Re: About volatile qualifier

From:

"Alex Blekhman" <tkfx.REMOVE@yahoo.com>

Newsgroups:

microsoft.public.vc.language

Date:

Wed, 18 Jun 2008 18:25:45 +0300

Message-ID:

<estEWeV0IHA.2384@TK2MSFTNGP02.phx.gbl>

"Krat" wrote:

"If we neglect to qualify the variable "flag" as "volatile" then
the compiler optimizes access to that variable by caching it in
a register. [...]"

As far as know, for this happen this two threads should run on
different processors or cores because of cache issues. Does this
happen on a single core? If yes what is the reason?

It happens all the same on a single core system. The author does
not speak about CPU cache here, but about abstract "caching"
optimization that compiler performs. Instead of reading the
variable from a memory on each access compiler reads it only once
and then stores the value in a register. It can be represented
with the following pseudocode:

; while statement
....
0001 mov ecx,byte ptr [flag]
flag_check:
0002 test ecx,ecx
0003 jne end_of_while
0004 ...
0005 jmp flag_check
end_of_while:
0006 ...

As you can see, after an execution of while's body the flow jumps
back to test the flag value without reading it from the memory. It
will behave exactly the same both on single and multi CPU system.

HTH
Alex