Does the behavior of a go channel change depending on how messages are sent to it?

php editor Xiaoxin In the go language, a channel is a mechanism used to transfer data between goroutines. The behavior of a channel depends on how messages are sent to it. When a channel is declared buffered, the goroutine sending the message will put the message into the channel's buffer and then continue execution. When the channel is declared unbuffered, the goroutine sending the message will block until other goroutine receives the message from the channel. Therefore, it can be said that the behavior of a channel changes depending on how messages are sent to it.

Question content

Why do these two pieces of code execute differently?

1. Running the following code will return a fatal error: all goroutines are sleeping - deadlock! error.

func main() {
    ch := make(chan int)
    ch <- 1
    fmt.println(<-ch)
}

Running the following code correctly will return 2 and 3 (one line each).

ch := make(chan int)
    go buffer(ch)
    fmt.Println(<-ch)
    fmt.Println(<-ch)
}

func buffer(ch chan int) {
    ch <- 2
    ch <- 3
}

#1 will only execute correctly if ch is defined as a buffered channel. Why? Why does buffer() in ch accept 2 values ​​without complaining?

Solution

Send operations on unbuffered channels will block until another goroutine reads data from it. This is why the first piece of code deadlocks: when you send to the channel, no other goroutine reads from it.

In the second case, you have a goroutine sending two values ​​to the channel, and the main goroutine reads from the channel twice while sending. The channel does not accept two sends, but one after the other, with the first send operation unblocked by the first read and the second send operation unblocked by the second read.

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