liekong1991
How do I cheat?
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Joined: 08 Feb 2020
Posts: 6
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 Posted: Sat Feb 08, 2020 5:35 am Post subject: About the "X18 XOR" Custom Type |
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I just dont really understand the code's meaning, "X18" means what?
| Code: |
alloc(ConvertRoutine,1024)
alloc(ConvertBackRoutine,1024)
alloc(TypeName,256)
alloc(ByteSize,4)
alloc(UsesFloat,1)
alloc(CallMethod,1)
alloc(PREFEREDALIGNMENT,4)
PREFEREDALIGNMENT:
dd 4
TypeName:
db 'X18 XOR',0
ByteSize:
dd 36
UsesFloat:
db 0 //Change to 1 if this custom type should be treated as a float
CallMethod:
db 1 //Remove or change to 0 for legacy call mechanism
//The convert routine should hold a routine that converts the data to an integer (in eax)
//function declared as: cdecl int ConvertRoutine(unsigned char *input, PTR_UINT address);
//Note: Keep in mind that this routine can be called by multiple threads at the same time.
ConvertRoutine:
//jmp dllname.functionname
[64-bit]
//or manual:
//parameters: (64-bit)
//rcx=address of input
//rdx=address
mov eax,[rcx] //eax now contains the bytes 'input' pointed to
xor eax,[rcx+18]
ret
[/64-bit]
[32-bit]
//jmp dllname.functionname
//or manual:
//parameters: (32-bit)
push ebp
mov ebp,esp
//[ebp+8]=address of input
//[ebp+c]=address
//example:
push ebx
mov ebx,[ebp+8] //place the address that contains the bytes into eax
mov eax,[ebx] //place the bytes into eax so it's handled as a normal 4 byte value
xor eax,[ebx+18]
pop ebx
pop ebp
ret
[/32-bit]
//The convert back routine should hold a routine that converts the given integer back to a row of bytes (e.g when the user wats to write a new value)
//function declared as: cdecl void ConvertBackRoutine(int i, PTR_UINT address, unsigned char *output);
ConvertBackRoutine:
//jmp dllname.functionname
//or manual:
[64-bit]
//parameters: (64-bit)
//ecx=input
//rdx=address
//r8=address of output
//example:
xor ecx,[r8+18]
mov [r8],ecx //place the integer at the 4 bytes pointed to by r8
ret
[/64-bit]
[32-bit]
//parameters: (32-bit)
push ebp
mov ebp,esp
//[ebp+8]=input
//[ebp+c]=address
//[ebp+10]=address of output
//example:
push eax
push ebx
mov eax,[ebp+8] //load the value into eax
mov ebx,[ebp+10] //load the output address into ebx
xor eax,[ebx+18]
mov [ebx],eax //write the value into the address
pop ebx
pop eax
pop ebp
ret
[/32-bit]
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[/code]
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