C++多线程原子加载/存储

Question

When I read the 5th chapter of the book CplusplusConcurrencyInAction , the example code as follows, multithread load/store some atomic values concurrently with the momery_order_relaxed.Three array save the value of x、y and z respectively at each round.

#include <thread>
#include <atomic>
#include <iostream>
​
std::atomic<int> x(0),y(0),z(0);  // 1
std::atomic<bool> go(false);  // 2
​
unsigned const loop_count=10;
​
struct read_values
{
  int x,y,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>* var_to_inc,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);
    var_to_inc->store(i+1,std::memory_order_relaxed);  // 4
    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* v)
{
  for(unsigned i=0;i<loop_count;++i)
  {
    if(i)
      std::cout<<",";
    std::cout<<"("<<v[i].x<<","<<v[i].y<<","<<v[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);
}

one of the valid output mentioned in this chapter:

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

The 3rd output of values1 is (2,0,0) ,at this point it reads x=2 ,and y=z=0 .It means when y=0 ,the x is already equals to 2, Why the 3rd output of the values2 it reads x=0 and y=2 ,which means x is the old value because x、y、z is increasing, so when y=2 that x is at least 2. And I test the code in my PC,I can't reproduce the result like that.

Answer 1

The reason is that reading via x.load(std::memory_order_relaxed) guarantees only that you never see x decrease within the same thread (in this example code). (It also guarantees that a thread writing to x will read that same value again in the next iteration.)

In general, different threads can read different values from the same variable at the same time. That is, there need not be a consistent "global state" that all threads agree on. The example output is supposed to demonstrate that: The first thread might still see y = 0 when it already wrote x = 4 , while the second thread might still see x = 0 when it already writes y = 2 . The standard allows this because real hardware may work that way: Consider the case when the threads are on different CPU cores, each with its own private L1 cache.

However, it is not possible that the second thread sees x = 5 and then later sees x = 2 - the atomic object always guarantees that there is a consistent global modification order (that is, all writes to the variable are observed to happen in the same order by all the threads).

But when using std::memory_order_relaxed there are no guarantees about when a thread finally does "see" those writes*, or how the observations of different threads relate to each other. You need stronger memory ordering to get those guarantees.

^{*In fact, a valid output would be all threads reading only 0 all the time, except the writer threads reading what they wrote the previous iteration to their "own" variable (and 0 for the others). On hardware that never flushed caches unless prompted, this might actually happen, and it would be fully compliant with the C++ standard!}

And I test the code in my PC,I can't reproduce the result like that.

The "example output" shown is highly artificial. The C++ standard allows for this output to happen. This means you can write efficient and correct multithreaded code even on hardware with no inbuilt guarantees on cache coherency (see above). But common hardware today (x86 in particular) brings a lot of guarantees that actually make certain behavior impossible to observe (including the output in the question).

Also, note that x , y and z are extremely likely to be adjacent (depends on the compiler), meaning they will likely all land on the same cache line. This will lead to massive performance degradation (look up "false sharing"). But since memory can only be transferred between cores at cache line granularity, this (together with the x86 coherency guarantees) makes it essentially impossible that an x86 CPU (which you most likely performed your tests with) reads outdated values of any of the variables. Allocating these values more than 1-2 cache lines apart will likely lead to more interesting/chaotic results.