在golang中对uint8的读写是原子的吗？

Question

As in the title, are read and write operations regarding uint8, atomic? Logically it must be a single cpu instruction obviously to read and write for a 8 bit variable. But in any case, two cores could simultaneously read and write from the memory, is it possible to create a stale data this way?

Answer 1

There's no guarantee that the access on native types are on any platform atomic. This is why there is sync/atomic. See also the advice in the memory model documentation.

Example for generic way of atomically setting a value (Play)

var ax atomic.Value // may be globally accessible

x := uint8(5)

// set atomically
ax.Store(x)

x = ax.Load().(uint8)

Probably more efficient solution for uint8 (Play):

var ax int64 // may be globally accessible

x := uint8(5)

atomic.StoreInt64(&ax, 10)

x = uint8(atomic.LoadInt64(&ax))

fmt.Printf("%T %v
", x, x)

Answer 2

No. If you want atomic operations, you can use the sync/atomic package.

If you mean "would 8bit operations be atomic even if I ignore the Go memory model?", then the answer is still, it depends probably not.

If the hardware guarantees atomicity of read/write operations, then it might be atomic. But that still doesn't guarantee cache coherence, or compiler optimizations from reordering operations. You need to serialize the operations somehow, with the primitives Go provides in the "atomic" package, and using the "sync" package and channels to coordinate between goroutines.