douzhuang2016 2015-09-20 18:23
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收到错误消息:所有goroutine都处于睡眠状态-死锁

How come something simple like this does not work?

c1 := make(chan string)
c1 <- "foo"
fmt.Println(<-c1)

But if I put it in a go routine it works?

c1 := make(chan string)
go func() {
    c1 <- "foo"
}()
fmt.Println(<-c1)

The question might seem simple and stupid, but I'm trying to understand why I can't do this and I don't know of anything better to ask in this case.

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1条回答 默认 最新

  • doo6568 2015-09-20 18:41
    关注

    Channel c1 is an unbuffered channel. Communication succeeds on an unbuffered channel only when both a sender and receiver are ready.

    The line c1 <- "foo blocks forever because no receiver is ready.

    The program with the goroutine works because the sending and receiving goroutines run to the point where the sender and receiver are both ready.

    This program will also work:

    c2 := make(chan string, 1) // <-- note channel size of one
    c2 <- "foo"
    fmt.Println(<-c2)
    

    Channel c2 is buffered in this program. The send c2 <- "foo" proceeds because the buffer is not full.

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