Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
327 views
in Technique[技术] by (71.8m points)

c++ - Why does memory_order_relaxed and memory_order_seq_cst make no difference?

I was playing with one of the examples in C++ Concurrency in Action which uses std::memory_order_relaxed for reading and writing 3 atomic variables from 5 different threads. The example program is as follows:

#include <thread>
#include <atomic>
#include <iostream>

std::atomic<int> x(0);
std::atomic<int> y(0);
std::atomic<int> z(0);
std::atomic<bool> go(false);

const unsigned int loop_count = 10;

struct read_values
{
   int x;
   int y;
   int z;
};

read_values values1[loop_count];
read_values values2[loop_count];
read_values values3[loop_count];
read_values values4[loop_count];
read_values values5[loop_count];

void increment( std::atomic<int>* v, read_values* values )
{
    while (!go)
       std::this_thread::yield();

    for (unsigned i=0;i<loop_count;++i)
    {
       values[i].x=x.load( std::memory_order_relaxed );
       values[i].y=y.load( std::memory_order_relaxed );
       values[i].z=z.load( std::memory_order_relaxed );
       v->store( i+1, std::memory_order_relaxed );
       std::this_thread::yield();
    }
}

void read_vals( read_values* values )
{

   while (!go)
      std::this_thread::yield();

   for (unsigned i=0;i<loop_count;++i)
   {
      values[i].x=x.load( std::memory_order_relaxed );
      values[i].y=y.load( std::memory_order_relaxed );
      values[i].z=z.load( std::memory_order_relaxed );
      std::this_thread::yield();
   }
}

void print( read_values* values )
{
   for (unsigned i=0;i<loop_count;++i)
   {
      if (i)
         std::cout << ",";
      std::cout << "(" << values[i].x <<","
                       << values[i].y <<","
                       << values[i].z <<")";
   }
   std::cout << std::endl;
}

int main()
{
   std::thread t1( increment, &x, values1);
   std::thread t2( increment, &y, values2);
   std::thread t3( increment, &z, values3);
   std::thread t4( read_vals, values4);
   std::thread t5( read_vals, values5);

   go = true;

   t5.join();
   t4.join();
   t3.join();
   t2.join();
   t1.join();

   print( values1 );
   print( values2 );
   print( values3 );
   print( values4 );
   print( values5 );

   return 0;
}

Every time I run the program I get exactly the same output:

(0,10,10),(1,10,10),(2,10,10),(3,10,10),(4,10,10),(5,10,10),(6,10,10),(7,10,10),(8,10,10),(9,10,10)
(0,0,1),(0,1,2),(0,2,3),(0,3,4),(0,4,5),(0,5,6),(0,6,7),(0,7,8),(0,8,9),(0,9,10)
(0,0,0),(0,1,1),(0,2,2),(0,3,3),(0,4,4),(0,5,5),(0,6,6),(0,7,7),(0,8,8),(0,9,9)
(0,0,0),(0,0,0),(0,0,0),(0,0,0),(0,0,0),(0,0,0),(0,0,0),(0,0,0),(0,0,0),(0,0,0)
(0,0,0),(0,0,0),(0,0,0),(0,0,0),(0,0,0),(0,0,0),(0,0,0),(0,0,0),(0,0,0),(0,0,0)

If I change from std::memory_order_relaxed to std::memory_order_seq_cst the program gives exactly the same output!

I would have expected different output from the 2 versions of the program. Why is there no difference between the output for std::memory_order_relaxed and std::memory_order_seq_cst?

Why does std::memory_order_relaxed always produce exactly the same results for every run of the program?

I am using: - 32bit Ubuntu installed as a virtual machine (under VMWare) - An INtel Quad Core processor - GCC 4.6.1-9

The code is compiled with: g++ --std=c++0x -g mem-order-relaxed.cpp -o relaxed -pthread

Note the -pthread is necessary, otherwise the following error is reported: terminate called after throwing an instance of 'std::system_error' what(): Operation not permitted

Is the behaviour I am seeing due to lack of support with GCC, or as a result of running under VMWare?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

How many processor cores do you have assigned to the VM? Assign multiple cores to the VM to let it take advantage of concurrency.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...