What is the "volatile" keyword used for?

Question

I read some articles about the volatile keyword but I could not figure out its correct usage. Could you please tell me what it should be used for in C# and in Java?

Answer 1

Consider this example:

int i = 5;
System.out.println(i);

The compiler may optimize this to just print 5, like this:

System.out.println(5);

However, if there is another thread which can change i, this is the wrong behaviour. If another thread changes i to be 6, the optimized version will still print 5.

The volatile keyword prevents such optimization and caching, and thus is useful when a variable can be changed by another thread.

Answer 2

For both C# and Java, "volatile" tells the compiler that the value of a variable must never be cached as its value may change outside of the scope of the program itself. The compiler will then avoid any optimisations that may result in problems if the variable changes "outside of its control".

Answer 3

Reads of volatile fields have acquire semantics. This means that it is guaranteed that the memory read from the volatile variable will occur before any following memory reads. It blocks the compiler from doing the reordering, and if the hardware requires it (weakly ordered CPU), it will use a special instruction to make the hardware flush any reads that occur after the volatile read but were speculatively started early, or the CPU could prevent them from being issued early in the first place, by preventing any speculative load from occurring between the issue of the load acquire and its retirement.

Writes of volatile fields have release semantics. This means that it is guaranteed that any memory writes to the volatile variable are guaranteed to be delayed until all previous memory writes are visible to other processors.

Consider the following example:

something.foo = new Thing();

If foo is a member variable in a class, and other CPUs have access to the object instance referred to by something, they might see the value foo change before the memory writes in the Thing constructor are globally visible! This is what "weakly ordered memory" means. This could occur even if the compiler has all of the stores in the constructor before the store to foo. If foo is volatile then the store to foo will have release semantics, and the hardware guarantees that all of the writes before the write to foo are visible to other processors before allowing the write to foo to occur.

How is it possible for the writes to foo to be reordered so badly? If the cache line holding foo is in the cache, and the stores in the constructor missed the cache, then it is possible for the store to complete much sooner than the writes to the cache misses.

The (awful) Itanium architecture from Intel had weakly ordered memory. The processor used in the original XBox 360 had weakly ordered memory. Many ARM processors, including the very popular ARMv7-A have weakly ordered memory.

Developers often don't see these data races because things like locks will do a full memory barrier, essentially the same thing as acquire and release semantics at the same time. No loads inside the lock can be speculatively executed before the lock is acquired, they are delayed until the lock is acquired. No stores can be delayed across a lock release, the instruction that releases the lock is delayed until all of the writes done inside the lock are globally visible.

A more complete example is the "Double-checked locking" pattern. The purpose of this pattern is to avoid having to always acquire a lock in order to lazy initialize an object.

Snagged from Wikipedia:

public class MySingleton {
    private static object myLock = new object();
    private static volatile MySingleton mySingleton = null;

    private MySingleton() {
    }

    public static MySingleton GetInstance() {
        if (mySingleton == null) { // 1st check
            lock (myLock) {
                if (mySingleton == null) { // 2nd (double) check
                    mySingleton = new MySingleton();
                    // Write-release semantics are implicitly handled by marking
                    // mySingleton with 'volatile', which inserts the necessary memory
                    // barriers between the constructor call and the write to mySingleton.
                    // The barriers created by the lock are not sufficient because
                    // the object is made visible before the lock is released.
                }
            }
        }
        // The barriers created by the lock are not sufficient because not all threads
        // will acquire the lock. A fence for read-acquire semantics is needed between
        // the test of mySingleton (above) and the use of its contents. This fence
        // is automatically inserted because mySingleton is marked as 'volatile'.
        return mySingleton;
    }
}

In this example, the stores in the MySingleton constructor might not be visible to other processors before the store to mySingleton. If that happens, the other threads that peek at mySingleton will not acquire a lock and they will not necessarily pick up the writes to the constructor.

volatile never prevents caching. What it does is guarantee the order in which other processors "see" writes. A store release will delay a store until all pending writes are complete and a bus cycle has been issued telling other processors to discard/writeback their cache line if they happen to have the relevant lines cached. A load acquire will flush any speculated reads, ensuring that they won't be stale values from the past.

Answer 4

To understand what volatile does to a variable, it's important to understand what happens when the variable is not volatile.

• Variable is Non-volatile

When two threads A & B are accessing a non-volatile variable, each thread will maintain a local copy of the variable in it's local cache. Any changes done by thread A in it's local cache won't be visible to the thread B.

• Variable is volatile

When variables are declared volatile it essentially means that threads should not cache such a variable or in other words threads should not trust the values of these variables unless they are directly read from the main memory.

So, when to make a variable volatile?

When you have a variable which can be accessed by many threads and you want every thread to get the latest updated value of that variable even if the value is updated by any other thread/process/outside of the program.

Answer 5

The volatile keyword has different meanings in both Java and C#.

Java

From the Java Language Spec :

A field may be declared volatile, in which case the Java memory model ensures that all threads see a consistent value for the variable.

C#

From the C# Reference (retrieved 2021-03-31):

The volatile keyword indicates that a field might be modified by multiple threads that are executing at the same time. The compiler, the runtime system, and even hardware may rearrange reads and writes to memory locations for performance reasons. Fields that are declared volatile are not subject to these optimizations. (...)

Answer 6

In Java, "volatile" is used to tell the JVM that the variable may be used by multiple threads at the same time, so certain common optimizations cannot be applied.

Notably the situation where the two threads accessing the same variable are running on separate CPU's in the same machine. It is very common for CPU's to cache aggressively the data it holds because memory access is very much slower than cache access. This means that if the data is updated in CPU1 it must immediately go through all caches and to main memory instead of when the cache decides to clear itself, so that CPU2 can see the updated value (again by disregarding all caches on the way).

Answer 7

When you are reading data that is non-volatile, the executing thread may or may not always get the updated value. But if the object is volatile, the thread always gets the most up-to-date value.

Answer 8

Volatile is solving concurrency problem. To make that value in sync. This keyword is mostly use in a threading. When multiple thread updating same variable.