Golang：永远的通道

Question

有个问题，这里发生了什么？

forever := make(chan bool)

log.Printf(" [*] Waiting for messages. To exit press CTRL+C")
<-forever

Answer 1

该代码创建一个无缓冲通道，并尝试从中接收。

而且因为没有人在上面发送任何东西，所以它本质上是一个永远阻塞的操作。

目的是为了防止 goroutine 结束/返回，很可能是因为有其他 goroutines 同时做一些工作或者他们等待某些事件或传入消息（就像您的日志消息所说）。

而需要的是，如果没有这个，应用程序可能会在不等待其他 goroutine 的情况下退出。也就是说，如果 main goroutine 结束，程序也会结束。引用自 Spec: Program execution:

Program execution begins by initializing the main package and then invoking the function main. When that function invocation returns, the program exits. It does not wait for other (non-main) goroutines to complete.

查看此答案以了解更多类似技巧：

有关频道的介绍，请参阅

Answer 2

问题中的代码可能来自RabbitMQ golang教程here。

这是其中的更扩展部分，其中包含我自己的一些建议：

    ...
    // create an unbuffered channel for bool types.
    // Type is not important but we have to give one anyway.
    forever := make(chan bool) 

    // fire up a goroutine that hooks onto msgs channel and reads
    // anything that pops into it. This essentially is a thread of
    // execution within the main thread. msgs is a channel constructed by
    // previous code.
    go func() {
        for d := range msgs {
            log.Printf("Received a message: %s", d.Body)
        }
    }()

    log.Printf(" [*] Waiting for messages. To exit press CTRL+C")
    // We need to block the main thread so that the above thread stays
    // on reading from msgs channel. To do that just try to read in from
    // the forever channel. As long as no one writes to it we will wait here.
    // Since we are the only ones that know of it it is guaranteed that
    // nothing gets written in it. We could also do a busy wait here but
    // that would waste CPU cycles for no good reason.
    <-forever

Golang：永远的通道

Golang: forever channel

channel

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