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Posted to dev@brpc.apache.org by GitBox <gi...@apache.org> on 2021/07/24 13:49:33 UTC

[GitHub] [incubator-brpc] charyxiao opened a new issue #1497: butex_wait & remind相关问题

charyxiao opened a new issue #1497:
URL: https://github.com/apache/incubator-brpc/issues/1497


   butex_wait 函数看代码有几点疑问
   1. wait_for_butex 为什么不原地执行？为什么要放到remind里
   2. wait_for_butex之后的sched如果切换的bthread上下文的代码断点在task_runner的thread_return = m->fn(m->arg)中，那就不会执行g->_last_context_remained的信息 相反会重新调用g->set_remained，将wait_for_butex冲掉
   ![image](https://user-images.githubusercontent.com/5407327/126870356-8dfec3bf-a3b2-4cf5-8881-2fb15381ea7c.png)
   
   
   
   


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[GitHub] [incubator-brpc] jenrryyou commented on issue #1497: butex_wait & remained相关问题

Posted by GitBox <gi...@apache.org>.
jenrryyou commented on issue #1497:
URL: https://github.com/apache/incubator-brpc/issues/1497#issuecomment-891484786


   有同样的疑惑。另外如果切换的后的bthread再次调用butex_wait，会不会覆盖task_group的remained回调？


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[GitHub] [incubator-brpc] charyxiao commented on issue #1497: butex_wait & remained相关问题

Posted by GitBox <gi...@apache.org>.
charyxiao commented on issue #1497:
URL: https://github.com/apache/incubator-brpc/issues/1497#issuecomment-894795553


   第二点 调用sched(&g) 不一定是调度新的bthread的 也可能是旧的bthread的 举个更极端的例子 如果没有新的bthread呢


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[GitHub] [incubator-brpc] jenrryyou commented on issue #1497: butex_wait & remained相关问题

Posted by GitBox <gi...@apache.org>.
jenrryyou commented on issue #1497:
URL: https://github.com/apache/incubator-brpc/issues/1497#issuecomment-891487709


   对第一点有同样的疑问；
   第二点我的理解是：wait_for_butex里直接调用的是`TaskGroup::sched(&g)`，会调度新的`bthread`，一定会从头执行`task_runner`，把之前设置的回调执行；
   ![image](https://user-images.githubusercontent.com/4698734/127952379-1fc7d2b3-af2f-48ef-9e2a-d23075c77afd.png)
   


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[GitHub] [incubator-brpc] jenrryyou removed a comment on issue #1497: butex_wait & remained相关问题

Posted by GitBox <gi...@apache.org>.
jenrryyou removed a comment on issue #1497:
URL: https://github.com/apache/incubator-brpc/issues/1497#issuecomment-891484786


   有同样的疑惑。另外如果切换的后的bthread再次调用butex_wait，会不会覆盖task_group的remained回调？


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[GitHub] [incubator-brpc] jenrryyou edited a comment on issue #1497: butex_wait & remained相关问题

Posted by GitBox <gi...@apache.org>.
jenrryyou edited a comment on issue #1497:
URL: https://github.com/apache/incubator-brpc/issues/1497#issuecomment-891487709


   对第一点有同样的疑问；
   第二点我的理解是：`butex_wait`里直接调用的是`TaskGroup::sched(&g)`，会调度新的`bthread`，一定会从头执行`task_runner`，把之前设置的回调执行；
   ![image](https://user-images.githubusercontent.com/4698734/127952379-1fc7d2b3-af2f-48ef-9e2a-d23075c77afd.png)
   


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[GitHub] [incubator-brpc] ZeroLiu2018 commented on issue #1497: butex_wait & remained相关问题

Posted by GitBox <gi...@apache.org>.
ZeroLiu2018 commented on issue #1497:
URL: https://github.com/apache/incubator-brpc/issues/1497#issuecomment-962740435


   > 第二点 调用sched(&g) 不一定是调度新的bthread的 也可能是旧的bthread的 举个更极端的例子 如果没有新的bthread呢
   
   我说一下我的理解，先说第二点：除了 task_runner ,在 sche_to 里面，jump_stack 完成返回后，也会执行 _last_context_remained,也就是说在真正切到另一个 bthread 跑逻辑前，都会执行 _last_context_remained。
   然后就是第一点，在第一点之前你可以看看切换 bthread 的逻辑，切换 bthread 之前有个 set_remained, 这个 remianed 的作用是将当前 bthread 放到队列。它也同样不是在原地执行，我想两者的原因是一样的(wait_for_butex 里面同样有将当前 bthread 入队的逻辑): 需要调度出新的 bthread 才能将旧 bthread 入列，否则，stack 切换可能很难处理。


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