Gameboy Development Forum

MEM[i] corresponds to the value stored at the address i. So basically it contains successively the value stored at 0x0134, 0x0135, etc...
If you prefer in C code, this would be something like :

unsigned char x=0;
unsigned short *p=0x0134;

for(i=0;i!=0x19;i++){
     x=x-p[i]-1;
}

And then you compare the value of x with the header checksum at address 0x014D.

Last edited by Xephyr (2015-12-03 03:46:42)

What is going on with the header checksum might make more sense if you analyze it bit by bit. To calculate the header checksum, simply invert the bits in each byte and add all the the inverted bytes up, starting with byte 134 and ending with byte 14C.

For example, lets say you wanted to calculate the checksum of the bytes 78 and 4C. These bytes would be represented in binary as:
0111 1000
0110 1100

So, first invert them:
1000 0111
1001 0011

And add them up:
0001 1010

So the resulting checksum would be 1A


Now, this is why.

The formula is start with zero, then subtract the value of the byte you are on, then subtract 1 from that result.
0 - 9 could be rewritten as 0 + (-9)
To get the twos compliment negative of a number, one can simply invert all the bits, and then add 1 to the result
So, instead of subtracting the byte, you could instead say the formula is start with zero, then add the inverse of the byte plus 1, and then subtract 1 from the result
You can simply to start with zero, then add the inverse of each byte.

I hope this might be more clear to someone else who is wondering what is going on here.