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gibberishh Cheater
Reputation: 1
Joined: 30 Aug 2021 Posts: 45
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Posted: Wed Mar 02, 2022 10:04 am Post subject: Choosing between mov and readmem() |
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Does anyone have opinions on which of the following is better? Both work exactly as expected. Looking for inputs on potential breakages, maintenance issues and/or optimization. Many of my tables have 50+ cheats and have been working without many complaints from users for approximately a year.
1) My current code for finding the base address for all cheats in a game. I was new to assembly and AA when I copied this from a helpful person.
Code: | aobscanmodule(xp,$process,8B 81 BC 01 00 00)
alloc(newmem,$100)
alloc(CharAdr,4) // CharAdr will store the base address for all pointers
label(code)
label(return)
registersymbol(xp)
registersymbol(CharAdr)
newmem:
mov [CharAdr],ecx
code:
mov eax,[ecx+000001BC] // ecx holds the base address in this instance. Large parts of the game itself use this kind of base+offset notation for values.
jmp return
xp:
jmp newmem
nop
return: |
Over time, I started looking for a different way to find the base address, one that wouldn't replace the code. Why? My tables are so freakin' large that I don't want to mess up working code by mistake. So all new development takes place in a separate development table. If the bytes are replaced, I can't use the same AOB. I either have to use define() and specify a hard-coded address, or search for a different AOB (there are plenty of viable AOBs to choose from, but it's not nice to have to keep track of several of them). So I used define() for a long time.
2) Now I have come up with this new code:
Code: | aobscanmodule(xp,$process,03 05 * * * * 0F BE 84 30)
alloc(CharAdr,4)
registersymbol(CharAdr)
CharAdr:
readmem(xp+2,4)
// add eax,[01F8FBC0] // eax is then used by the game for offsets for various values. |
In the first approach, I use CharAdr as the pointer for offsets.
In the second approach, I have to use [CharAdr] as the pointer instead.
Is there any downside to this? Since no bytes are replaced, I can open as many instances of the table as my computer will allow and all of them can attach simultaneously to the game. Of course, that's only theoretical. The most I have ever needed to attach is 3 instances.
Thanks all
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It's not cheating. It's playing by my rules. |
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TheyCallMeTim13 Wiki Contributor
Reputation: 51
Joined: 24 Feb 2017 Posts: 976 Location: Pluto
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Posted: Wed Mar 02, 2022 10:46 am Post subject: |
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The only real issue is that it won't work for a 64 bit process, because of how addresses are stored in instructions using the instruction's address as part of how the address is calculated. But for 32 bit processes it works fine.
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ParkourPenguin I post too much
Reputation: 152
Joined: 06 Jul 2014 Posts: 4702
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Posted: Wed Mar 02, 2022 12:29 pm Post subject: |
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Those are two different types of instructions: the first is using a register and the second is using a hard-coded address. Both of your methods for finding the address being accessed are fine for their respective instructions.
You can get a hard-coded address in a 64-bit process, but you might have to account for RIP-relative addressing using Lua.
gibberishh wrote: | If the bytes are replaced, I can't use the same AOB. | Many people combine scripts into one and use flags to toggle individual options.
And please don't try to attach multiple instances of CE to the same process. I can't think of any good reason to do that.
gibberishh wrote: | In the first approach, I use CharAdr as the pointer for offsets.
In the second approach, I have to use [CharAdr] as the pointer instead. | That distinction is meaningless as it only pertains to the different semantics of the instructions accessing data. In general, both scripts do the same thing: store the address being accessed in memory you allocated.
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sbryzl Master Cheater
Reputation: 6
Joined: 25 Jul 2016 Posts: 252
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Posted: Wed Mar 02, 2022 4:33 pm Post subject: |
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You don't even have to allocate any memory. You already have a base pointer in xp as soon as the aob scan is complete. You can just create a pointer with a base of xp+2.
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