muduo网络库源码学习————互斥锁

muduo源码的互斥锁源码位于muduo/base，Mutex.h，进行了两个类的封装，在实际的使用中更常使用MutexLockGuard类，因为该类可以在析构函数中自动解锁，避免了某些情况忘记解锁。代码如下所示：

// Use of this source code is governed by a BSD-style license// that can be found in the License file.//// Author: Shuo Chen (chenshuo at chenshuo dot com)//互斥锁#ifndef MUDUO_BASE_MUTEX_H#define MUDUO_BASE_MUTEX_H#include <muduo/base/CurrentThread.h>#include <boost/noncopyable.hpp>#include <assert.h>#include <pthread.h>namespace muduo{class MutexLock : boost::noncopyable//继承自noncopyable，表示不可拷贝{ public:  MutexLock()  : holder_(0)//构造函数，将holder初始化为0，表示该锁没有被任何线程拥有  {    int ret = pthread_mutex_init(&mutex_, NULL);//初始化互斥锁    assert(ret == 0); (void) ret;  }  ~MutexLock()//析构函数  {    assert(holder_ == 0);//断言该锁没有被任何线程占用，才可以销毁    int ret = pthread_mutex_destroy(&mutex_);    assert(ret == 0); (void) ret;  }  bool isLockedByThisThread()//是否当前线程拥有该锁  {    return holder_ == CurrentThread::tid();//只需判断当前线程的tid是否等于holder_  }  void assertLocked()//断言当前线程拥有该锁  {    assert(isLockedByThisThread());  }  // internal usage  void lock()//加锁  {    pthread_mutex_lock(&mutex_);    holder_ = CurrentThread::tid();//将当前线程的tid保存至holder_  }  void unlock()//解锁  {    holder_ = 0;//holder_清零    pthread_mutex_unlock(&mutex_);  }//获取threadMutex对象  pthread_mutex_t* getPthreadMutex() /* non-const */  {    return &mutex_;  } private:  pthread_mutex_t mutex_;//变量保存  pid_t holder_;//当前使用该锁的线程id};//MutexLockGuard类使用RAII技法封装，在实际应用中这个类更常用class MutexLockGuard : boost::noncopyable{ public:    //explicit只能显式调用  explicit MutexLockGuard(MutexLock& mutex): mutex_(mutex)  {//构造函数获取资源    mutex_.lock();  }//在对象生存期结束的时候利用析构函数可以实现自动解锁  ~MutexLockGuard()  {//析构函数释放资源    mutex_.unlock();  } private:  MutexLock& mutex_;//整个对象结束的时候mutex_并没有结束(引用)};}// Prevent misuse like:// MutexLockGuard(mutex_);// A tempory object doesn't hold the lock for long!#define MutexLockGuard(x) error "Missing guard object name"#endif  // MUDUO_BASE_MUTEX_H

测试程序分别使用集中不同的方式往向量中插入数据，从中也可以看出锁的开销，测试代码如下所示：

//互斥锁测试代码#include <muduo/base/CountDownLatch.h>#include <muduo/base/Mutex.h>#include <muduo/base/Thread.h>#include <muduo/base/Timestamp.h>#include <boost/bind.hpp>#include <boost/ptr_container/ptr_vector.hpp>#include <vector>#include <stdio.h>using namespace muduo;using namespace std;MutexLock g_mutex;//声明锁对象vector<int> g_vec;//int动态数组(向量)const int kCount = 10*1000*1000;//常量1千万void threadFunc(){  for (int i = 0; i < kCount; ++i)  {    MutexLockGuard lock(g_mutex);//使用锁    g_vec.push_back(i);//往向量中插入1000w个整数  }}int main(){  const int kMaxThreads = 8;//最多8个线程  g_vec.reserve(kMaxThreads * kCount);//预留8千万个整数(这个所占的内存空间有300多M)  Timestamp start(Timestamp::now());//当前时间戳  for (int i = 0; i < kCount; ++i)  {    g_vec.push_back(i);//往向量中插入1000w个整数  }//输出插入这么多个数的时间  printf("single thread without lock %f\n", timeDifference(Timestamp::now(), start));  start = Timestamp::now();//更新当前时间戳  threadFunc();//调用上面的函数  //和上面一样，计算下插入这么多个数的时间  printf("single thread with lock %f\n", timeDifference(Timestamp::now(), start));  for (int nthreads = 1; nthreads < kMaxThreads; ++nthreads)  {//ptr_vector指针的vector    boost::ptr_vector<Thread> threads;    g_vec.clear();//先清除g_vec向量    start = Timestamp::now();//更新当前时间戳    for (int i = 0; i < nthreads; ++i)    {      threads.push_back(new Thread(&threadFunc));//创建线程      threads.back().start();//启动线程    }    for (int i = 0; i < nthreads; ++i)    {      threads[i].join();    }    //分别输出1到8个线程执行插入操作的时间    printf("%d thread(s) with lock %f\n", nthreads, timeDifference(Timestamp::now(), start));  }}

单独编译后运行结果如下：