About the order of channels

Question

I ran the code below several times and sometimes "test" was printed after "test2".
Why is that?
I think "true" is sent to "test" first.

package main

import "log"
import "time"

func main() {
    test := make(chan bool, 1)
    test2 := make(chan bool, 1)

    go func() {
        for {
            select {
            case <-test:
                log.Println("test")
            case <-test2:
                log.Println("test2")
            }
        }
    }()

    time.Sleep(3 * time.Second)
    test <- true
    test2 <- true
    time.Sleep(1)
}

Edited 2014/7/27:
After reading the article, I think that each of "test <- true" and "test2 <- true" could be "happens-before" or "happens-after" or also "happens-concurrently".

possible duplicate of [IF there is a small moment between the first iteration and the second iteration](http://stackoverflow.com/questions/24973935/if-there-is-a-small-moment-between-the-first-iteration-and-the-second-iteration) — OneOfOne, Jul 26 '14 at 18:15
You could have edited the first question, please don't repost the same question within 10 minutes. — OneOfOne, Jul 26 '14 at 18:16

score 1 · Answer 1 · answered Jul 26 '14 at 18:39

1

"If one or more of the communications can proceed, a single one that can proceed is chosen via a uniform pseudo-random selection." You are seeing a uniform pseudo-random selection.

The Go Programming Language Specification

Select statements

Execution of a "select" statement proceeds in several steps:

1) For all the cases in the statement, the channel operands of receive operations and the channel and right-hand-side expressions of send statements are evaluated exactly once, in source order, upon entering the "select" statement. The result is a set of channels to receive from or send to, and the corresponding values to send. Any side effects in that evaluation will occur irrespective of which (if any) communication operation is selected to proceed. Expressions on the left-hand side of a RecvStmt with a short variable declaration or assignment are not yet evaluated.

2) If one or more of the communications can proceed, a single one that can proceed is chosen via a uniform pseudo-random selection. Otherwise, if there is a default case, that case is chosen. If there is no default case, the "select" statement blocks until at least one of the communications can proceed.

3) Unless the selected case is the default case, the respective communication operation is executed.

4) If the selected case is a RecvStmt with a short variable declaration or an assignment, the left-hand side expressions are evaluated and the received value (or values) are assigned.

5) The statement list of the selected case is executed.

Since communication on nil channels can never proceed, a select with only nil channels and no default case blocks forever.

answered Jul 26 '14 at 18:39

peterSO

158,998
31
281
276

I don't know why two communications (test <- true and test2 <-true) happen at the same time in this code. I added "time.Sleep(3 * time.Second)" before "test <- true", so I suppose that the goroutine becomes "waiting a value mode" before "test <- true" is runned for sure. But still sometimes "test" is printed after "test2". – Nigiri Jul 26 '14 at 18:55
I suppose, the execution order is 1."select " 2."test <- true" 3."<-test' . – Nigiri Jul 26 '14 at 18:58
@TaichiSato: test <- true; test2 <- true; select. – peterSO Jul 26 '14 at 19:02
I've added "time.Sleep(3 * time.Second)" by the way. – Nigiri Jul 26 '14 at 19:07
@TaichiSato: No. Try changing [runtime.GOMAXPROCS](http://golang.org/pkg/runtime/#GOMAXPROCS). – peterSO Jul 26 '14 at 19:08
@TaichiSato: goroutine; select (block) ; test <- true; test2 <- true; select (execute). – peterSO Jul 26 '14 at 19:12
"select (execute)" is always after "test2 <- true" ? – Nigiri Jul 26 '14 at 19:13
Sending and receiving on a channel is not atomic? – Nigiri Jul 26 '14 at 19:18

About the order of channels

1 Answers1