我试图测量 NUMA 的非对称内存访问效果，但失败了。
本实验
在 Intel Xeon X5570 @ 2.93GHz、2 个 CPU、8 个内核上执行。
在固定到核心 0 的线程上，我使用 numa_alloc_local 在核心 0 的 NUMA 节点上分配大小为 10,000,000 字节的数组x。然后我遍历数组x 50 次并读取和写入数组中的每个字节。测量执行 50 次迭代所用的时间。
然后，在我的服务器中的每个其他内核上，我固定一个新线程并再次测量经过的时间，以对数组x中的每个字节进行 50 次读取和写入迭代。
数组x很大以最小化缓存效应。我们想要测量 CPU 必须一直到 RAM 进行加载和存储时的速度，而不是在缓存有帮助的时候。
我的服务器中有两个 NUMA 节点，因此我希望在分配数组x的同一节点上具有亲和力的内核具有更快的读/写速度。我没有看到。
为什么？
正如我在其他地方看到的那样，也许 NUMA 只与具有 > 8-12 个内核的系统相关？
http://lse.sourceforge.net/numa/faq/
numatest.cpp
#include <numa.h>
#include <iostream>
#include <boost/thread/thread.hpp>
#include <boost/date_time/posix_time/posix_time.hpp>
#include <pthread.h>
void pin_to_core(size_t core)
{
cpu_set_t cpuset;
CPU_ZERO(&cpuset);
CPU_SET(core, &cpuset);
pthread_setaffinity_np(pthread_self(), sizeof(cpu_set_t), &cpuset);
}
std::ostream& operator<<(std::ostream& os, const bitmask& bm)
{
for(size_t i=0;i<bm.size;++i)
{
os << numa_bitmask_isbitset(&bm, i);
}
return os;
}
void* thread1(void** x, size_t core, size_t N, size_t M)
{
pin_to_core(core);
void* y = numa_alloc_local(N);
boost::posix_time::ptime t1 = boost::posix_time::microsec_clock::universal_time();
char c;
for (size_t i(0);i<M;++i)
for(size_t j(0);j<N;++j)
{
c = ((char*)y)[j];
((char*)y)[j] = c;
}
boost::posix_time::ptime t2 = boost::posix_time::microsec_clock::universal_time();
std::cout << "Elapsed read/write by same thread that allocated on core " << core << ": " << (t2 - t1) << std::endl;
*x = y;
}
void thread2(void* x, size_t core, size_t N, size_t M)
{
pin_to_core(core);
boost::posix_time::ptime t1 = boost::posix_time::microsec_clock::universal_time();
char c;
for (size_t i(0);i<M;++i)
for(size_t j(0);j<N;++j)
{
c = ((char*)x)[j];
((char*)x)[j] = c;
}
boost::posix_time::ptime t2 = boost::posix_time::microsec_clock::universal_time();
std::cout << "Elapsed read/write by thread on core " << core << ": " << (t2 - t1) << std::endl;
}
int main(int argc, const char **argv)
{
int numcpus = numa_num_task_cpus();
std::cout << "numa_available() " << numa_available() << std::endl;
numa_set_localalloc();
bitmask* bm = numa_bitmask_alloc(numcpus);
for (int i=0;i<=numa_max_node();++i)
{
numa_node_to_cpus(i, bm);
std::cout << "numa node " << i << " " << *bm << " " << numa_node_size(i, 0) << std::endl;
}
numa_bitmask_free(bm);
void* x;
size_t N(10000000);
size_t M(50);
boost::thread t1(boost::bind(&thread1, &x, 0, N, M));
t1.join();
for (size_t i(0);i<numcpus;++i)
{
boost::thread t2(boost::bind(&thread2, x, i, N, M));
t2.join();
}
numa_free(x, N);
return 0;
}
输出
g++ -o numatest -pthread -lboost_thread -lnuma -O0 numatest.cpp
./numatest
numa_available() 0 <-- NUMA is available on this system
numa node 0 10101010 12884901888 <-- cores 0,2,4,6 are on NUMA node 0, which is about 12 Gb
numa node 1 01010101 12874584064 <-- cores 1,3,5,7 are on NUMA node 1, which is slightly smaller than node 0
Elapsed read/write by same thread that allocated on core 0: 00:00:01.767428
Elapsed read/write by thread on core 0: 00:00:01.760554
Elapsed read/write by thread on core 1: 00:00:01.719686
Elapsed read/write by thread on core 2: 00:00:01.708830
Elapsed read/write by thread on core 3: 00:00:01.691560
Elapsed read/write by thread on core 4: 00:00:01.686912
Elapsed read/write by thread on core 5: 00:00:01.691917
Elapsed read/write by thread on core 6: 00:00:01.686509
Elapsed read/write by thread on core 7: 00:00:01.689928
对数组x进行50 次迭代读取和写入大约需要 1.7 秒，无论哪个内核正在执行读取和写入。
更新：
我的 CPU 上的缓存大小为 8Mb，因此 10Mb 数组x可能不足以消除缓存效果。我尝试了 100Mb 数组x，并尝试在最里面的循环中使用 __sync_synchronize() 发出完整的内存围栏。它仍然没有显示 NUMA 节点之间的任何不对称性。
更新 2：
我尝试使用 __sync_fetch_and_add()读取和写入数组x 。依然没有。