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I am confused as to how memory is stored when declaring variables in assembly language. I have this block of sample code:

val1  db  1,2
val2  dw  1,2
val3  db  '12'

From my study guide, it says that the total number of bytes required in memory to store the data declared by these three data definitions is 8 bytes (in decimal). How do I go about calculating this?

It also says that the offset into the data segment of val3 is 6 bytes and the hex byte at offset 5 is 00. I'm lost as to how to calculate these bytes and offsets.

Also, reading val1 into memory will produce 0102 but reading val3 into memory produces 3132. Are apostrophes represented by the 3 or where does it come from? How would val2 be read into memory?

raphnguyen
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2 Answers2

5

You have two bytes, 0x01 and 0x02. That's two bytes so far.

Then you have two words, 0x0001 and 0x0002. That's another four bytes, making six to date.

The you have two more bytes making up the characters of the string '12', which are 0x31 and 0x32 in ASCII (a). That's another two bytes bringing the grand total to eight.

In little-endian format (which is what you're looking at here based on the memory values your question states), they're stored as:

offset  value
------  -----
     0   0x01
     1   0x02
     2   0x01
     3   0x00
     4   0x02
     5   0x00
     6   0x31
     7   0x32

(a) The character set you're using in this case is the ASCII one (you can follow that link for a table describing all the characters in that set).

The byte values 0x30 thru 0x39 are the digits 0 thru 9, just as the bytes 0x41 thru 0x5A represent the upper-case alpha characters. The pseudo-op:

db '12'

is saying to insert the bytes for the characters '1' and '2'.

Similarly:

db 'Pax is a really cool guy',0

would give you the hex-dump representation:

addr  +0 +1 +2 +3 +4 +5 +6 +7 +8 +9 +A +B +C +D +E +F  +0123456789ABCDEF
0000  50 61 78 20 69 73 20 61 20 72 65 61 6C 6C 79 20   Pax is a really 
0010  63 6F 6F 6C 20 67 75 79 00                        cool guy.
paxdiablo
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  • +1 for the table. but the byte at offset 1 should be 2 I believe. – cyco130 Sep 22 '11 at 08:54
  • @paxdiablo Thanks for the helpful table! So words are always read into memory backwards? And can you help explain why '12' is represented as 0x31 and 0x32? – raphnguyen Sep 22 '11 at 09:02
  • @raphnguyen there are two methods for representing words. this backwards order is called big endian and the other is called little endian. '12' is a string. 0x31 and 0x32 are the ASCII codes for '1' and '2'. – cyco130 Sep 22 '11 at 09:09
  • @raphnguyen: see the update, hopefully that explains it better. – paxdiablo Sep 22 '11 at 09:18
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val1 is two consecutive bytes, 1 and 2. db means "direct byte". val2 is two consecutive words, i.e. 4 bytes, again 1 and 2. in memory they will be 1, 0, 2, 0, assuming you're on a big endian machine. val3 is a two bytes string. 31 and 32 in are 49 and 50 in hexadecimal notation, they are ASCII codes for the characters "1" and "2".

cyco130
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