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Editorial Team
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Asked: 2026-05-26T12:12:14+00:00

I am working through some indirect addressing problems and I am not sure how

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I am working through some indirect addressing problems and I am not sure how to properly count bytes. We are given this code:

.data
v1  db  9,7,5,3,1
v2  dw  0
v3  dw  -1
v4  db  '$'

mov  dx,offset v2
mov  ah,9
int  21h

The question asks how many bytes will have been written to the standard output device after these instructions have been executed and the answer is 4.

For this problem, I set it up like so:

offset  0  1  2  3  4  5  6  7  8  9
data    09 07 05 03 01 00 00 FF FF 24

We are moving 5 into dx, writing two bytes 00 05. We then set the dos code to write it out, so our output writes out the two bytes making four? Please correct me if my logic is wrong.

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1 Answer

  1. Editorial Team

    DOS function 9 writes starting at the offset in DX until it reaches a $. You’ve loaded the offset of V2 into DX. You’ve defined V2 and V3 as two bytes apiece (none of which will contain a “$”), and those are followed by V4 (containing the $). Therefore, it writes the four bytes of V2 and V3, then stops.

    Edit: I should add that contrary to the title question, none of the code you’ve shown actually does any indirect addressing (though DOS function 9 undoubtedly does use indirect addressing, reading from the address loaded into dx).

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