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NanoByte
Expert Cheater
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Joined: 13 Sep 2013
Posts: 222
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 Posted: Sat May 31, 2014 5:28 pm Post subject: call and ret = return a value to the same spot :D |
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This will make alot of people alot happy 
i got alot of offsets that i have to edit
but don't what to push and pop and etc to get the maximum amount so i can set it
so i was think something like this
| Code: | newmem:
push rdx
mov rdx,[rdi+123]
call maxxi //dont know how this works just guessing from my c# background
//now i want to send rdx to maxxi to add +4
//now i got the value back with +4
mov [rdi+123],rdx
i got more then 20 offsets that i have to edit
would be alot easier if i could say mov [rdi+123],[rdi+123+4] but i know i cant :(
maxxi: //
push eax
Mov eax,[rdx+4] //add 4 to the rdx to get the maxximum nr
mov rdx,[eax]
pop eax
ret // dont know how this works!! -- return this value to where it was called
originalcode: |
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Daijobu
Master Cheater
 Reputation: 13
Joined: 05 Feb 2013
Posts: 301
Location: the Netherlands
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| Posted: Sat May 31, 2014 5:54 pm Post subject: |
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Should work. As far as I'm aware you can call a label and use ret.
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justa_dude
Grandmaster Cheater
 Reputation: 23
Joined: 29 Jun 2010
Posts: 892
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| Posted: Sat May 31, 2014 7:11 pm Post subject: Re: call and ret = return a value to the same spot :D |
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| NanoByte wrote: | | // dont know how this works!! -- return this value to where it was called |
call is basically shorthand for "push instruction pointer (eip/rip), jump target"
ret is basically shorthand for "add esp/rsp address, jump [esp-4/rsp-8]'
This might be good to know if your function is working on data in the stack, since you'll need to use a different offset inside the function.
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Gniarf
Grandmaster Cheater Supreme
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Joined: 12 Mar 2012
Posts: 1285
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| Posted: Sat May 31, 2014 8:19 pm Post subject: |
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@NanoByte: If I understand correctly what you want to do, line 3 needs to be "lea rdx,[rdi+123]" instead of a mov (you want rdx=rdi+123, not rdx=the data at rdi+123, right?).
There is also a problem in maxxi:
| Code: | Mov eax,[rdx+4] //puts the 4 bytes at rdx+4 into eax, so for example now eax=100 hitpoints
mov rdx,[eax] //puts the 4 bytes at ADDRESS 100 hitpoints into rdx->crash |
What you want to do is probably: | Code: | Mov eax,[rdx+4]
mov edx,eax //copies eax into rdx, so now edx=100 hitpoints |
Which you could write in one line: | Code: | | Mov edx,[rdx+4] //now edx=100 hitpoints and you don't need the push/pop eax anymore |
...This or what you want to do is more complex than "mov [rdi+123],[rdi+123+4]"
Seeing that you code is going to look like:
| Code: | newmem:
push rdx
//repeat and adjust this block 20 times
lea rdx, [rdi+123]
call maxxi
//mov [rdi+123],rdx
mov [rdi+123],edx //edx if your variable is a "4 bytes" in cheat engine, rdx is it is a "8 bytes"
//...
pop rdx
originalcode:
//...
jmp returnhere
maxxi:
Mov edx,[rdx+4]
ret |
You might want to write it that way: | Code: | newmem:
//repeat and adjust this block 20 times
push dword [rdi+123+4] //saves the 4 bytes (=a dword) at rdi+???+4 into a new temporary variable
//btw I think that cheat engine understands that 123+4=127
pop dword [rdi+123] //copies the dword from the temporary variable into rdi+??? and frees the temp variable
//...
originalcode:
//...
jmp returnhere |
Unless this is exactly what you wanted to avoid when you said "but don't what to push and pop and etc to get the maximum amount so i can set it", I'm not sure I understand very well what you meant there.
--------------------- EDIT ---------------------
Thinking again about it, this one would be even better (don't know why I didn't think about it the first time): | Code: | newmem:
push eax
//repeat and adjust those 2 lines 20 times
mov eax, dword [rdi+123+4]
mov dword [rdi+123], eax
//...
pop eax
originalcode:
//...
jmp returnhere |
Making functions for something you repeat over and over may seem a good idea, but I think the operation is so small this time that one more function does not improve readability (and definitely lowers performance, even if you don't notice the difference).
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