还是互斥锁的有关问题

还是互斥锁的问题

C/C++ code

#include<pthread.h>#include<stdlib.h>#include<unistd.h>#include<stdio.h>#include<errno.h>pthread_mutex_t mutex=PTHREAD_MUTEX_INITIALIZER;int lock_var=0;time_t end_time;int sum;void pthread1(void *arg);void pthread2(void *arg);void pthread3(void *arg);int main(int arg,char *argv[]){    pthread_t id1,id2,id3;    pthread_t mon_th_id;    int ret;    sum=10;    end_time=time(NULL)+10;    //pthread_mutex_init(&mutex,NULL);    ret=pthread_create(&id1,NULL,(void *)pthread1,NULL);    if(ret!=0)        perror("pthread cread1");    ret=pthread_create(&id2,NULL,(void *)pthread2,NULL);    if(ret!=0)        perror("pthread cread2");    ret=pthread_create(&id3,NULL,(void *)pthread3,NULL);    if(ret!=0)        perror("pthread cread3");    pthread_join(id1,NULL);    pthread_join(id2,NULL);    pthread_join(id3,NULL);    exit(0);}void pthread1(void *arg){    int i;    printf("come in thread1\n");    while(time(NULL)<end_time)    {        if(pthread_mutex_lock(&mutex)!=0)//lock        {            perror("pthread_mutex_lock 55 line");        }        else        {            printf("pthread1:pthread1 lock the variable\n");            for(i=0;i<2;i++)            {                sleep(2);                lock_var++;                printf("in thread1 and lock_var=%d\n",lock_var);            }        }        if(pthread_mutex_unlock(&mutex)!=0)//unlock        {            perror("pthread_mutex_unlock");        }        else        {            printf("pthread1:pthread1 unlock the variable\n");            sleep(1);        }    }}void pthread2(void *arg){    printf("come in thread2\n");    int nolock=0;    int ret;    while(time(NULL)<end_time)    {        ret=pthread_mutex_trylock(&mutex);//try lock        if(ret==EBUSY)            printf("pthread2:the vaiable is locked by pthread1\n");        else        {            if(ret!=0)            {                perror("pthread_mutex_trylock 93 line");                exit(1);            }            else                printf("pthread2:pthread2 got lock.The variable is %d\n",lock_var);            if(pthread_mutex_unlock(&mutex)!=0)//unlock            {                perror("pthread_mutex_unlock");            }            else                printf("pthread2:pthread2 unlock the variable\n");        }        sleep(1);    }}void  pthread3(void *arg){    printf("come in thread3 \n");    int nolock=0;    int ret;    while(time(NULL)<end_time)    {        ret=pthread_mutex_trylock(&mutex);        if(ret!=EBUSY)            printf("pthread3:variable is locked by pthread1 or 2\n");        else        {            if(ret!=0)            {                perror("pthread_mutex_trylock 123 line");                exit(1);            }            else                 printf("pthread3:pthread3 got lock.The variable is %d\n",lock_var);                        if(pthread_mutex_unlock(&mutex)!=0)                perror("pthread_mutex_unlock");            else                 printf("pthread3:pthread2 unlock the variable\n");        }        sleep(3);        }}

我的机子上的运行结果为
come in thread3 
pthread3:variable is locked by pthread1 or 2
come in thread2
pthread2:the vaiable is locked by pthread1
come in thread1
pthread2:the vaiable is locked by pthread1
pthread2:the vaiable is locked by pthread1
pthread_mutex_trylock 123 line: Success

我怎么感觉很奇怪啊 不是应该先进入thread1 吗 如果没有进入thread1枷锁的话，那thread3又怎么会测试得到变量已经被threa1 或threa2 加过锁了？、

很奇怪测试结果怎么是这样的
居然是最后进入thread1的

球高手赐教

[解决办法]
好像是你无法保证那个线程先执行。

可以设定线程优先级

[解决办法]
1. thread3 里为什么写成 if(ret!=EBUSY)
2. come in thread3 在第一行打印是有可能的, 实际上 printf("come in thread1\n"); printf("come in thread2\n");都可能, 因为创建线程后只是创建, 先执行谁不一定
3. pthread2:the vaiable is locked by pthread1
come in thread1
先打印2, 再1也是可能的, 虽然先在1里lock但print的顺序不一定的
正常的语句是顺序执行的, 但2个线程的print就不一定了, 因为print是不可重入的,因为输出的只有一个硬件
楼主可以试试一个 printf, 一个perror也许会按顺序走,因为他们从软件上是输出到2个设备
 

查看更多 下一篇