golang chan cannot block

The channel in Go language is a very useful data structure, which can realize data sharing and synchronization in concurrent programming, and is very efficient. However, there is one thing that requires special attention when using channels, and it is also a mistake that many Go language beginners often make, that is, chan cannot block.

In the Go language, data transmission and synchronization between multiple goroutines can be achieved through channels, thus avoiding the cumbersome synchronization locks and inevitable deadlock problems. When we use channels to send and receive data between two or more goroutines, we often use the following code:

ch := make(chan int)

go func() {
    ch <- 1
}()

value := <-ch
fmt.Println(value)

In this code, we create a channel of type int, and An integer 1 is sent to the channel in a new goroutine. Then, call <-ch in the main goroutine to accept the data in the channel and print it out. This example is simple, but it demonstrates using channels to synchronize data between two goroutines.

In the above code, we may find that this channel is not explicitly closed and it is not cached. So, in this case, what happens when we read an unclosed and uncached channel?

In this case, if the channel is empty, then we will block when reading it until a goroutine writes a new value or closes the channel. However, if the channel remains empty, our program will be blocked permanently. This is a very dangerous situation and often leads to deadlocks and other problems in programs in practical applications.

So how can we avoid this situation? It's actually very simple, we just need to make sure that the channel does not block when using it. As long as we can ensure that the channel will not remain empty when using it, we can avoid blocking problems.

A common way is to determine whether the status of the following channel is empty before goroutine writes the value, or to give it a capacity when caching the channel to ensure that writing the value will not cause blocking. For example, in the following example we use a cached channel:

ch := make(chan int, 1)

ch <- 1

value := <-ch
fmt.Println(value)

Here, we specify a capacity of 1 when creating the channel, so after writing a value, even if we do not read this immediately value, the program will still not be blocked. This avoids the problems mentioned above.

In addition to using cached channels to avoid blocking, we can also use select statements to handle channel read and write operations. The select statement can monitor multiple channels at the same time. Once one of the channels receives a value, the corresponding operation will be performed immediately. For example, the following example:

ch := make(chan int)

timer := time.NewTicker(time.Second)

select {
case ch <- 1:
    fmt.Println("value sent")
case <-timer.C:
    fmt.Println("timeout")
}

Here, we create a new Ticker that triggers once every second. Then we listen to two channels in the select statement. If ch can be written, "value sent" is output, otherwise "timeout" is output after one second.

In summary, although channel is a very useful data structure, special care needs to be taken when using it to avoid blocking. As long as we can ensure that the channel does not block when using it, we can make full use of this tool to achieve efficient concurrent programming.

The above is the detailed content of golang chan cannot block. For more information, please follow other related articles on the PHP Chinese website!