英文:
Read from one channel and write to another within one select-case
问题
考虑以下这段Go代码，它会立即调用worker.DoWork()，然后每分钟调用一次：
triggerChan := make(chan time.Time, 1)
triggerChan <- time.Now() // 我们不想等一分钟才调用worker.DoWork()的第一次调用。
ticker := time.NewTicker(time.Minute)
defer ticker.Stop()
for {
select {
case triggerChan <- <-ticker.C:
case <-triggerChan:
worker.DoWork()
}
}
在第一个select之前，一个值被写入triggerChan，以便将其定向到case <-triggerChan，从而立即调用worker.DoWork()。然而，如果ticker.C已经有一个可用的值（如果我们使用time.Microsecond而不是time.Minute，这种情况并不太可能），会发生什么？运行时会注意到triggerChan <- <-ticker.C会阻塞吗？还是它会快乐地软锁定，不知道对<-ticker.C的结果要做什么处理？它是否意识到这种"从通道到通道"的情况？
英文:
Consider this Go code which shall call worker.DoWork() immediately and then once every minute:
triggerChan := make(chan time.Time, 1)
triggerChan <- time.Now() // We don't want to wait a minute for the first call of worker.DoWork() to happen.
ticker := time.NewTicker(time.Minute)
defer ticker.Stop()
for {
select {
case triggerChan <- <-ticker.C:
case <-triggerChan:
worker.DoWork()
}
}
A value is written to triggerChan before the first select to direct it into case <-triggerChan to have worker.DoWork() called immediately. However, in the case that ticker.C already had a value available (not all that unlikely if we used something like time.Microsecond instead of time.Minute), what would happen? Would the runtime see that triggerChan <- <-ticker.C would block, or would it happily soft-lock unaware of what is trying to be done with the result of <-ticker.C? Is it aware of this "from channel into channel"-situation?
答案1
得分: 3
select 根据 triggerChan 的状态进行决策，但是它首先评估发送操作的右侧表达式。因此，在进入时，它会等待 <-ticker.C 返回。然后它会从通道中读取第一个值，然后写入它...
英文:
The select decides based on the state of triggerChan, however it first evaluates the right hand side expressions for send operations.
So on entry, it will wait until <-ticker.C returns. Then it will read the first value from the channel, and then write to it...
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