Golang canals keep order

Question

Do Go channels have any order guarantees?

For example: you have two goroutines A and B that use the channel. A clicks data on the channel, and B reads it. Are you sure B will read the data. In the same order that A placed it in the channel?

I understand that if there are several producers or consumers, the order can be non-deterministic, but I specifically ask about the presence of only one manufacturer and 1 consumer.

EDIT: I ask about buffered channels

thanks

Answer 1

You can see the channel idea illustrated in " The nature of channels in transit" : it shows how order or reading / writing was respected.
See Also Channels :

Receivers are always blocked until data is received.

• If the channel is unbuffered, the sender blocks until the receiver receives a value.
• If the channel has a buffer, the sender is blocked only until the value is copied to the buffer; if the buffer is full, it means until a receiver receives a value.

Unbuffered channels

Buffered channel

Answer 2

I read The Nature of Channels In Go 's blog, but I still confuse why the code below outputs -5 17 (I expect it outputs 17 -5)

package main import "fmt" func sum(s []int, c chan int) { sum := 0 for _, v := range s { sum += v } c <- sum // send sum to c } func main() { s := []int{7, 2, 8, -9, 4, 0} c := make(chan int) go sum(s[:len(s)/2], c) go sum(s[len(s)/2:], c) x := <-c y := <-c fmt.Println(x, y) } 

and the codes below: 17 -5 (as expected)

 package main import "fmt" func sum(s []int, c chan int) { sum := 0 for _, v := range s { sum += v } c <- sum // send sum to c } func main() { s := []int{7, 2, 8, -9, 4, 0} c := make(chan int) go sum(s[:len(s)/2], c) x := <-c // Only difference here go sum(s[len(s)/2:], c) y := <-c fmt.Println(x, y) } 

This code is from Walk along the way