Comments (3)
LittlePeng
commented on May 20, 2024
看设计是1ms 是为了检查 timeout用的。
其实可以这样：在 loop 前计算一下所有请求中的最小超时时间min_timeout ，epoll_wait 超时指定为 min_timeout 就能解决问题
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cauhn
commented on May 20, 2024
一般这个timeout值都是固定的，可以根据自身情况设置。
如果在loop前计算所有请求的最小超时时间，则需要查询Epoll_t->pTimeout里最小的timeout时间，这个时间复杂度是o(n) n=60*1000，这个还是不太划算的。
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yplusplus
commented on May 20, 2024
比较折中的方案是检查Xms内有没有到期的定时器，有的话wait到最近到期的定时器；没有的话直接wait Xms
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