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Would a thread handling a subsequent network request be guaranteed to see the value of a volatile variable written durin


By : Pablo Motta
Date : August 01 2020, 11:00 AM
Any of those help If the first request completes then it means that someFlag = 1 has been executed by some thread. At that point the value of someFlag is guaranted to be visible by any other threads performing read. So when the second request comes you can be sure that it will see value of 1.
code :


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Is a volatile boolean switch written to by only one thread thread-safe?


By : Mandeep Brar
Date : March 29 2020, 07:55 AM
seems to work fine
However this thread is the only thread that ever writes to the variable... Is the variable being volatile enough to make sure that all threads read the right value from it.
code :
private final AtomicBoolean isEvenTick = new AtomicBoolean();
...
boolean currentValue;
do {
  currentValue = isEvenTick.get();
} while (!isEvenTick.compareAndSet(currentValue, !currentValue);

volatile in c++: should I define a variable which may changes by several thread as volatile?


By : henry chen
Date : March 29 2020, 07:55 AM
With these it helps If you have suitable synchronization in place, then there is no need for volatile.
If you do not have suitable synchronization in place, then volatile does not supply suitable synchronization, although in some C++ implementations it might appear to help to a greater or lesser extent.

Are assignments to non-volatile member variables in one thread guaranteed to be seen in another thread?


By : Emir Septian Sori
Date : March 29 2020, 07:55 AM
it helps some times that can be modified by another thread but this is not gurranteed. also this is not thread safe either. To make a variable gurranteed to see from another thread you need to do any of the following

Are values written by one thread guaranteed to be seen by another thread without locking on that variable?


By : user3181989
Date : March 29 2020, 07:55 AM
hop of those help? It's complicated, but in your particular case, this is guaranteed to work across all C++ implementations.
The reason is that the default memory order for operations on std::atomic is std::memory_order_seq_cst, which does guarantee release/acquire semantics.

Guaranteed write a member variable value to memory without marking it volatile


By : Cesar Solis
Date : March 29 2020, 07:55 AM
will be helpful for those in need No, it's not correct. The reader of the value would have to synchronize on the same object to read the value, and then, the happens-before principle would apply.
From the documentation:
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