Why Does My Go Routine Print Unexpected Values Due to Memory Reordering?

Go Routine Issue: Incorrect Synchronization

The issue you've encountered in GoLang stems from its memory model. As per the documentation, within a single goroutine, reads and writes may be reordered as long as it doesn't affect the behavior within that goroutine.

Consider this code:

var a, b int

func f() {
    a = 1
    b = 2
}

func g() {
    print(b)
    print(a)
}

func main() {
    go f()
    g()
}

Due to the memory model's reordering, you may observe that g() prints 2 and then 0. Even though a's assignment comes before b's in f(), the compiler may optimize the code by assigning b first for efficiency reasons.

Within the same goroutine, the order of assignments is guaranteed, as the compiler cannot reorder them if it would cause issues. However, in the given example, there is no synchronization between the two goroutines. Therefore, the compiler is not required to consider the potential impact of reordering on the other goroutine.

If you introduce synchronization between the goroutines, such as a WaitGroup or a mutex, the compiler will ensure that there are no inconsistencies at the synchronization point. In this case, you would expect g() to print 2 and 1, as both assignments will be completed before the print() calls.

Understanding this aspect of GoLang's memory model is crucial for writing concurrent programs and avoiding potential race conditions or other issues that could arise due to unexpected reordering.

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