When using channels, what is the order of execution of goroutines?

php editor Baicao is here to answer a common question: "When using channels, what is the execution order of goroutines?" In the Go language, goroutines are lightweight Threads can be executed concurrently. When using channels for communication between coroutines, the channel's receive and send operations are blocking, that is, they wait for the operations of other coroutines to complete. Therefore, when multiple goroutines operate a channel at the same time, their execution order is uncertain and depends on the scheduling of each coroutine. This means that it is impossible to determine which goroutine will be executed first and which will be executed later. This is determined by the scheduler of the Go language.

Question content

What is the execution order of goroutines when using channels? I think writing or reading the channel stops the current goroutine. But my test code doesn't follow this rule:

func main() {
    ch := make(chan int)
    go sum(ch, 3)
    fmt.Println("Write number: 10")
    ch <- 10
    fmt.Println("Write number: 20")
    ch <- 20
    fmt.Println("Write number: 30")
    ch <- 30

    fmt.Println("Finish main")
}

func sum(ch chan int, len int) {
    fmt.Println("Func 'sum' start")

    sum := 0
    for i := 0; i < len; i++ {
       fmt.Println("For start")
       num := <-ch
       fmt.Printf("Read from ch: %d\n", num)
       sum += num
       fmt.Println("For finish")
    }

    fmt.Printf("Sum: %d\n", sum)
}

How I think this program works:

1.Create channel

2. Create a goroutine (not started, only initialized)

3. Print: write down the number: 10

4. Record to channel 10. Lock key functions.

5. The most important thing is to be blocked. Start sum goroutine.

6. Print summation function: Func 'sum' start

7. The summation function runs in a loop and prints: "To begin"

8. Read the number 10 from "ch" and print: "Read from ch: 10"

9.Next step. Print: "Done" and continue with the next iteration.

10. Print: "starts with" and try writing "with". But the channel is empty. Stop the quota and enter the main line

...again and again.

Then I want to see:

Write number: 10
Func 'sum' start
For start
Read from ch: 10
For finish
For start
Write number: 20
Read from ch: 20
For finish
For start
Write number: 30
Read from ch: 30
For finish
Sum: 60
Finish main

But, I saw:

Write number: 10
Func 'sum' start
For start
Read from ch: 10
For finish
For start
Write number: 20
Write number: 30
Read from ch: 20
For finish
For start
Read from ch: 30
For finish
Sum: 60
Finish main

How can this be? The main function writes to the channel twice without reading.

Also, if you change the number of calls in for:

go sum(ch, 2)

I don't get the error. But no one read the last message

Example: before this message.

Solution

Goroutines run concurrently. On systems with multiple cores, they can run in parallel. The details depend on the scheduler implementation in the Go runtime. For all intents and purposes, with the exception of synchronous operations like channel communication, things happen in random order.

This is not the case and is not what is happening (as long as the channel is not buffered). The Println call occurs before the send operation, and the main Goroutine blocks after printing until the sum Goroutine is ready to receive.

Whether you see "Read from ch: 30" printed is also random. The corresponding receive operation must occur because main blocks until it does. However, main may return before Println after the receive has occurred, and the program will terminate immediately when main returns, regardless of the presence of any other goroutines. If the channel is buffered, the likelihood of this happening increases.

But in fact, it's not. If there are only two receivers it will always lead to a deadlock: https://go.dev/play/p/ qFVh529mkqR

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