


Go Concurrency and Channel Confusion
In Go, concurrency allows multiple tasks to execute concurrently using goroutines. Channels facilitate communication between these goroutines. However, understanding concurrency can be challenging, especially when dealing with channels.
Consider the following code snippet:
<code class="go">package main import "fmt" func display(msg string, c chan bool) { fmt.Println("display first message:", msg) c <p>In this code, we create two goroutines: display and sum. The display goroutine prints a message, sends a signal to the channel, and then waits for a response. The sum goroutine performs a long computation, prints the result, and also sends a signal to the channel. In the main goroutine, we block until a signal is received from the channel.</p> <p>The expected output of the code is:</p> <pre class="brush:php;toolbar:false">display first message: hello
However, we observe that the program prints both the message and the sum of the computation:
display first message: hello 10000000000
Understanding the Issue
The issue arises due to the nondeterministic nature of goroutine scheduling. The scheduler in Go chooses freely between the goroutines that are not blocked. In this example, the scheduler can execute any of the goroutines at any given time.
One possible execution order is:
- main creates the goroutines.
- The scheduler selects display, which prints the message and waits for a response.
- The scheduler switches to sum, which executes for a long time.
- The scheduler switches back to display, which sends the signal.
- The scheduler switches to main, which prints the signal and exits.
In this scenario, the sum is printed before display sends the signal, resulting in the unexpected output.
Solution
To ensure that the program prints only the message and exits before the sum is computed, we can use a different approach:
<code class="go">func main() { result := make(chan string) go display("hello", result) go sum(result) fmt.Println(<p>In this revised version, the result channel carries a single value, the message from the display goroutine. The main goroutine now prints the value from the channel, ensuring that it receives the message before exiting.</p></code>
The above is the detailed content of Why does the Go program print the sum of the computation before the message in the given code snippet, even though the main goroutine blocks until a signal is received from the channel?. For more information, please follow other related articles on the PHP Chinese website!

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