Linux 互斥锁的实现

Question

互斥锁是实现线程同步的一种机制，只要在临界区前后对资源加锁就能阻塞其他进程的访问。

#include <iostream>  
#include <pthread.h>  
  
using namespace std;  
  
#define NUM_THREADS 5  
  
int sum = 0; //定义全局变量，让所有线程同时写，这样就需要锁机制  
pthread_mutex_t sum_mutex; //互斥锁  
  
void* say_hello( void* args )  
{  
    cout << "hello in thread " << *(( int * )args) << endl;  
    pthread_mutex_lock( &sum_mutex ); //先加锁，再修改 sum 的值，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 ); //释放锁，供其他线程使用  
    pthread_exit( 0 ); 
    return NULL;  
}  
  
int main()  
{  
    pthread_t tids[NUM_THREADS];  
    int indexes[NUM_THREADS];  
      
    pthread_attr_t attr; //线程属性结构体，创建线程时加入的参数  
    pthread_attr_init( &attr ); //初始化  
    pthread_attr_setdetachstate( &attr, PTHREAD_CREATE_JOINABLE ); //是设置你想要指定线程属性参数，这个参数表明这个线程是可以 join 连接的，join 功能表示主程序可以等线程结束后再去做某事，实现了主程序和线程同步功能  
    pthread_mutex_init( &sum_mutex, NULL ); //对锁进行初始化      
  
    for( int i = 0; i < NUM_THREADS; ++i )  
    {  
        indexes[i] = i;  
        int ret = pthread_create( &tids[i], &attr, say_hello, ( void* )&( indexes[i] ) ); //5 个进程同时去修改 sum  
        if( ret != 0 )  
        {  
            cout << "pthread_create error:error_code=" << ret << endl;  
        }  
    }   
    pthread_attr_destroy( &attr ); //释放内存   
    void *status;  
    for( int i = 0; i < NUM_THREADS; ++i )  
    {  
        int ret = pthread_join( tids[i], &status ); //主程序 join 每个线程后取得每个线程的退出信息 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  
    return 0;
}  

测试结果:

jack@jack:~/coding/muti_thread$ ./pthread_chap5 
hello in thread hello in thread hello in thread 410  
before sum is hello in thread 0 in thread 4  
after sum is 4 in thread 4hello in thread   
  
  
2  
3  
before sum is 4 in thread 1  
after sum is 5 in thread 1  
before sum is 5 in thread 0  
after sum is 5 in thread 0  
before sum is 5 in thread 2  
after sum is 7 in thread 2  
before sum is 7 in thread 3  
after sum is 10 in thread 3  
finally sum is 10  

可知，sum 的访问和修改顺序是正常的，这就达到了多线程的目的了，但是线程的运行顺序是混乱的，混乱就是正常？