curious correct Go program

correct.go

package main

import (
        "fmt"
        "time"
)

func main() {
        c := make(chan struct{},1)
        go func() {
                time.Sleep(100 * time.Millisecond)                                                  
                fmt.Printf("0 ")
                c <- struct{}{}
                fmt.Printf("1 ")
                c <- struct{}{}
                fmt.Printf("2 ")
                c <- struct{}{}
                fmt.Printf("3 ")
                return
        }()
        <- c
        fmt.Printf("done ")
}

This program runs and exist properly although our intuition might say it will fail.

The program doesn't deadlock and it prints "done" even though it appears to be wrong at first glance.

The chan is of length 1, how is it we can put more than one struct{}{} into it? We can't and we don't. The goroutine is launched while the outer scope blocks on polling the chan. The goroutine enqueues a struct{}{}, and the outer scope dequeues it. While that is happening, the goroutine now can (and typically does) enqueue another struct{}{} before the final "done" message is printed. At this point, our goroutine should be "stuck" waiting to enqueue a new struct{}{}, but it doesn't matter, the program exits and the goroutine itself is stopped.

This shows the need for proper synchronization without cleverness.