Why Does This Go Program Print 'ping' Then 'hello' Despite Concurrent Channel Writes?

Understanding Channel Output Order in Golang

Consider the following Golang program:

func main() {
  messages := make(chan string)
  go func() { messages <p>This program involves two goroutines asynchronously writing to a channel and the main routine reading from the same channel. Despite the unbuffered nature of the channel, the output consistently prints "ping" and then "hello."</p><p>To understand the reasoning behind this output, it's crucial to grasp that channel output order is not based on the order in which the goroutines are created. Instead, it's determined by the scheduler, which decides on the execution order of goroutines in a non-deterministic manner.</p><p>When the program executes, the two goroutines responsible for sending messages to the channel run concurrently. Since the channel is unbuffered, both goroutines block until a receiver becomes available.</p><p>When the main routine attempts to read from the channel using msg := </p><p>Subsequently, when the main routine tries to read from the channel again using msg2 := </p><p>In this particular case, the scheduler consistently selects the goroutine sending "ping" as the first to send its message. However, it's important to note that this outcome is non-deterministic and can change based on factors such as system load and execution environment.</p><p>To summarize, the output order in Golang channels is not guaranteed and is subject to the scheduler's execution order for goroutines. Hence, despite the simplicity of this example, the output may vary on different executions.</p>

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