the Dining Philosopher Problem - here any chopstick can be used, so the five chopstick are represented báy a counting semaphore initialized with a value of 5.
main.cpp
- Committer:
- cspista
- Date:
- 2022-03-17
- Revision:
- 0:3def6e6a9e93
File content as of revision 0:3def6e6a9e93:
#include "mbed.h"
#include "rtos.h"
Semaphore s(5); // a pool of 5 chopsticks
Timer mytime;
void notify(const char* name) {
printf("%s acquired two chopsticks %8.1f\n\r", name,mytime.read());
}
void test_thread(void const* args) {
while (true) {
Thread::wait(1000+rand()%500); //Thinking
s.wait();
s.wait();
notify((const char*)args);
Thread::wait(500+rand()%500); //Eating
s.release();
s.release();
}
}
int main (void) {
mytime.start();
Thread t2(test_thread, (void *)"Philosopher 2");
Thread t3(test_thread, (void *)"Philosopher 3");
Thread t4(test_thread, (void *)"Philosopher 4");
Thread t5(test_thread, (void *)"Philosopher 5");
test_thread((void *)"Philosopher 1");
}