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在5.3 mutex一节中,开头有这么一句话
在正常模式中,等待者按照 FIFO 的顺序排队获取锁
我注意到这个是官方注释 在下文的unlockSlow方法中,正常唤醒逻辑的代码是runtime_Semrelease(&m.sema, false, 1),您的注释是
unlockSlow
runtime_Semrelease(&m.sema, false, 1)
唤醒一个阻塞的 goroutine,但不是唤醒第一个等待者
这两个说法是不是冲突了?
The text was updated successfully, but these errors were encountered:
抱歉回复晚了, 刚才看了一下, 这个注释确实是有问题的.
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问题描述
在5.3 mutex一节中,开头有这么一句话
我注意到这个是官方注释
在下文的
unlockSlow
方法中,正常唤醒逻辑的代码是runtime_Semrelease(&m.sema, false, 1)
,您的注释是这两个说法是不是冲突了?
The text was updated successfully, but these errors were encountered: