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So...

I have a question, about just how bad or not bad someting that I am doing is.

I have a data structure - simple one, just a freelist.

The freelist has a state structure, one per...

Lists are easy to implement, but lock-free lists are not easy to implement. It might be potentially misleading to write the former immediately after URLs for lock-free code =-)

Thankyou BTW! :-)

*facepalm*

So there's a sent folder, and when you send a PM there's no message that it's been sent, and after sending - and just seeing the message disappear with no ack - the sent folder is...

Hi.

I received a PM. I wrote a reply in the quick reply form. I hit submit.

The page reloaded, the message disappeared and was not sent. I was still logged in.

I've just tried this again....

This is the reader thread compiled on ARM64 with -O0.



reader_start:
{
local_c0 = c0;
28c: 90000000 adrp x0, 0 <main>
290: 91000000 add x0, x0, #0x0
294: f9400000 ldr x0, [x0]

This is the reader thread compiled on ARM64 with -O0.



reader_start:
{
local_c0 = c0;
28c: 90000000 adrp x0, 0 <main>
290: 91000000 add x0, x0, #0x0
294: f9400000 ldr x0, [x0]

Mir, do you have any sources you can let me know of?

I'm been looking for confirmation of this (from sources I trust) and I can't find any.

I've not found anything which disproves it, either.
...

I think you are right. I'm in the process of checking.



Yes. That was not the misunderstanding; rather, I had forgotten that loads could down below a load barrier. I was treating it as if it...

When I say "got hold of real values", I mean to say "got hold of values which are correctly ordered".

This brings us to the work done with local_c0 and local_c1 (and will after this bring us back to the point actually in hand);



reader_start:
{
local_c0 = c0;
local_c1 = c1;
...

Too many typos. Tired!

Here is is, corrected;

If what I have read it true, then in the example you give above, the underlying object being referenced should be declared as volatile (because it...

If a variable - an int, say, is declared as volatile, then the compilers knows it is volatile.

That variable can be accessed through pointers. Those pointers may or may not have casts. From what...

As an aside, the method described is a user-mode mutex. I may be wrong, but I think it is not lock-free and will not scale. It would be interesting to think of a lock-free method to ensure two...

Mir - it's okay; we're not actually trying to get two values to update atomically :-)

There's been some discussion of the nature of memory barriers and atomic ops, and I had written some code to...

So I think this can happen (now I get all the fun of writing two threads vertically in text :-);



writer reader
====== ======
store c0
store c1 ...

I may be completely wrong, but I think what's being said is that the second thread with the acquire knows not to execute until the release in the first thread is complete.

Is that accurate?

One small note, since there is a small chance it might reveal a difference in our views which otherwise is hidden, and so allow us to reconcile our views : the acquire fence will only affect the...

In other words, it prevents loads/stores from moving before the Acquire.
[/QUOTE]

I may be completely missing something here - I hope not :-) but that does not, according to my mental picture of...

I may well be wrong, but I think it is not so. (I have not gone on to consider the other changes you've mentioned because I think this first matter affects them, they do not fix this first matter,...

So, I had a go on ARM.

The core of the code looks like this;



/***** statics *****/
int long long unsigned static volatile __attribute__( (aligned(128)) )
c0 = 0,
c1 = 1;

Another way to approach this has occurred to me, from looking at the language of the Intel manual. I think it is something in fact you yourself have already spotted.

The language says "every...

I think the method I've described isn't quite right; if we imagine the writer thread, it could increment the first counter, pause, and then during the pause the reader thread will see the two...

To be "globally visible" implies the store completed - what else could it mean?[/quote]

I may be wrong, but I think you are reading it to mean "preceeding stores will be completed and AS SUCH...

If memory barriers do not cause stores to complete, then the latency would matter.



I may be wrong, but I think you are mis-reading the documentation.

The documentation states and only...