如何在不使用任何同步方法的情况下同步线程？

Question

I was asked in an interview that it is required to print numbers 1 to 3 in a sequential order by 3 threads without using any synchronisation techniques.

I think that we can access the thread id to find out which thread it is and keep a flag associated with this thread in a structure which can be set to true this thread in currently accessing the data.

Interviewer laughed at this idea. So obviously it is wrong.

What can be the way to achieve this task?

Answer 1

You use a single-core scheduler with lock-free ICPP scheduling strategy . This is not covered by the C++ stdlib, but POSIX has an API for it.

Since you are on a single core, only one thread runs at any time, and ICPP guarantees that it will never be preempted while holding a resource by another thread that wants to access that same resource. Therefore, you don't need thread synchronization.

Answer 2

Create thread A
Thread A prints '1'
Thread A creates thread B and exits
Thread B prints '2'
Thread B creates thread C and exits
Thread C prints '3' and
exits

No explicit synchro required, prints 1,2,3 in order using three threads, is a totally pointless exercise.

Answer 3

My first thought, but context switch might happen between std::count and ++i .

class Solution{
  private:
     atomic<int> i(0);
  public:

// func 1 2 3 are identical.
     void func1(){
         std::cout<< ++i;
     }

     void func2(){
         std::cout<< ++i;
     }

     void func3(){
         std::cout<< ++i;
     }

}

Second thought, using atomic with spin lock, so execution order should be fixed.

class Solution{
  private:
     atomic<int> i(1);
  public:
     void func1(){
         std::cout<< i;
         ++i;
     }

     void func2(){
         while( i != 2);
         std::cout<< i;
         ++i;
     }

     void func3(){
         while( i != 3);
         std::cout<< i;
         ++i;
     }

}

Answer 4

for fun:

Emit them by time?

for example

print 1 at 00:01
print 2 at 00:02
print 3 at 00:03

If you set the time interval big enough that the chance they run out-of-order is lower than a bit flip in the system, then you are good.