Are golang channels based on LIFO?

Question

I was wondering regarding the order of elements in golang channels. After running a few examples, it seems that the order in which the elements come off the channel is "last in first out". Am I right?

The following snippet is the example which I used. After running the code, the output is 20 10, while 10 was sent to the channel first and 20 last.

package main

import "fmt"

func multiply(c chan int, num int) {
    c <- num * 10
}

func main() {
    c := make(chan int)

    go multiply(c, 1)
    go multiply(c, 2)

    v1 := <-c
    v2 := <-c
    fmt.Println(v1, v2)
}

Answer 1

Golang channels are not LIFO.

Channels act as first-in-first-out queues. For example, if one goroutine sends values on a channel and a second goroutine receives them, the values are received in the order sent.

Values sent on the channel will be received whenever the receiver of the channel is ready. And if not then it will block. For managing that you can go for buffered channels.

Below code will check if the values are available to be received from the channel.

package main

import "fmt"

func multiply(c chan int, num int) {
    c <- num * 10
}

func main() {
    c := make(chan int, 3)

    go multiply(c, 1)
    go multiply(c, 2)
    go multiply(c, 3)

    for i:=0;i<3;i++{
         foo, ok := <- c
         if !ok {
                fmt.Println("done")
                return
         }
         fmt.Println(foo)
    }

}

Working code on Go playground

Buffered Channels

Channels can be buffered. Provide the buffer length as the second argument to make to initialize a buffered channel:

make(chan int, 100)

The capacity, in number of elements, sets the size of the buffer in the channel. If the capacity is zero or absent, the channel is unbuffered and communication succeeds only when both a sender and receiver are ready. Otherwise, the channel is buffered and communication succeeds without blocking if the buffer is not full (sends) or not empty (receives). A nil channel is never ready for communication.

In your case, it depends on which go routine will send the value on the channel first. The values that you are printing completely depends on the go routines.

For more information go through Golang Channels