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本帖最后由 chuwangmvp 于 2012-4-11 16:14 编辑
#include <stdio.h>
#include <pthread.h>
#include <unistd.h>
void *clean(void *arg)
{
printf("cleanup :%s \n",(char *)arg);
return (void *)0;
}
void *thr_fn1(void *arg)
{
printf("thread 1 start \n");
pthread_cleanup_push( (void*)clean,"thread 1 first handler");
pthread_cleanup_push( (void*)clean,"thread 1 second hadler");
printf("thread 1 push complete \n");
if(arg)
{
return((void *)4);
}
pthread_cleanup_pop(0);
pthread_cleanup_pop(0);
return (void *)4;
}
void *thr_fn2(void *arg)
{
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(arg)
{
pthread_exit((void *)2);
}
pthread_cleanup_pop(0);
pthread_cleanup_pop(0);
pthread_exit((void *)2);
}
int main(void)
{
int err;
pthread_t tid1,tid2;
void *tret;
err=pthread_create(&tid1,NULL,thr_fn1,(void *)1);
if(err!=0)
{
printf("error .... \n");
return -1;
}
err=pthread_create(&tid2,NULL,thr_fn2,(void *)1);
if(err!=0)
{
printf("error .... \n");
return -1;
}
err=pthread_join(tid1,&tret);
if(err!=0)
{
printf("error .... \n");
return -1;
}
printf("thread 1 exit code %d \n",(int)tret);
err=pthread_join(tid2,&tret);
if(err!=0)
{
printf("error .... ");
return -1;
}
printf("thread 2 exit code %d \n",(int)tret);
return 1;
}
运行结果:
thread 2 start
thread 2 push complete
clean up: thread 2 second handler
clean up: thread 2 first handler
thread 1 start
thread 1 push complete
thread 1 exit code 4
thread 2 exit code 2
我当时猜想的是:
thread 1 start
thread 1 push complete
thread 2 start
thread 2 push complete
clean up: thread 2 second handler
clean up: thread 2 first handler
thread 1 exit code 4
thread 2 exit code 2
问题:
为什么是thread 2 先执行了的。。。。 |
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