"A reference to a volatile field will not be treated as volatile" implications
The following codeusing System.Threading;class Test{ volatile int counter = 0; public void Increment() { Interlocked.Increment(ref counter); }}Raises the following compiler warning:"A reference to a...
View ArticleAnswer by Kent Boogaart for "A reference to a volatile field will not be...
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...
View ArticleAnswer by Jon Skeet for "A reference to a volatile field will not be treated...
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...
View ArticleAnswer by Darin Dimitrov for "A reference to a volatile field will not be...
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...
View ArticleAnswer by Robert Fraser for "A reference to a volatile field will not be...
Use this:#pragma warning disable 420if(Interlocked.CompareExchange(ref isLoaded, 1, 0) != 0) return;#pragma warning restore 420
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