c++ - memory_order_relaxed and visibility

Question

Consider two threads, T1 and T2, that store and load an atomic integer a_i respectively. And let's further assume that the store is executed before the load starts being executed. By before, I mean in the absolute sense of time.

T1                                    T2
// other_instructions here...         // ...
a_i.store(7, memory_order_relaxed)    // other instructions here
// other instructions here            // ...
                                      a_i.load(memory_order_relaxed)
                                      // other instructions here

Is it guaranteed that T2 sees the value 7 after the load?

Answer 1

Is it guaranteed that T2 sees the value 7 after the load?

Memory order is irrelevant here; atomic operations are atomic. So long as you have ensured that the write "happens-before" the read (which you stated to be true in the premise of your question), and there are no other intervening operations, T2 will read the value which was written by T1. This is the nature of atomic operations, and memory orders do not modify this.

What memory orders control is if T2 sees 7 (whether "happens-before" is ensured or not), whether or not it can access other data modified by T1 before it stored 7 into the atomic. And with relaxed memory ordering, T2 has no such guarantees.

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Note: you changed your question from being about a situation where the load "happens after" the store, when the store is explicitly "synchronized" with the load, into a situation that is more nebulous. There is no "absolute time" as far as the C++ object model is concerned. All atomic operations on a particular atomic object happen in an order, but unless there is something which explicitly creates a "happens before/after" relationship between the two loads, then what value gets loaded cannot be known. It will be one of the two possibilities, but which one cannot be known.