与参数0 pthread_cleanup_pop？ [英] pthread_cleanup_pop with argument 0?

问题描述

在APUE书2E

我学习线程

我想pthread_cleanup_pop功能是通过pthread_cleanup_push设置推（）函数将被执行与否。因此，如果参数为0，它不执行和不为零，执行。

不过，我看了看code图11.5 APUE。这是...

 的#includeapue.h
＃包括LT＆;＆pthreads.h中GT;空虚
清理（​​无效* ARG）
{
    的printf（清理：％S \\ n，（字符*）ARG）;
}无效*
thr_fn1（无效* ARG）
{
    的printf（线程1开始\\ n）;
    pthread_cleanup_push（清理，线程1先处理程序）;
    pthread_cleanup_push（清理，线程1第二处理程序，）;
    的printf（线程1推完成\\ n）;
    如果（ARG）
        收益率（（无效*）1）;
    pthread_cleanup_pop（0）;
    pthread_cleanup_pop（0）;
    收益率（（无效*）1）;
}无效*
thr_fn2（无效* ARG）
{
    的printf（线程2开始\\ n）;
    pthread_cleanup_push（清理，线程2先处理程序）;
    pthread_cleanup_push（清理线程2第二处理程序，）;
    的printf（线程2推完成\\ n）;
    如果（ARG）
        了pthread_exit（（void *的），2）;
    pthread_cleanup_pop（0）;
    pthread_cleanup_pop（0）;
    了pthread_exit（（void *的），2）;
}INT
主要（无效）
{
    INT犯错;
    的pthread_t TID1，TID2;
    无效* TRET;    ERR =在pthread_create（安培; TID1，NULL，thr_fn1，（无效*）1）;
    如果（ERR！= 0）
        err_quit（不能创建线程1：％S \\ n字符串错误（ERR））;
    ERR =在pthread_create（安培; TID2，NULL，thr_fn2，（无效*）1）;
    如果（ERR！= 0）
        err_quit（不能创建线程2：％S \\ n字符串错误（ERR））;
    ERR =在pthread_join（TID1，＆安培; TRET）;
    如果（ERR！= 0）
        err_quit（不能用螺纹联接1％S \\ n字符串错误（ERR））;
    的printf（线程1退出code％d个\\ N（INT）TRET）;
    ERR =在pthread_join（TID2，＆安培; TRET）;
    如果（ERR！= 0）
        err_quit（不能用线加入2％S \\ n字符串错误（ERR））;
    的printf（线程2退出code％d个\\ N（INT）TRET）;
    出口（0）;
}
 

在该程序中，虽然流行的功能具有零作为参数，则推功能excuted（线程2先处理程序，线程2第二处理程序，，由清理功能打印）

为什么流行音乐作品，即使把0参数？

我才拿到pthread_cleanup_pop错了??

解决方案

  

我才拿到pthread_cleanup_pop错了？

没有，你没有。

您code执行清除处理程序的原因，那是你永远的控制达到 pthread_cleanup_pop（0）。相反，你总是执行在 thr_fn1 收益和了pthread_exit 在 thr_fn2 。

尝试建立与 pthread_create的线程（安培; TID1，NULL，thr_fn1，（无效*）0）; 而不是 1 。当我这样做，我得到（预期）：

 螺纹1日开始
线程1完全推
线程2开始
线程2完全推
线程1退出code 1
线程2退出code 2
 

I am studying thread in APUE book 2e

I thought pthread_cleanup_pop functions is for setting the pushed function by pthread_cleanup_push() to be executed or not. So if the argument is zero, it's not executed and non-zero, executed.

But I looked at the code Figure 11.5 in APUE. That is...

#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);
}

in that program, although pop functions has zero as an argument, the pushed functions are excuted( "thread 2 first handler", "thread 2 second handler", are printed by cleanup function)

why pop works even though put 0 for argument?

did I get pthread_cleanup_pop wrong??

解决方案

did I get pthread_cleanup_pop wrong?

No, you didn't.

The reason your code executed cleanup handlers, is that your control never reaches pthread_cleanup_pop(0). Instead, you always execute return in thr_fn1, and pthread_exit in thr_fn2.

Try creating the threads with pthread_create(&tid1, NULL, thr_fn1, (void *)0); instead of 1. When I do that, I get (expected):

thread 1 start
thread 1 push complete
thread 2 start
thread 2 push complete
thread 1 exit code 1
thread 2 exit code 2

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