为什么这个Google I / O 2012并发示例无法正常工作？

I'm trying to follow along with Rob Pike's Google I/O 2012 talk called "Go Concurrency". I'm trying the example where channels are multiplexed so "Ann" and "Joe" don't talk in lock-step. But using the code below, they still are lock-stepping. Where am I going wrong?

Video: http://www.youtube.com/watch?v=f6kdp27TYZs&feature=player_detailpage#t=1025s

package main

import (
    "fmt"
    "time"
    "math/rand"
    )

func fanIn(input1, input2 <-chan string) <-chan string {
    c := make(chan string)
    go func() { for {c <- <-input1 } }()
    go func() { for {c <- <-input2 } }()
    return c
}

func main() {
    c := fanIn(boring("Joe"), boring("Ann"))
    for i:=0; i<10; i++ {
        fmt.Println(<-c)
    }
    fmt.Printf("You're both boring, I'm leaving...
")
}

func boring(msg string) <-chan string {
    c := make(chan string)
    go func() { // launch goroutine from inside the fn
        for i:=0; ; i++ {
            c <- fmt.Sprintf("%s %d", msg, i)
            time.Sleep(time.Duration(rand.Intn(1e3)) * time.Millisecond )
        }
    }()
    return c
}

And the output of this (go version go1.0.2 on Ubuntu 10.04 LTS)

Joe 0
Ann 0
Joe 1
Ann 1
Joe 2
Ann 2
Joe 3
Ann 3
Joe 4
Ann 4
You're both boring, I'm leaving...

Where did I go wrong? Thanks!

Your code is fine; it just tends to take a little more than that to get them out of sync. Just loop more times and you should see them get out of lock-step:

for i := 0; i < 20; i++ { // Going up to 20 is enough to usually see it
    fmt.Println(<-c)
}

I got this output:

Joe 0
Ann 0
Joe 1
Ann 1
Joe 2
Ann 2
Joe 3
Ann 3
Joe 4
Ann 4
Joe 5
Ann 5
Joe 6
Ann 6
Ann 7
Joe 7
Joe 8
Joe 9
Ann 8
Ann 9