linux多线程优先顺序
本帖最后由 cyd411 于 2013-04-03 16:52:15 编辑 输出代码
this is main thread
thread 2 start
thread 2 push complete.
cleanup: thread 2 second handler
cleanup: thread 2 first handler
thread 1 start
thread 1 push complete.
thread1 exit code 1
thread2 exit code 2
这里不明白了，为什么不是先输出
thread 1 start
thread 1 push complete
这个呢？下面我用pthread_join 等待线程1了，不是应该先输出thread 1 start 再输出线程2的相关打印吗？ 为什么先输出的是2呢？
代码如下：
多线程
#include <stdio.h>
#include <pthread.h>
void *Clean(void *pArg)
{
printf("cleanup: %s \n" ,((char *)pArg));
return (void *)0;
}
void *Thread1(void *pArg)
{
printf("thread 1 start \n");
pthread_cleanup_push((void *)Clean, "thread 1 first handler");
pthread_cleanup_push((void *)Clean, "thread 1 second handler");
printf("thread 1 push complete. \n");
if(pArg)
{
/* 这里使用的是return 返回 所以不会调用 pthread_cleanup_push 指定的Clean*/
return (void *)1;
}
pthread_cleanup_pop(0);
pthread_cleanup_pop(0);
return (void *)1;
}
void *Thread2(void *pArg)
{
printf("thread 2 start \n");
pthread_cleanup_push((void *)Clean, "thread 2 first handler");
pthread_cleanup_push((void *)Clean, "thread 2 second handler");
printf("thread 2 push complete. \n");
if(pArg)
{
/*
void pthread_exit(void *rval_ptr)
功能:终止调用线程
rval_ptr:线程退出返回值的指针*/
pthread_exit((void *)2);
}
pthread_cleanup_pop(0);
pthread_cleanup_pop(0);
return (void *)2;
}
int main(void)
{
pthread_t tThreadID1, tThreadID2;
int iRst;
void *tRst;
printf("this is main thread\n");
iRst = pthread_create(&tThreadID1, NULL, (void *)Thread1, (void *)1);
if(iRst!=0)
{
printf("create thread1 error .... \n");
return -1;
}
iRst = pthread_create(&tThreadID2, NULL, (void *)Thread2, (void *)1);
if(iRst!=0)
{
printf("create thread2 error .... \n");
return -1;
}
iRst = pthread_join(tThreadID1, &tRst);
if(iRst != 0)
{
printf("pthread_join thread1 error .... \n");
return -1;
}
else
{
printf("thread1 exit code %d \n", (int)tRst);
}
iRst = pthread_join(tThreadID2, &tRst);
if(iRst != 0)
{
printf("pthread_join thread2 error .... \n");
return -1;
}
else
{
printf("thread2 exit code %d \n", (int)tRst);
}
return 0;
}
