Your example has a single store, there is no reordering. There is no value before 8, it doesn't require any barrier. If you make and additional store to the array, that's would be equivalent to my example.
I said there was a value before 8, I just didn't describe it well enough.
First off your example looks a little under-constrained (What if reader:0 happens before writer:0?), so let's assume all the writes of 0 at the start of every program happen before any other instructions.
Let there be a third thread, "builder". It writes 0 to every address in the array, then fills it with new values including an 8.
The "writer" thread loops repeatedly over the array until it finds an 8, then stores the address in X.
The "reader" thread waits for X to be nonzero then loads the value at [X].
In the toy coherent computer, reader will always load an 8.
But in a very weak memory model, one that doesn't enforce dependent loads, reader is allowed to get a 0.
The write to X and the write of 8 don't have to show up in any particular order.
But in that situation I would say that X is "more recent" than 8, because of causality. And if you asked me yesterday I would have called the 0 "stale".
