文档中关于semaphore的两个例子在连续执行第二次的时候都会报错

发布于 2021-01-16 15:13:04 浏览：1205 订阅该版

文档中关于semaphore的两个例子在连续执行第二次的时候都会报错：
```
(obj != object) assertion failed at function:rt_object_init, line number:267
```
请问如果正确的回收信号量，或都说正确使用信号量
```c
#include <rtthread.h>

/*
 * Copyright (c) 2006-2020, RT-Thread Development Team
 *
 * SPDX-License-Identifier: Apache-2.0
 *
 * Change Logs:
 * Date           Author       Notes
 * 2021-01-16     Administrator       the first version
 *
 * //RT-Thread 编程指南 65RT-Thread 编程指南 5.1 节 信号量
 */

#define THREAD_PRIORITY 25
#define THREAD_TIMESLICE 5

/* 指 向 信 号 量 的 指 针 */
static rt_sem_t dynamic_sem = RT_NULL;

/**********thread1*************/
ALIGN(RT_ALIGN_SIZE)
static char thread1_stack[1024];
static struct rt_thread thread1;

static void rt_thread1_entry(void *parameter)
{

static rt_uint32_t count = 0;
    while (1)
    {
        if (count <= 100)
        {
            count++;
        }
        else{
            rt_sem_delete(dynamic_sem);
            return;
        }

/* count 每 计 数 10 次， 就 释 放 一 次 信 号 量 */
        if (0 == (count % 10))
        {
            rt_kprintf("t1 release a dynamic semaphore.\n");
            rt_sem_release(dynamic_sem);
        }
    }
}

/**********thread2*************/
ALIGN(RT_ALIGN_SIZE)
static char thread2_stack[1024];
static struct rt_thread thread2;

static void rt_thread2_entry(void *parameter)
{
    static rt_err_t result;
    static rt_uint32_t number = 0;
    while(1)
    {
        /* 永 久 方 式 等 待 信 号 量， 获 取 到 信 号 量， 则 执 行 number 自 加 的 操 作 */
        result = rt_sem_take(dynamic_sem, RT_WAITING_FOREVER);
        //信号量被删除的时候返回RT_ERROR，到时候可以做相应的操作

if (result != RT_EOK)
        {
            rt_kprintf("t2 take a dynamic semaphore, failed.\n");
            return;
        }
        else
        {
            number++;
            rt_kprintf("t2 take a dynamic semaphore. number = %d\n", number);
        }
    }
}

/* 信 号 量 示 例 的 初 始 化 */
int semaphore_sample(void)
{
    /* 创 建 一 个 动 态 信 号 量， 初 始 值 是 0 */
    dynamic_sem = rt_sem_create("dsem", 0, RT_IPC_FLAG_FIFO);

if (dynamic_sem == RT_NULL)
    {
        rt_kprintf("create dynamic semaphore failed.\n");
        return -1;
    }
    else
    {
        rt_kprintf("create done. dynamic semaphore value = 0.\n");
    }

rt_thread_init(
            &thread1,
            "thread1",
            rt_thread1_entry,
            RT_NULL,
            &thread1_stack[0],
            sizeof(thread1_stack),
            THREAD_PRIORITY,
            THREAD_TIMESLICE);

rt_thread_startup(&thread1);

rt_thread_init(
            &thread2,
            "thread2",
            rt_thread2_entry,
            RT_NULL,
            &thread2_stack[0],
            sizeof(thread2_stack),
            THREAD_PRIORITY-1,
            THREAD_TIMESLICE);

rt_thread_startup(&thread2);

return 0;
}

/* 导 出 到 msh 命 令 列 表 中 */
MSH_CMD_EXPORT(semaphore_sample, semaphore sample);

/* res
create done. dynamic semaphore value = 0.
t1 release a dynamic semaphore.
t2 take a dynamic semaphore. number = 1
t1 release a dynamic semaphore.
t2 take a dynamic semaphore. number = 2
t1 release a dynamic semaphore.
t2 take a dynamic semaphore. number = 3
t1 release a dynamic semaphore.
t2 take a dynamic semaphore. number = 4
t1 release a dynamic semaphore.
t2 take a dynamic semaphore. number = 5
t1 release a dynamic semaphore.
t2 take a dynamic semaphore. number = 6
t1 release a dynamic semaphore.
t2 take a dynamic semaphore. number = 7
t1 release a dynamic semaphore.
t2 take a dynamic semaphore. number = 8
t1 release a dynamic semaphore.
t2 take a dynamic semaphore. number = 9
t1 release a dynamic semaphore.
t2 take a dynamic semaphore. number = 10
*/
```