8

我有一个应用程序，我必须在多线程方法中增加一些统计计数器。递增必须是线程安全的，所以我决定使用 gcc atomic 内置__sync_add_and_fetch()函数。为了了解它们的影响，我做了一些简单的性能测试，发现这些函数比简单的前/后递增要慢得多。

这是我创建的测试程序：

#include <iostream>
#include <pthread.h>
#include <time.h>

using namespace std;

uint64_t diffTimes(struct timespec &start, struct timespec &end)
{
  if(start.tv_sec == end.tv_sec)
  {
    return end.tv_nsec - start.tv_nsec;
  }
  else if(start.tv_sec < end.tv_sec)
  {
    uint64_t nsecs = (end.tv_sec - start.tv_sec) * 1000000000;
    return nsecs + end.tv_nsec - start.tv_nsec;
  }
  else
  {
    // this is actually an error
    return 0;
  }
}

void outputResult(const char *msg, struct timespec &start, struct timespec &end, uint32_t numIterations, uint64_t val)
{
  uint64_t diff = diffTimes(start, end);
  cout << msg << ": "
       << "\n\t iterations: " << numIterations
       << ", result: " << val
       << "\n\t times [start, end] =  [" << start.tv_sec << ", " << start.tv_nsec << "]"
       << "\n\t [" << end.tv_sec << ", " << end.tv_nsec << "]"
       << "\n\t [total, avg] = [" << diff
       << ", " << (diff/numIterations) << "] nano seconds"
       << endl;
}

int main(int argc, char **argv)
{
  struct timespec start, end;
  uint64_t val = 0;
  uint32_t numIterations = 1000000;

  //
  // NON ATOMIC pre increment
  //
  clock_gettime(CLOCK_PROCESS_CPUTIME_ID, &start);
  for(uint32_t i = 0; i < numIterations; ++i)
  {
    ++val;
  }
  clock_gettime(CLOCK_PROCESS_CPUTIME_ID, &end);

  outputResult("Non-Atomic pre-increment", start, end, numIterations, val);
  val = 0;

  //
  // NON ATOMIC post increment
  //
  clock_gettime(CLOCK_PROCESS_CPUTIME_ID, &start);
  for(uint32_t i = 0; i < numIterations; ++i)
  {
    val++;
  }
  clock_gettime(CLOCK_PROCESS_CPUTIME_ID, &end);

  outputResult("Non-Atomic post-increment", start, end, numIterations, val);
  val = 0;

  //
  // ATOMIC add and fetch
  //
  clock_gettime(CLOCK_PROCESS_CPUTIME_ID, &start);
  for(uint32_t i = 0; i < numIterations; ++i)
  {
    __sync_add_and_fetch(&val, 1);
  }
  clock_gettime(CLOCK_PROCESS_CPUTIME_ID, &end);

  outputResult("Atomic add and fetch", start, end, numIterations, val);
  val = 0;

  //
  // ATOMIC fetch and add
  //
  clock_gettime(CLOCK_PROCESS_CPUTIME_ID, &start);
  for(uint32_t i = 0; i < numIterations; ++i)
  {
    __sync_fetch_and_add(&val, 1);
  }
  clock_gettime(CLOCK_PROCESS_CPUTIME_ID, &end);

  outputResult("Atomic fetch and add", start, end, numIterations, val);
  val = 0;

  //
  // Mutex protected post-increment
  //
  pthread_mutex_t mutex;
  pthread_mutex_init(&mutex, NULL);
  clock_gettime(CLOCK_PROCESS_CPUTIME_ID, &start);
  for(uint32_t i = 0; i < numIterations; ++i)
  {
    pthread_mutex_lock(&mutex);
    val++;
    pthread_mutex_unlock(&mutex);
  }
  clock_gettime(CLOCK_PROCESS_CPUTIME_ID, &end);

  outputResult("Mutex post-increment", start, end, numIterations, val);
  val = 0;

  //
  // RWlock protected post-increment
  //
  pthread_rwlock_t rwlock;
  pthread_rwlock_init(&rwlock, NULL);
  clock_gettime(CLOCK_PROCESS_CPUTIME_ID, &start);
  for(uint32_t i = 0; i < numIterations; ++i)
  {
    pthread_rwlock_wrlock(&rwlock);
    val++;
    pthread_rwlock_unlock(&rwlock);
  }
  clock_gettime(CLOCK_PROCESS_CPUTIME_ID, &end);

  outputResult("RWlock post-increment", start, end, numIterations, val);
  val = 0;

