workqueue.c 中 rt_workqueue_submit_work bug

RT-Thread 版本：4.1.x，5.1.0 ，其他版本没验证 应该也存在
问题描述
rt_workqueue_submit_work 延迟任务执行，如果在定时到达之前，提前重新调用并传参tick=0，会导致定时到达之后出现bug。

问题复现代码，bsp：qemu-vexpress-a9
main.c 如下

#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <rtthread.h>

struct rt_work test_work;

void work_func(struct rt_work *work, void *work_data)
{
    printf("work_func %lu\n", rt_tick_get_millisecond());
}

int main(void)
{
    rt_thread_mdelay(200);
    printf("Hello RT-Thread!\n");

    rt_work_init(&test_work, work_func, NULL);

    printf("main %lu\n", rt_tick_get_millisecond());
    rt_work_submit(&test_work, rt_tick_from_millisecond(1000));
    rt_thread_mdelay(200);
    rt_work_submit(&test_work, 0);

    return 0;
}

msh输出如下：

 \ | /
- RT -     Thread Operating System
 / | \     5.1.0 build Jan  9 2024 10:46:20
 2006 - 2022 Copyright by RT-Thread team   
[I/SDIO] SD card capacity 65536 KB.
[I/SDIO] switching card to high speed failed!
[I/FileSystem] file system initialization done!

msh />Hello RT-Thread!
main 1220
work_func 1420
(queue != RT_NULL) assertion failed at function:_delayed_work_timeout_handler, line number:176 
backtrace:
please use: addr2line -e rtthread.elf -a -f 6003fe10 6004c848 60057fb4 60053454 6005361c       

问题分析：

_workqueue_submit_work 函数当 ticks传参等于0 的时候，没有停止上次的定时器。导致任务触发后，定时还在继续。当定时器触发之后任务已经执行完成。

解决办法：
修改 _workqueue_submit_work 函数，

    if (ticks == 0)
    {
        rt_list_insert_after(queue->work_list.prev, &(work->list));
        work->flags |= RT_WORK_STATE_PENDING;
        work->workqueue = queue;

        if (work->flags & RT_WORK_STATE_SUBMITTING)
        {
            rt_timer_stop(&(work->timer));
            rt_timer_detach(&(work->timer));
            work->flags &= ~RT_WORK_STATE_SUBMITTING;
        }
.
.
.

Start in workqueue.c at _workqueue_submit_work and reproduce the issue on the qemu-vexpress-a9 BSP using the provided main.c sequence. Verify that resubmitting delayed work with tick=0 does not leave the old timer active and that the later timeout produces no assertion.