[go-nuts] question about Andrew Gerrand's chat talk

Here is your answer.
select will choose from the operations which can proceed it doesn't
act on one until it can actually proceed. If one goroutine is writing
then the other can't be writing too because that would block. The only
other operation that could proceed is to receive.

I believe the output is as expected.

agent 1 started...
agent 2 started...
agent 2 : read id 1.
agent 1 : written it's own id.

1. agent 1 starts
2. agent 2 starts
3. agent 1 gets to the select statement and select determines that comm
does not have a value that can be received, but comm is ready to be written
to...
4. since the channel is non-buffered, it sits and waits at line 11
5. agent 2 comes along and select sees that not only is comm not ready
to be written to (because agent 1 is trying to do so), but comm has a value
that can be received
6. agent 2 pulls the value off the channel, and runs the fmt.Printf
function
7. agent 1 can now move on as well since the value it was putting on the
channel has been pulled off the channel and so runs the fmt.Printf function

There isn't really a chicken and egg problem. Given a choice (a.k.a.
select statement), if a non-buffered channel has nothing on it or nothing
being put on it, then it is ready for writing. Once a value is placed on
the channel, the select operation that can proceed would be the
read/receive.

At least, that's how I understand it.

What Jesse said made more sense.

The scheduler chooses, not the goroutines. The scheduler knows about
all waiting operations, and matches send operations with receive
operations. There cannot be a send without a corresponding receive, so
the situation you describe (where both goroutines choose to send
simultaneously) cannot happen.

Andrew

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