Cheat Engine

[Ullisses]

I've been studying ASM, and i'm having a hard time understanding this.

[Zanzer]

EBP+08 equals a pointer address.
EBX now equals that address.
So the value at that address is changed to 110.
Adding 2 to EBX then moves that pointer up 2 bytes.
This means you are overwriting the last 2 bytes of the previous 4-byte '110' value you just moved.

So, whatever address that was, you just wrote the following 6-bytes:
6E 00 7B 00 00 00

You should also note that "ebp+10" is a hex 10, decimal 16.
So by adding 2, you should have expected "ebp+a".

[Stylo]

If we take a look at the second code snippet you posted (the one which sets ebp-4 to zero and than check its value against 100).
you're having problem to understand it because it's not the whole code that you're looking at.
the compiler usually does that kind of thing when a loop is about to happen, if you think about it, the ebp-4 is a local variable that represent the loop counter (for example the variable i in high level language), it starts from 0 and than immediately compared to 100 to see if it passed that value.
what you cannot see which is further that code, is the rest of the code that happen inside the loop (you can notice the jnl <some address> line which jump if it is lower than 100)

the part of