What is 8 bytes ASROM in cheat engine 6.1?

Answer 1

alloc(TypeName,256)

alloc(ByteSize,4)

alloc(ConvertRoutine,1024)

alloc(ConvertBackRoutine,1024)

TypeName:

db 'Flash *8 type',0

ByteSize:

dd 4

//The convert routine should hold a routine that converts the data to an nteger (in eax)

//function declared as: stdcall int ConvertRoutine(unsigned char *input);

//Note: Keep in mind that this routine can be called by multiple threads at the same time.

ConvertRoutine:

[32-bit]

push ebp

mov ebp,esp

push ecx

mov ecx,[ebp+8]

[/32-bit]

//at this point ecx contains the address where the bytes are stored

//put the bytes into the eax register

mov eax,[ecx] //second fun fact, addressing with 32-bit registers doesn't work in 64-bit, it becomes a 64-bit automatically (most of the time)

shr eax,3 //shift right by 3 bits (divide by 8)

//and now exit the routine

[64-bit]

ret

[/64-bit]

[32-bit]

pop ecx

pop ebp

ret 4

[/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: stdcall void ConvertBackRoutine(int i, unsigned char *output);

ConvertBackRoutine:

[32-bit]

push ebp

mov ebp,esp

push edx //save the registers

push ecx

mov edx,[ebp+0c]

mov ecx,[ebp+08]

[/32-bit]

//at this point edx contains the address to write the value to

//and ecx contains the value

push eax

push edx

mov edx,[edx] //edx now contains the original value

and edx,7 //only save the first 3 bits

mov eax,ecx //eax gets the user input value

shl eax,3 //shift left by 3 bits (multiply by 8)

or eax,edx //add the bits of the original value

pop edx

mov [edx],eax //write the new value into the old value

pop eax

[64-bit]

//everything is back to what it was, so exit

ret

[/64-bit]

[32-bit]

//cleanup first

pop ecx

pop edx

pop ebp

ret 8

[/32-bit]