CODEWITHSUNDEEP
assemblyx86-16emu8086memory-segmentation
David Ilic
David Ilic

Reputation: 165

Assembly 8086 - Finding instruction results

The table contains parts of memory:

enter image description here

The registers have these contents:

DS: 0726
CS: 0624
SS: 0727
ES: 0626
AX: 0003
BX: 0042
BP: 0036
SP: 002B

What is the result after these instructions:

a) mov ax, 10011b     ;ax = ____h 
b) mov ah, 4[BX]     ;ah = ____h 
c) mul bl     ;ax = ____h 
d) pop cx     ;cx =____ h 
e) mov ax, [BP]    ;ax = ________h

Keep in mind, the instructions are separate/sandboxed from each other. One instruction doesn't influence the next one

For context, this is not homework. I'm prepping for exams and I don't know how to solve this. I've just begun studying the 8086.

My results:

a) ax = 0013h
b) ah = 03h (Probably not, since I'm not sure what the 4 is doing in 4[BX]. Multiplication?)
c) ax = 00C6h 
d) cx = ? (I know how to get the address of the top of the stack, it's 729Bh, but I assume the value is in the table above? I don't know how to find it)
e) ax = F6 (Tried to make some sense of the table above. Probably not correct)

Upvotes: 0

Views: 326

Answers (1)

vitsoft
vitsoft

Reputation: 5775

When BP or SP is used in addressing, the default segment register is SS, otherwise it's DS. Rewrite the first column of memory dump table with linear address, i.e. instead of seg:offs calculate 16*seg+offs. This gives addresses

07260
07270
07280
07290
072A0

Ad d): pop cx loads a word from SS:SP, which is 0727:002B, which corresponds to linear address 0729B. At B-th column of fourth line we can see 00 12, so the answer is cx = 1200h.

Ad e): mov ax,[BP] loads a word from SS:BP, which is 0727:0036, which corresponds with linear address 072A6. At 6-th position in the last line we can see 18 24, so the answer is ax = 2418h.

Upvotes: 4

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