我正在学习 APUE book 2e 中的线程
我认为 pthread_cleanup_pop 函数用于设置 pthread_cleanup_push() 推送的函数是否执行。因此,如果参数为零,则不执行,非零,执行。
但是我在 APUE 中查看了代码图 11.5。那是...
#include "apue.h"
#include <pthread.h>
void
cleanup(void *arg)
{
printf("cleanup: %s\n", (char *)arg);
}
void *
thr_fn1(void *arg)
{
printf("thread 1 start\n");
pthread_cleanup_push(cleanup, "thread 1 first handler");
pthread_cleanup_push(cleanup, "thread 1 second handler");
printf("thread 1 push complete\n");
if (arg)
return((void *)1);
pthread_cleanup_pop(0);
pthread_cleanup_pop(0);
return((void *)1);
}
void *
thr_fn2(void *arg)
{
printf("thread 2 start\n");
pthread_cleanup_push(cleanup, "thread 2 first handler");
pthread_cleanup_push(cleanup, "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)
err_quit("can't create thread 1: %s\n", strerror(err));
err = pthread_create(&tid2, NULL, thr_fn2, (void *)1);
if (err != 0)
err_quit("can't create thread 2: %s\n", strerror(err));
err = pthread_join(tid1, &tret);
if (err != 0)
err_quit("can't join with thread 1: %s\n", strerror(err));
printf("thread 1 exit code %d\n", (int)tret);
err = pthread_join(tid2, &tret);
if (err != 0)
err_quit("can't join with thread 2: %s\n", strerror(err));
printf("thread 2 exit code %d\n", (int)tret);
exit(0);
}
在该程序中,虽然弹出函数的参数为零,但推送的函数被执行(“线程 2 第一个处理程序”,“线程 2 第二个处理程序”,由清理函数打印)
为什么即使将 0 作为参数,pop 也能工作?
我得到 pthread_cleanup_pop 错了吗?