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Does Interlocked provide visibility in all threads?

Suppose I have a variable "counter", and there are several threads accessing and setting the value of "counter" by using Interlocked, i.e.:

int value = Interlocked.Increment(ref counter);

and

int value = Interlocked.Decrement(ref counter);

Can I assume that, the change made by Interlocked will be visible in all threads?

If not, what should I do to make all threads synchronize the variable?

EDIT: someone suggested me to use volatile. But when I set the "counter" as volatile, there is compiler warning "reference to volatile field will not be treated as volatile".

When I read online help, it said, "A volatile field should not normally be passed using a ref or out parameter".

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Chaowlert Chaisrichalermpol Avatar asked Feb 05 '09 17:02

Chaowlert Chaisrichalermpol


3 Answers

InterlockedIncrement/Decrement on x86 CPUs (x86's lock add/dec) are automatically creating memory barrier which gives visibility to all threads (i.e., all threads can see its update as in-order, like sequential memory consistency). Memory barrier makes all pending memory loads/stores to be completed. volatile is not related to this question although C# and Java (and some C/C++ compilers) enforce volatile to make memory barrier. But, interlocked operation already has memory barrier by CPU.

Please also take a look my another answer in stackoverflow.

Note that I have assume that C#'s InterlockedIncrement/Decrement are intrinsic mapping to x86's lock add/dec.

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minjang Avatar answered Oct 02 '22 14:10

minjang


Can I assume that, the change made by Interlocked will be visible in all threads?

This depends on how you read the value. If you "just" read it, then no, this won't always be visible in other threads unless you mark it as volatile. That causes an annoying warning though.

As an alternative (and much preferred IMO), read it using another Interlocked instruction. This will always see the updated value on all threads:

int readvalue = Interlocked.CompareExchange(ref counter, 0, 0);

which returns the value read, and if it was 0 swaps it with 0.

Motivation: the warning hints that something isn't right; combining the two techniques (volatile & interlocked) wasn't the intended way to do this.

Update: it seems that another approach to reliable 32-bit reads without using "volatile" is by using Thread.VolatileRead as suggested in this answer. There is also some evidence that I am completely wrong about using Interlocked for 32-bit reads, for example this Connect issue, though I wonder if the distinction is a bit pedantic in nature.

What I really mean is: don't use this answer as your only source; I'm having my doubts about this.

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Roman Starkov Avatar answered Oct 02 '22 15:10

Roman Starkov


Actually, they aren't. If you want to safely modify counter, then you are doing the correct thing. But if you want to read counter directly you need to declare it as volatile. Otherwise, the compiler has no reason to believe that counter will change because the Interlocked operations are in code that it might not see.

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MSN Avatar answered Oct 02 '22 16:10

MSN