🚀 go-pugleaf

Thanks Very Much Andrue! (and to all!) 

Your points here will help me a lot, thanks very much for 
checking it for me, i appreciate it very much!

By the way i was sitting on my application today and i developed
it some more, and maybe i will not have to use so much memory 
as i was thinking that i will need! :-) 

Anyway, After i write this -

 DWORD arraySize = physRAMAvail * 0.8;
 BYTE * wibble = new BYTE[ arraySize ];
 memset( wibble, arraySize, 0 );

and if i don't minimized my application - can i be sure that 
all the cells of the arraySize[] array will be locked into the 
physical memory without disk swapping? even if i don't read or
write the buffer or to some of it's cells for a while?

You said that i have to allocate not more then 80% of the 
avalible physical memory (RAM), does this code gives me the
value of AVAILABLE (free) physical RAM? 

GlobalMemoryStatus( &ms );
physRAMAvail = ms.dwAvailPhys;

Well, i got to go working on my application now :-)

I'm very excited!!  

Dani.



"Andrue Cope [TeamB]" <no.spam@not.a.valid.address> wrote:
>JD wrote:
>
>> Re-read Alan's posts to me.
>
>Actually that was me :)
>
>I've carried out some research this morning on Win2K and it has
>produced some definitive and interesting results.
>
>If you use new[] to ask for up to 80% of memory then:
>
>  1.Until you populate that memory nothing much happens.
>  Windows is using lazy allocation as I expected. The VM (Virtual
>Memory) size increases but that's all. No physical RAM is consumed.
>
>  2.As you write to the array physical RAM is consumed to back the VM
>store.
>  This is very important because of the next point;
>
>  3.You will only be given physical RAM up to the 80% limit of what's
>available as/when you access your address space.
>  Actually you do get a bit more but it's temporary and rapidly gets
>reclaimed.
>
>  4.You get to keep whatever physical RAM you were given.
>  As long as you *don't minimize your application* you get to keep the
>physical RAM being used to back your single allocation.
>
>What does this mean for the Deniel?
>
>It means that you want to have something like this at program startup:
>
>  DWORD arraySize=physRAMAvail*0.8;
>  BYTE * wibble=new BYTE[ arraySize ];
>
>  memset( wibble, arraySize, 0 );
>
>This will give you what you want - 'physRAMAvail' bytes locked into
>memory without disk swapping.
>
>An important point however:If your application is minimized all bets
>are off and from my tests Windows will /severely/ trim your working set
>and page stuff out to disk.
>
>Some points:
>1.'80%' is our guess. We didn't bother calculating it exactly.
>2.I'm not sure what happens to the RAM you are given in scenario (3). I
>don't know if you get to keep 'the most you can be given' or if Windows
>stops trying to help you out.
>
>Basic conclusion:The Windows memory manager is actually trying to
>cooperate with Deniel's scenario. It has a pretty good idea what you
>are trying to do and as long as you make your intention clear from the
>start you stand a good chance of achieving what you want.
>
>I'd like to know how Linux handles this myself. It implies Windows
>Memory Manager has some intelligence but I still have a gut feeling
>that the older Unix strategy of taking RAM purely on demand is better.
>-- 
>Andrue Cope [TeamB]
>[Bicester, Uk]
>http://info.borland.com/newsgroups/guide.html