Why Does My Go Unbuffered Channel Print Messages in an Unexpected Order?

Understanding Channel Output Order in Go

The given Go code snippet involves the use of an unbuffered channel to communicate between goroutines. While the code is expected to print "hello" and then "ping" due to blocking behavior, it consistently prints "ping" followed by "hello." This raises questions about the order in which messages are written to and read from the channel.

As the code suggests, unbuffered channels block both the sender and receiver until data is available. This means that when the go routines attempts to write to the channel, they will block until the message is read.

Upon execution, the code forks two go routines that attempt to write "hello" and "ping" to the channel. Since no receiver is immediately available, both routines block.

When the first read operation (msg := <-messages) occurs, the program will not arbitrarily select a waiting goroutine. Instead, it will select the one that has already written to the channel. This explains why "ping" is always assigned to msg, as it is always written before "hello."

Therefore, the order of printed messages is determined by the order of goroutine execution, which is non-deterministic. By adding print statements to the goroutines that write to the channel, one can verify that the order of writing corresponds to the order of messages read from the channel.

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