  return 0;
}

结果如下：

# ./atomicVsNonAtomic
Non-Atomic pre-increment:
         iterations: 1000000, result: 1000000
         times [start, end] =  [0, 1585375]
         [0, 1586185]
         [total, avg] = [810, 0] nano seconds
Non-Atomic post-increment:
         iterations: 1000000, result: 1000000
         times [start, end] =  [0, 1667489]
         [0, 1667920]
         [total, avg] = [431, 0] nano seconds
Atomic add and fetch:
         iterations: 1000000, result: 1000000
         times [start, end] =  [0, 1682037]
         [0, 16595016]
         [total, avg] = [14912979, 14] nano seconds
Atomic fetch and add:
         iterations: 1000000, result: 1000000
         times [start, end] =  [0, 16617178]
         [0, 31499571]
         [total, avg] = [14882393, 14] nano seconds
Mutex post-increment:
         iterations: 1000000, result: 1000000
         times [start, end] =  [0, 31526810]
         [0, 68515763]
         [total, avg] = [36988953, 36] nano seconds
RWlock post-increment:
         iterations: 1000000, result: 1000000
         times [start, end] =  [0, 68547649]
         [0, 110877351]
         [total, avg] = [42329702, 42] nano seconds

这是 gcc 编译：

g++ -o atomicVsNonAtomic.o -c -march=i686 -O2 -I. atomicVsNonAtomic.cc
g++ -o atomicVsNonAtomic atomicVsNonAtomic.o -lrt -lpthread

以及相关信息和版本：

# gcc --version
gcc (GCC) 4.3.2
Copyright (C) 2008 Free Software Foundation, Inc.
This is free software; see the source for copying conditions.  There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.

# uname -a
Linux gtcba2v1 2.6.32.12-0.7-default #1 SMP 2010-05-20 11:14:20 +0200 i686 i686 i386 GNU/Linux

现在是实际问题:) 原子操作这么慢是正常的吗？

一百万次迭代的差异是：

  • 简单的后增量：431 纳秒
  • 原子获取和添加操作：14882393 纳秒

当然我知道原子操作应该更昂贵，但这似乎被夸大了。为了完整起见，我还检查了 pthread 互斥锁和 rwlock。至少原子操作比 pthread 操作快，但我仍然想知道我是否做错了什么。我无法在不指定选项的情况下链接它-march=i686，也许这有影响？

更新：

我拿出了-O2编译器优化，能够得到更连贯的结果，如下所示：

# ./atomicVsNonAtomic
Non-Atomic pre-increment:
         iterations: 1000000, result: 1000000
         times [start, end] =  [0, 1647303]
         [0, 4171164]
         [total, avg] = [2523861, 2] nano seconds
Non-Atomic post-increment:
         iterations: 1000000, result: 1000000
         times [start, end] =  [0, 4310230]
         [0, 7262704]
         [total, avg] = [2952474, 2] nano seconds
Atomic add and fetch:
         iterations: 1000000, result: 1000000
         times [start, end] =  [0, 7285996]
         [0, 25919067]
         [total, avg] = [18633071, 18] nano seconds
Atomic fetch and add:
         iterations: 1000000, result: 1000000
         times [start, end] =  [0, 25941677]
         [0, 44544234]
         [total, avg] = [18602557, 18] nano seconds
Mutex post-increment:
         iterations: 1000000, result: 1000000
         times [start, end] =  [0, 44573933]
         [0, 82318615]
         [total, avg] = [37744682, 37] nano seconds
RWlock post-increment:
         iterations: 1000000, result: 1000000
         times [start, end] =  [0, 82344866]
         [0, 125124498]
         [total, avg] = [42779632, 42] nano seconds
4

3 回答 3

20

答案是 GCC 优化了你的非原子增量。当它看到如下循环时：

for (int i=0; i<N; i++) x++;

它将其替换为：

x += N;

这可以在生成的程序集中看到，其中包含：

call    clock_gettime
leal    -32(%ebp), %edx
addl    $1000000, -40(%ebp)     <- increment by 1000000
adcl    $0, -36(%ebp)
movl    %edx, 4(%esp)
movl    $2, (%esp)
call    clock_gettime

所以你不是在衡量你认为自己是什么。

您可以设置变量volatile来防止这种优化。在我的电脑上，这样做之后，非原子访问的速度大约是原子访问的 8 倍。当使用 32 位变量而不是 64 位（我正在编译为 32 位）时，差异下降到大约 3 倍。

于 2012-07-23T08:40:18.923 回答
6

我猜 gcc 正在将您的非原子增量操作优化为类似

val += numIterations;

您说 10^6 增量需要 431 纳秒，每次循环迭代需要 0.000431 ns。在 4 GHz 处理器上，时钟周期为 0.25 ns，因此很明显循环正在被优化掉。这解释了您所看到的巨大性能差异。

编辑：您测量了一个原子操作需要 14 ns - 再次假设一个 4 GHz 处理器，它可以工作 56 个周期，这相当不错！

于 2012-07-23T08:40:38.847 回答
1

任何同步机制的慢度都无法用单个线程来衡量。像 POSIX 互斥锁/Windows 临界区这样的单进程同步对象只有在它们被竞争时才真正花费时间。

您将不得不引入几个线程 - 执行反映实际应用程序时间的其他工作 - 以便同步方法真正了解它需要多长时间。

于 2012-07-23T09:35:10.870 回答