c/c++: 多线程编程基础讲解（五）

http://blog.csdn.net/lzx_bupt/article/details/6913151

最近喜欢听大学听到的老歌，deutschland 德国世界杯时候流行的，据说不是主题曲但是比主题曲还要火。

本篇进入难点了，mutex互斥锁概念，mutex=mutual exclusion的缩写，顺便说一句：以前老师都爱用缩写，也不跟同学说全称，这尼玛能理解深刻么！下文是用法：

#include <iostream>
#include <pthread.h>//按规矩不能少

using namespace std;

#define NUM_THREADS 5

int sum = 0;//定义个全局变量，让所有线程进行访问，这样就会出现同时写的情况，势必会需要锁机制；
pthread_mutex_t sum_mutex;

void* say_hello(void* args)
{
    cout << "hello in thread " << *((int *)args) << endl;
    pthread_mutex_lock (&sum_mutex);//修改sum就先加锁，锁被占用就阻塞，直到拿到锁再修改sum；
    cout << "before sum is " << sum << " in thread " << *((int *)args) << endl;
    sum += *((int *)args);
    cout << "after sum is " << sum << " in thread " << *((int *)args) << endl;
    pthread_mutex_unlock (&sum_mutex);//完事后解锁，释放给其他线程使用;

    pthread_exit(0);//退出随便扔个状态码
}

int main()
{
    pthread_t tids[NUM_THREADS];
    int indexes[NUM_THREADS];
    //下三句是设置线程参数没啥可说的
    pthread_attr_t attr;
    pthread_attr_init(&attr);
    pthread_attr_setdetachstate(&attr, PTHREAD_CREATE_JOINABLE);

    pthread_mutex_init (&sum_mutex, NULL);//这句是对锁进行初始化，必须的；

    for(int i = 0; i < NUM_THREADS; ++i)
    {
        indexes[i] = i;
        int ret = pthread_create( &tids[i], &attr, say_hello, (void *)&(indexes[i]) );//5个进程去你们去修改sum吧哈哈；
        if (ret != 0)
        {
           cout << "pthread_create error: error_code=" << ret << endl;
        }
    }

    pthread_attr_destroy(&attr);//删除参数变量

    void *status;
        for (int i = 0; i < NUM_THREADS; ++i)
    {
                int ret = pthread_join(tids[i], &status);
        if (ret != 0)
        {
            cout << "pthread_join error: error_code=" << ret << endl;
        }
    }

    cout << "finally sum is " << sum << endl;

    pthread_mutex_destroy(&sum_mutex);//注销锁，可以看出使用pthread内置变量神马的都对应了销毁函数，估计是内存泄露相关的吧；
}

惯例：g++ -lpthread -o ex_mutex ex_mutex.cpp

运行：

hello in thread 4
before sum is 0 in thread 4
after sum is 4 in thread 4
hello in thread 3
before sum is 4 in thread 3
after sum is 7 in thread 3
hello in thread 2
before sum is 7 in thread 2
after sum is 9 in thread 2
hello in thread 1
before sum is 9 in thread 1
after sum is 10 in thread 1
hello in thread 0
before sum is 10 in thread 0
after sum is 10 in thread 0
finally sum is 10

发现个现象，thread4先运行，很诡异吧而i是从0递增的，所以呢多线程的顺序是混乱的，混乱就是正常；只要sum访问及修改是正常的，就达到多线程的目的了，运行顺序不能作为参照；