Are bootable anti-virus/malware scans as good as the "normal" ones?

Bogdanov89

Junior Member
Jan 27, 2015
13
0
66
I recently started using the various bootable anti-virus/malware scanners, like the Kaspersky's Rescue Disk.

However i am really interested in how do these bootable scans compare to the "normal" Windows run scans.

For example, would a maximum scan from Kaspersky Rescue Disk find as many malware/viruses as a max scan from Kaspersky Total Security?

Assuming both anti-virus/malware scanners are up to date (latest versions and databases), which is expected to reliably find and clean more viruses/malware?

Thank you for helping out
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John Connor

Lifer
Nov 30, 2012
22,757
618
121
I'm guessing the bootaable version would work better since the malware can't see if your runnig an anti-virus and some malware won't get removed via Windows. Thus a live CD is needed sometimes. But this doesn't mean you don't run an anti-virus on your computer.

I run Sandboxie for the broswer which creates a virtual environment so nothing touches the computer. Add to that an addon called NoScript and Bitdefender Free.
 

smakme7757

Golden Member
Nov 20, 2010
1,487
1
81
They are better at finding and removing virus on an infected system.

That's because they scan the hard drive while it's offline and not in use. Which gives a lot of advantages.
 

PliotronX

Diamond Member
Oct 17, 1999
8,883
107
106
I haven't found a live CD that can find anything Malwarebytes from normal or safe mode cannot. It is also inconvenient to have the media laying around, booting (sometimes not possible with secure boot for example) to it, a 200MB definition update, extremely lengthy scans, etc. For expedience, if MBAM, Emsisoft, and Adwcleaner don't fully clean it out, a reload is in order.
 

corkyg

Elite Member | Peripherals
Super Moderator
Mar 4, 2000
27,370
239
106
In theory, bootable AV/Am ware should work as well, if not better except for the fact that unless it can update its definitions on line with each use, it may be working with stale data.