


Does Go Guarantee Message Order Preservation in Channels with Non-Zero Capacity When Blocked?
Channel Order Preservation When Blocked
The ability of channels to preserve message order when blocked is a crucial consideration in concurrent programming. In the scenario presented, a slice of channels is used to receive messages from a single channel, and the sender wants to ensure that the order of the messages sent to each channel is maintained even if the recipients are consuming them at different rates.
According to the Go specification, "If the capacity is greater than zero, the channel is asynchronous: communication operations succeed without blocking if the buffer is not full (sends) or not empty (receives), and elements are received in the order they are sent."
However, the specification does not explicitly address the case where multiple goroutines are blocked on writing to a channel with a non-zero capacity. The question at hand is whether there are any guarantees about the order of sends after the channel becomes unblocked.
The answer, unfortunately, is no. There are no guarantees. The specification does not provide any specific ordering guarantees in this scenario. Even if the channel is not full, there is no guarantee that the goroutine that was started first to send a message would actually execute first. Therefore, you cannot rely on the messages arriving in order in this case.
This conclusion is in line with the general principle that the execution order of goroutines in Go is non-deterministic. While the runtime system may make efforts to optimize goroutine scheduling for performance reasons, there is no guarantee about the order in which goroutines will run. Consequently, you should not rely on the order of execution to ensure the correct behavior of your code.
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