测量NUMA（非统一内存访问）。没有明显的不对称性。为什么？

  • Z时代
  • 2024-01-10
  • 分类：问答

我尝试测量NUMA的非对称内存访问效果，但失败了。

回答：

在2.93GHz，2个CPU，8核的Intel Xeon X5570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节点上分配了大小为10,000,000字节的数组

。然后我遍历数组 50次，并读取和写入数组中的每个字节。测量经过的时间以执行50次迭代。

然后，在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 每个字节进行50次读写操作。

数组 很大，可以最大程度地减少缓存影响。我们要在CPU必须一直到RAM进行加载和存储的时候来衡量速度，而不是在缓存有帮助的时候来衡量。

我的服务器中有两个NUMA节点，因此我希望在分配了数组 的同一节点上具有亲和力的内核具有更快的读写速度。我没看到

为什么？

正如我在其他地方看到的那样，也许NUMA仅与具有8-12个以上内核的系统有关？

http://lse.sourceforge.net/numa/faq/

回答：

#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;
}

回答：

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

不管哪个内核在进行读写，在数组 进行50次迭代读取和写入大约需要1.7秒。

回答：

我的CPU上的缓存大小为8Mb，因此10Mb阵列 可能不足以消除缓存效果。我尝试了100Mb数组

，并且尝试在最里面的循环中使用__sync_synchronize（）发出完整的内存隔离。它仍然没有揭示NUMA节点之间的任何不对称性。

回答：

我尝试使用__sync_fetch_and_add（）读写数组 。依然没有。

回答：

啊哈！神秘主义者是对的！硬件预取以某种方式优化了我的读/写。

如果这是缓存优化，那么强制使用内存屏障将使优化失败：

c = __sync_fetch_and_add(((char*)x) + j, 1);

但这没有任何区别。确实有所作为的是，将我的迭代器索引乘以质数1009来破坏预取优化：

*(((char*)x) + ((j * 1009) % N)) += 1;

有了这一更改，NUMA的不对称性就清楚地显示出来了：

numa_available() 0
numa node 0 10101010 12884901888
numa node 1 01010101 12874584064
Elapsed read/write by same thread that allocated on core 0: 00:00:00.961725
Elapsed read/write by thread on core 0: 00:00:00.942300
Elapsed read/write by thread on core 1: 00:00:01.216286
Elapsed read/write by thread on core 2: 00:00:00.909353
Elapsed read/write by thread on core 3: 00:00:01.218935
Elapsed read/write by thread on core 4: 00:00:00.898107
Elapsed read/write by thread on core 5: 00:00:01.211413
Elapsed read/write by thread on core 6: 00:00:00.898021
Elapsed read/write by thread on core 7: 00:00:01.207114

至少我认为这是正在发生的事情。

感谢Mysticial！

对于只看一下这篇文章以大致了解NUMA性能特征的任何人，根据我的测试，这是底线：

对非本地NUMA节点的内存访问的延迟约为对本地节点的内存访问的延迟的1.33倍。

以上是 测量NUMA（非统一内存访问）。没有明显的不对称性。为什么？ 的全部内容， 来源链接： utcz.com/qa/419976.html

回到顶部