Why is the output of my Go program always 1 when a concurrent goroutine is incrementing a shared variable?

Is this because the go compiler optimized the code?

In the provided Go code, the variable i is shared between the main goroutine and a concurrent goroutine created using go func() { ... }(). The purpose of the concurrent goroutine is to increment i indefinitely. However, when running the program, you noticed that the output is always 1, despite expecting a much larger number since the concurrent goroutine has ample time to increment i many times over.

To understand this behavior, we need to consider the Go Memory Model and the compiler's optimizations.

Go Memory Model

The Go Memory Model specifies the conditions under which reads of a variable in one goroutine can be guaranteed to observe values produced by writes to the same variable in a different goroutine.

According to the Go Memory Model, for changes to shared variables to be visible across goroutines, they must be followed by a synchronization event, such as a channel operation or locking a mutex.

Compiler Optimizations

In your example, the assignment to i within the concurrent goroutine is not followed by any synchronization event. As a result, the assignment is not guaranteed to be observed by the main goroutine, and the compiler is free to optimize the code as it sees fit.

The aggressive compiler might optimize the code by eliminating the increment operation entirely, effectively reducing the concurrent goroutine to an infinite loop that does nothing. This optimization would explain why the output is always 1, as the main goroutine never observes the increments performed by the concurrent goroutine.

To ensure that updates to shared variables are correctly propagated across goroutines, it is important to synchronize access to those variables using channels, mutexes, or other synchronization primitives provided by Go's sync and sync/atomic packages.

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