A pointer of the form volatile int* p; is a pointer to an int that the compiler will treat as volatile. This means that the compiler will assume that it is possible for the variable that p is pointing at to have changed even if there is nothing in the source code to suggest that … Read more
The effect of the volatile keyword is approximately that each individual read or write operation on that variable is made atomically visible to all threads. Notably, however, an operation that requires more than one read/write — such as i++, which is equivalent to i = i + 1, which does one read and one write … Read more
Any x86 instruction that has lock prefix has full memory barrier. As shown Abel’s answer, Interlocked* APIs and CompareExchanges use lock-prefixed instruction such as lock cmpxchg. So, it implies memory fence. Yes, Interlocked.CompareExchange uses a memory barrier. Why? Because x86 processors did so. From Intel’s Volume 3A: System Programming Guide Part 1, Section 7.1.2.2: For … Read more
Declaring a volatile Java variable means: The value of this variable will never be cached thread-locally: all reads and writes will go straight to “main memory”. Access to the variable acts as though it is enclosed in a synchronized block, synchronized on itself. Just for your reference, When is volatile needed ? When multiple threads … Read more
first, let’s see what says the fox =WHATTHEFOXSAY() is a unique easter egg google sheets function (discovered by @kishkin) that randomly generates a pre-set string of text on user demand which is a huge deal because while the generation is random, the recalculation is not affected by onEdit, onChange nor onOpen events, so with some … Read more
Whether it is optimized out depends entirely on compilers and what they choose to optimize away. The C++98/03 memory model does not recognize the possibility that x could change between the setting of it and the retrieval of the value. The C++11 memory model does recognize that x could be changed. However, it doesn’t care. … Read more
Yes, you can read from it and write to it safely – but you can’t do anything compound such as incrementing it safely, as that’s a read/modify/write cycle. There’s also the matter of how it interacts with access to other variables. The precise nature of volatile is frankly confusing (see the memory model section of … Read more
Yes, it is guaranteed that thread 2 will print “done” . Of course, that is if the write to b in Thread 1 actually happens before the read from b in Thread 2, rather than happening at the same time, or earlier! The heart of the reasoning here is the happens-before relationship. Multithreaded program executions … Read more
Volatile is unnecessary. Well, sort of** volatile is used to create a memory barrier* between reads and writes on the variable. lock, when used, causes memory barriers to be created around the block inside the lock, in addition to limiting access to the block to one thread. Memory barriers make it so each thread reads … Read more
On Intel an un-contended volatile read is quite cheap. If we consider the following simple case: public static long l; public static void run() { if (l == -1) System.exit(-1); if (l == -2) System.exit(-1); } Using Java 7’s ability to print assembly code the run method looks something like: # {method} ‘run2’ ‘()V’ in … Read more