3 ms·
If you want the write to be a synchronization point after which all threads will observe only the new value, it's only possible with shared mutex. Of course you
by surajrmal 8mo ago
If you want the write to be a synchronization point after which all threads will observe only the new value, it's only possible with shared mutex. Of course you can use a barrier to accomplish that instead but using something like hazard pointers or rcu doesn't synchronize by itself.
- loeg 8mo agoThis is true, but it is a subset of designs using shared_mutex.
- secondcoming 8mo agoI think it's possible with an atomic<shared_ptr> too (C++20)? A shared_mutex comes in useful when you can't really have multiple copies of the shared data due to perhaps memory usage, so readers fail to acquire it when the writer is updating it.
- foldr 8mo agoNot an expert, but can’t you get synchronization like this just by using release/acquire memory order with C11 atomic stores and loads?
- jeffbee 8mo agoYes. But if you are tempted to do this in most cases you should just use a mutex anyway.
- jpc0 8mo agoFrom my knowledge RCU/epoch/Hazard pointers are useful in data structures and algorithms where raw atomics cannot be used but you still nees lock free or in some cases wait free semantics. If you can use an atomic then these are overkill and you should just be using an atomic, but many times things that are atomic does not make it lock free, if there's no hardware support the compiler will add a mutex.