Going back to Avast 6?

I’ve normally updated my Avast Free version as new ones come out, but after getting to 7.0.1426 I read on this forum that Avast 7, in theory, does not support Windows 2000, which I still use. I don’t have a big problem with my desktop machine, 2.4 gig Pentium 4 with 1GB memory, but my laptop seems to have got much slower, and has regular appearance of the “your system is low on virtual memory” message. This is a 850 meg Pentium 3 with only 256MB memory and no real possibility of increasing this.

It appears that 6.0.1270 is the latest version that supports W2k. Do I lose anything of significance by going back there? Can I uninstall 7 and install 6 without having to re-register?

Thanks, Steve.

latest 7.0.1474 runs on W2000

http://filehippo.com/download_avast_antivirus/tech/

Please note that avast! Free Antivirus runs only on PCs with Windows 2000 Professional Service Pack 4 and newer.

in theory… what do you mean?
in practice, latest 7.0.1474 works on Win2k (version 8 will be also supported for Win2k OS), so don’t worry. A lot of new features (improved behav. shield, virtualization, networking) won’t work on Win2k, because this OS just doesn’t support these new technologies – also, we don’t test avast on Win2k at all, so if there’s a problem between avast and your OS, you should let us know on forum. Minimum OS for testing here is WinXP.

Thanks, all.

I’m not sure exactly where I read that W2k was not supported by Avast 7, but it was definitely a thread a few months back on this forum, and some searches I did elsewhere this morning seemed to confirm it. Maybe I didn’t follow the thread to its conclusion. I’m using 7 on W2k (Pro SP4), so I do know it works in practice.

It’s just that at some time the old laptop seemed to get much slower, and I can’t think of anything that had changed except the switch to Avast 7. I hadn’t been getting low virtual memory problems since I upgraded from 128MB memory(!).

Hopefully, I’ll be changing soon to a P4 laptop with 1GB (once I’ve found a power unit), which should solve the problem.

Steve.