"A reference to a volatile field will not be treated as volatile" implications

Jader_Dias_48465 0
Viewed 9663

The following code

using System.Threading;

class Test
{
    volatile int counter = 0;
    public void Increment()
    {
        Interlocked.Increment(ref counter);
    }
}

Raises the following compiler warning:

"A reference to a volatile field will not be treated as volatile"

Am I doing something wrong here to raise this warning? Why does the compiler me warn about this?

volatile multithreading .net c#
4 Answers

You are not doing anything wrong. According to the documentation:

A volatile field should not normally be passed using a ref or out parameter, since it will not be treated as volatile within the scope of the function. There are exceptions to this, such as when calling an interlocked API.

Darin_294070

Basically the warning is that when you pass a volatile field by reference, the calling code doesn't know to treat it in a volatile manner. For Interlocked.Increment that probably doesn't matter, due to the nature of the method - but then you don't need the variable to be volatile anyway if you're using Interlocked.

In general, I think I'd avoid mixing the two - if you're using Interlocked, do it everywhere (using Interlocked.CompareExchange(ref counter, 0, 0) to read it). I can't say I use volatile very often, personally. For simple counters I might use Interlocked, but I'm more likely to use a lock for most tasks.

user_226560

You're getting the error because you're passing the field by reference. I think what this means is that the target method has no idea the field is marked as volatile, and therefore will not treat it as such.

user_53800
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