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How to remove addresses without accessed by Opcodes?

Posted: Tue Oct 11, 2022 4:57 am
by longsers
when search for pointers,How to remove addresses without accessed by Opcodes?

when Find the pointer manually, multi-level pointer,some time ,it will have Multiple results when searching for pointers .Most of them have no Opcodes access.It's almost impossible to do if checking each result manually.

How to use lua to remove address in the search result which without Opcode access in the result?

Check the address. If access is skipped, wait two seconds for no access, and then select Delete


The code to select the scan result is,This is enough

fl.Items[i].Selected = true



Search results displayed in the ce window

local fl = getMainForm().Foundlist3
print(fl.Items.Count) --The number of scan results
print(fl.Items[0].Caption) -- The first address
print(fl.Items[0].SubItems[0]) --The first value
print(fl.Items[0].SubItems[1]) --The first previous value


Check for access breakpoints


function debugger_onBreakpoint()
debug_continueFromBreakpoint(co_run)
ckAccess=1
debug_removeBreakpoint(fl.Items[i].Caption)
end
debug_setBreakpoint(fl.Items[i].Caption, 1,bptAccess)


The countdown


cdjs = createTimer()
local maincount = fl.Items.Count
local jsct = 3
function countdown()
if(jsct > 0 and maincount>0 )then
print(jsct)
jsct = jsct - 1
elseif (jsct == 0 and maincount>0) then
jsct = 3
maincount = maincount - 1
print("end")-- do some thing
else
print("realend") -- do some thing
cdjs.destroy()
end
end
cdjs.interval = 1000
cdjs.OnTimer = countdown


A correct logical code seems to can do this

Re: How to remove addresses without accessed by Opcodes?

Posted: Sun Oct 16, 2022 9:03 am
by longsers
Two answers that seem to do it

[Link]

[Link]

Re: How to remove addresses without accessed by Opcodes?

Posted: Sun Oct 16, 2022 12:49 pm
by Cas
longsers wrote:
Sun Oct 16, 2022 9:03 am

Please don’t bypass the forum systems, they are in place for a reason. Thanks