系统中使用了
System V IPC 方式的进程锁,今天在压测的时候发现在单进程的情况下,依然有很多semop的系统调用,在单进程程的情况下,不存在锁冲突,经过了futex优化,应该不会有semop系统调用的,就仔细研究了一把这个问题,最终发现 System V IPC下的锁是没有经过 futex优化的,不管有没有锁争用,都必然有semop的系统调用(加锁和解锁各一次),而经过futex优化的NPTL实现在无冲突的时候是不会导致系统调用的,直接在应用层就解决了,两者在性能上差异较大。
为了验证这个差异,特意写了个简单的程序做了一个测试,每个测试线程均完成100万次的简单运算,多个线程之间共享一个锁,测试代码如下:
- #include <sys/types.h>
- #include <sys/ipc.h>
- #include <sys/sem.h>
- #include <pthread.h>
- #include <semaphore.h>
- #include <sys/time.h>
- #include <errno.h>
- #include <stdio.h>
- #include <stdlib.h>
- int g_iLockType = -1;
- int g_iSemID = -1;
- sem_t g_sem;
- long g_result = 1;
- int createSemMutex(key_t iKey)
- {
- #if defined(__GNU_LIBRARY__) && !defined(_SEM_SEMUN_UNDEFINED)
- /* union semun is defined by including <sys/sem.h> */
- #else
- /* according to X/OPEN we have to define it ourselves */
- union semun
- {
- int val; /* value for SETVAL */
- struct semid_ds *buf; /* buffer for IPC_STAT, IPC_SET */
- unsigned short *array; /* array for GETALL, SETALL */
- /* Linux specific part: */
- struct seminfo *__buf; /* buffer for IPC_INFO */
- };
- #endif
- int iSemID;
- union semun arg;
- u_short array[2] = { 0, 0 };
- //生成信号量集, 包含两个信号量
- if ( (iSemID = semget( iKey, 2, IPC_CREAT | IPC_EXCL | 0666)) != -1 )
- {
- arg.array = &array[0];
- //将所有信号量的值设置为0
- if ( semctl( iSemID, 0, SETALL, arg ) == -1 )
- {
- return -1;
- }
- }
- else
- {
- //信号量已经存在
- if ( errno != EEXIST )
- {
- return -2;
- }
- //连接信号量
- if ( (iSemID = semget( iKey, 2, 0666 )) == -1 )
- {
- return -3;
- }
- }
- return iSemID;
- }
- int SemMutexLock(int iSemID)
- {
- struct sembuf sops[3] = { {0, 0, SEM_UNDO}, {1, 0, SEM_UNDO}, {0, 1, SEM_UNDO} };
- size_t nsops = 3;
- int ret = -1;
- do
- {
- ret=semop(iSemID,&sops[0],nsops);
- } while ((ret == -1) &&(errno==EINTR));
- return ret;
- }
- int SemMutexUnLock(int iSemID)
- {
- struct sembuf sops[1] = { {0, -1, SEM_UNDO} };
- size_t nsops = 1;
- int ret = -1;
- do
- {
- ret=semop(iSemID,&sops[0],nsops);
- } while ((ret == -1) &&(errno==EINTR));
- return ret;
- }
- void* threadFunc(void* ptr)
- {
- int iThread = (long)(ptr);
- printf("thread %d start.\n",iThread);
- if (g_iLockType == 1)
- {
- for (int i = 0 ; i < 1000000 ; ++i)
- {
- if (0 != SemMutexLock(g_iSemID))
- {
- printf("SemMutexLock fail in thread %d.\n",iThread);
- return NULL;
- }
-
- for (int j = 0 ; j < 100 ; ++j)
- {
- int tmp = iThread * i * i * 997 % (i + 7);
- g_result = iThread * tmp * tmp * 997 % (i + 31);
- g_result = g_result * 1997 % 91;
- }
-
- if (0 != SemMutexUnLock(g_iSemID))
- {
- printf("SemMutexUnLock fail in thread %d.\n",iThread);
- return NULL;
- }
- }
- }
- else
- {
- for (int i = 0 ; i < 1000000 ; ++i)
- {
- if (0 != sem_wait(&g_sem))
- {
- printf("sem_wait fail in thread %d.\n",iThread);
- return NULL;
- }
-
- for (int j = 0 ; j < 100 ; ++j)
- {
- int tmp = iThread * i * i * 997 % (i + 7);
- g_result = iThread * tmp * tmp * 997 % (i + 31);
- g_result = g_result * 1997 % 91;
- }
-
- if (0 != sem_post(&g_sem))
- {
- printf("sem_post fail in thread %d.\n",iThread);
- return NULL;
- }
- }
- }
- return NULL;
- }
- int main(int argc, char* argv[])
- {
- if (argc != 3)
- {
- printf("usuage:%s lockType threadCount\n",argv[0]);
- return -1;
- }
-
- g_iLockType = atoi(argv[1]);
- int iThreadCount = atoi(argv[2]);
- printf("locktype=%d ,threadCount=%d\n",g_iLockType,iThreadCount);
- key_t semKey = ftok("sem_mutex_nptl_test",100);
- g_iSemID = createSemMutex(semKey);
- if (g_iSemID < 0)
- {
- printf("createSemMutex fail.return %d\n",g_iSemID);
- return 2;
- }
-
- if (0 != sem_init(&g_sem, 1, 1))
- {
- printf("sem_init fail.\n");
- return 3;
- }
-
- pthread_t pt[1000];
-
- struct timeval start;
- struct timeval end;
-
- gettimeofday(&start,NULL);
- for (int i = 0 ; i < iThreadCount ; ++i)
- {
- if (0 != pthread_create(pt+i,0,threadFunc,(void*)i))
- {
- printf("create thread %d fail.",i);
- return 3;
- }
- }
-
- for (int i = 0 ; i < iThreadCount ; ++i)
- {
- pthread_join(pt[i],NULL);
- }
-
- gettimeofday(&end,NULL);
-
- printf("time cost = %f ms\n", ((end.tv_sec - start. tv_sec) * 1000000 + (end.tv_usec - start.tv_usec)) / 1000.0);
- return 0;
- }
测试结果如下:
锁类型
竞争线程数
耗时(毫秒)
System V
IPC
1
9593.426
NPTL 1
9161.838
System V
IPC
2
24292.156
NPTL 2
21991.775
System V IPC
3
38749.863
NPTL 3
33610.102
System V
IPC
4
50919.030
NPTL 4
45871.873
System V IPC
5
61978.562
NPTL 5
56515.495
System V
IPC
10
126402.621
NPTL 10
112752.429
通过strace观察,发现NPTL的锁是没有semop调用的,只有futex_*** 调用,性能上NPTL比System V IPC 高出10%以上,具体的性能差异与加解锁之间的业务代码的复杂程度相关,建议在使用进程锁的时候,选用NPTL的锁实现替代旧的System V IPC,以提升效率。
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