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According to this online book , the volatile keyword in C# does not protect against reordering Write operations followed by Read operations. It gives this example in which both a and b can end up being set to 0 , despite x and y being volatile : class IfYouThinkYouUnderstandVolatile { volatile int x, y; void Test1() // Executed on one thread { x = 1; // Volatile write (release-fence) int a = y; // Volatile read (acquire-fence) ... } void Test2() // Executed on another thread { y = 1; // Volatile write (release-fence) int b = x; // Volatile read (acquire-fence) ... } } This seems to fit with what the specification says in 10.5.3: A read of a volatile field is called a volatile read. A volatile read has “acquire semantics”; that is, it is guaranteed to occur prior to any references to memory that occur after it in the instruction sequence. A write of a volatile field is called a volatile write. A volatile write has “release semantics”; that is, it is guaranteed to happen after any memory references prior to the write instruction in the instruction sequence. What is the reason for this? Is there a use case in which we don't mind Write-Read operations being reordered?

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