[GitHub] [dubbo] mxh-kkk commented on issue #12481: 为什么 dubbo 的netty workerGroup 线程数默认是 Math.min(Runtime.getRuntime().availableProcessors() + 1, 32)

["mxh-kkk (via GitHub)" <gi...@apache.org>]

mxh-kkk commented on issue #12481:
URL: https://github.com/apache/dubbo/issues/12481#issuecomment-1583850376

   > 大部分应用是IO密集。Runtime.getRuntime().availableProcessors()*2 这个一般是boss线程吧，work线程会比较多，比如tomcat默认200+
   
   我不太理解您说的，boss线程数不是1吗？Runtime.getRuntime().availableProcessors()*2 比 Runtime.getRuntime().availableProcessors() + 1 数量多吧？


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