Dark Byte
Site Admin
 Reputation: 470
Joined: 09 May 2003
Posts: 25806
Location: The netherlands
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 Posted: Mon Apr 17, 2017 5:37 am Post subject: |
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here is a custom type that does the *8+whatever for you
| Code: |
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]
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