>How can this be if we are going forward 4 bytes ( 0,1,2,3 ?) at a time we would arrive at >location 2017 ( the last byte of the 4th word) ??
You are referencing the multiplier value 4 in two places. word size (4 bytes) and 4th word, when it is really the 3rd word, since the index is 3. The multiplier designates how big our movements are, and the index designates how many movements to make. When index = 3 and multiplier equals 4, we move 4 bytes 3 times for a total of 12. On Wed, Oct 5, 2011 at 8:09 PM, Kenny Hegeland <[email protected]> wrote: > I'm wondering how you worked out 2017. As far as I know it's similar to the > addition addressing. If we were just counting numbers it would look like > this: > > 2002 + (3 * 4) = > 2002 + 12 = > 2014 > > > On Oct 5, 2011, at 10:07 PM, michael osullivan <[email protected]> wrote: > > This was discussed back in 2010-03 > > http://lists.nongnu.org/archive/html/pgubook-readers/2010-03/index.html > > but as i didn't understand it fully from reading those posts, and it was > over a year ago here is my own question: > > -------------------------------------------------------------------------------------------------------------------------- > Hi, > i have a question after reading the beginning of the book. On page 15 we > read that : > "In the indexed addressing mode, the instruction contains a memory address > to > access, and also specifies an index register to offset that address." > > Now this is all easy to follow: a memory address offset by the number > contained in an index register. So with the starting address being 2002, > with an offset of 3 we arrive at the memory location 2005. > > Then we have the next concept: the multiplier for the index, allowing us to > access memory one byte at a time , as above, or a word/4 bytes at a time > with a value of 4. > > So starting at 2002 and loading index register with 3 ( zero would mean the > location we specified in the first place ? ) , specifying a multiplier of 1 > (one byte at a time ) cycling through the the next 3 bytes we come to 2005. > > Now the next example we start at 2002 again but want to cycle 4 words > forward so we have > > 2002 index 3 + multiplier 4 (one word) arriving at memory location 2014 > > How can this be if we are going forward 4 bytes ( 0,1,2,3 ?) at a time we > would arrive at location 2017 ( the last byte of the 4th word) ?? > > > Any help greatly apreciated > > Mike > > _______________________________________________ > Pgubook-readers mailing list > [email protected] > https://lists.nongnu.org/mailman/listinfo/pgubook-readers > > > _______________________________________________ > Pgubook-readers mailing list > [email protected] > https://lists.nongnu.org/mailman/listinfo/pgubook-readers > >
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