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    <p>The exact output of your program is not defined and depends on the scheduler. The scheduler can choose freely between all goroutines that are currently not blocked. It tries to run those goroutines concurrently by switching the current goroutine in very short time intervals so that the user gets the feeling that everything happens simultanously. In addition to that, it can also execute more than one goroutine in parallel on different CPUs (if you happen to have a multicore system and increase <a href="http://golang.org/pkg/runtime/#GOMAXPROCS" rel="nofollow">runtime.GOMAXPROCS</a>). One situation that might lead to your output is:</p> <ol> <li><code>main</code> creates two goroutines</li> <li>the scheduler chooses to switch to one of the new goroutines immediately and chooses <code>display</code></li> <li><code>display</code> prints out the message and is blocked by the channel send (<code>c &lt;- true</code>) since there isn't a receiver yet.</li> <li>the scheduler chooses to run <code>sum</code> next</li> <li>the sum is computed and printed on the screen</li> <li>the scheduler chooses to not resume the <code>sum</code> goroutine (it has already used a fair amount of time) and continues with <code>display</code></li> <li><code>display</code> sends the value to the channel</li> <li>the scheduler chooses to run main next</li> <li>main quits and all goroutines are destroyed</li> </ol> <p>But that is just one possible execution order. There are many others and some of them will lead to a different output. If you want to print just the first result and quit the program afterwards, you should probably use a <code>result chan string</code> and change your <code>main</code> function to print <code>fmt.Println(&lt;-result)</code>.</p>
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    1. COThanks for the clarification. In step 3, display is blocked by the channel send since there isn't a receiver. Eh...When is the receiver ready?
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    2. COThe channel is synchronous which means that both, the receiver and the sender must be ready in order to transmit the value. If a goroutine executes `c <- true` first, it will be blocked until another goroutine executes `<-c`. But it is also possible that a goroutine that executes `<-c` is blocked until there is a matching `c <- true` in another goroutine.
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    3. COJust a comment. As far as I know, the Go scheduler in current version is not preemptive. Once a goroutine has the CPU, it won't switch to another one until it is blocked (due to I/O, channel, mutex, etc.). So if sum is executed first, it will use the CPU exclusively until blocked by the channel (I'm not sure if fmt.Println may produce a CPU switch). A preemptive scheduler is planned for Go 1.2, I read. Of course, if GOMAXPROCS is not 1, the other goroutine will use another SO-level process and will be scheduled by the SO. More info here: http://dominik.honnef.co/go-tip/2013-08-15/
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