I am very new to assembly, and try to learn it by understanding a disassembly of an old 16-bit dos game (disassembly generated by IDA Free).
There are 2 things I read in that code and I think, guessed what it does. Nevertheless I am not really sure if I'm correct, so I wanted to check (just shortened example code here btw):
1)
lds di, some_adress ; (eg: ds = 0012h, di=BAF6h afterwards)
xor cx, cx
mov [di], cx ; <- what segment is used here
I guess its using the ds as some magic default segment, to apply the offset and calculate the physical adress.
2)
assume ds:dseg (e.g. 0012h)
mov ax, 0BAF6h ; <- why is the leading 0 here btw
push ds
push ax
so my stack is looking like:
... ...
02 ds (0012)
00 ax (BAF6) <- sp
then:
mox bx, sp
les di, ss:[bx]
I guess the registers are now es=0012h and di=BAF6h, which would make sense when looking at the rest of the games code, but since my stack is looking like BAF6 0012 ..., this would mean that the first word is put into di while the second word is put into es. This confuses me a bit, since it's kinda reversing the order of the two words (from my point of view).
ds is the default segment for all other other 16-bit memory addressing modes except [bp+immediate], [bp+si+immediate] and [bp+di+immediate]. So basically, unless you use bp in indirect addressing, the default segment is ds. If you use bp in indirect addressing, then the default segment is ss.
A leading zero in hexadecimal numbers is a convention used by many disassemblers and assembler syntaxes to separate hexadecimal numbers from symbols (some other disassemblers and assemblers use 0x instead of 0). Leading zeros do not affect the numeric value of any number, and it's the same in all number systems (binary, decimal, hexadecimal, etc.).
Just as you assume, lesdi, ss:[bx] loads the first word into di and and the second word into es, so it's equivalent to:
mov di,ss:[bx]
mov es,ss:[bx+2]