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
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