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gawron25
How do I cheat?
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Joined: 04 Jun 2011
Posts: 3
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 Posted: Sun Nov 25, 2012 3:51 pm Post subject: aa to c++ |
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Hey
Can any one tell me how to convert this
| Code: | [ENABLE]
alloc(blah,2024)
alloc(blah3,2024)
label(blah2)
008D265E:
JMP blah
blah:
jmp blah2
blah2:
MOV BX,WORD PTR DS:[EDI+10]
->MOV EAX,blah3<-
MOV DWORD PTR DS:[EAX],E7395B84 |
MOV EAX,blah3
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SteveAndrew
Master Cheater
 Reputation: 30
Joined: 02 Sep 2012
Posts: 323
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| Posted: Tue Nov 27, 2012 12:23 am Post subject: |
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Well I think I see what your having trouble with...
In CE's auto assembler not including brackets in your instruction '[' and ']' means to use the address of the variable rather than the value of it.
In C++ inline assembler it doesn't matter whether you include them or not it gets interpreted the same exact way. (also MASM and likely other assemblers as well)
example:
Is the same exact thing as
In inline assembler...
In CE's autoassembler they are different!
So what you can do is either use 'lea' (load effective address) instead of 'mov' or use a pointer to blah3.
So something like this: (Does blah3 really need to be 2024 bytes? or what is it?) Also make sure to put an 0x before a hex number as it defaults to decimal when not writing a CE auto assembler script [like inline assembler instead])
| Code: |
BYTE blah3[2024];
void __declspec(naked) blah2()
{
__asm
{
mov bx,[edi+0x10]
lea eax,blah3
mov [eax],0xE7395B84
}
}
//OR
BYTE blah3[2024];
BYTE *blah3pointer = &blah3;
void __declspec(naked) blah2()
{
__asm
{
mov bx,[edi+0x10]
mov eax,blah3pointer
mov [eax],0xE7395B84
}
}
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the instruction 'lea eax,blah3' is the same as 'lea eax,[blah3]' here. brackets or not when dealing with variables in inline assembler it always will have brackets [ ] whether you put them or not. That's only variables though registers can have them or not have them normally as shown even just this little code.
If you had allocated the memory for blah3 dynamically you probably wouldn't have encountered this problem, as blah3 would have already been a pointer: (Then you could've kept the mov)
| Code: |
BYTE *blah3 = new BYTE[2024]
void __declspec(naked) blah2()
{
__asm
{
mov bx,[edi+0x10]
mov eax,blah3
mov [eax],0xE7395B84
}
}
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Any memory you allocate though make sure to keep track of it until you free it, no one likes memory leaks! And in C you have to manually manage your memory!
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n0 m3rcY
Cheater
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Joined: 18 Jun 2012
Posts: 42
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| Posted: Tue Nov 27, 2012 8:22 pm Post subject: |
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In masm syntax mov eax, esi+10 gives the value of esi+10 wheres mov eax, [esi+10] gives you the address (effectively lea).
Of course, this is just what I remember about masm being reverse fasm in the bracket-access of pointers, so...
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