Memory Addressing

0.0(0)
Studied by 0 people
call kaiCall Kai
learnLearn
examPractice Test
spaced repetitionSpaced Repetition
heart puzzleMatch
flashcardsFlashcards
GameKnowt Play
Card Sorting

1/3

encourage image

There's no tags or description

Looks like no tags are added yet.

Last updated 12:10 PM on 5/10/26
Name
Mastery
Learn
Test
Matching
Spaced
Call with Kai

No analytics yet

Send a link to your students to track their progress

4 Terms

1
New cards
How is memory addressing resolved and what are the standard instruction data sizes?
Memory Addressing:
* Memory addresses are always in byte resolution.
* You cannot access less than 1 byte.

Instruction data sizes:
* Byte = 1B
* Half Word (HW) = 2B
* Word (W) = 4B
* Double Word (DW) = 8B
2
New cards
What is the rule for Memory Alignment and how is it calculated?
Memory Alignment:
* An object of size (S bytes) at address (A) is aligned if:
AmodS=0A \bmod S = 0

Alignment rules:
* e.g. 4 bytes word can only start at addresses 0, 4, 8, 12...
* Last few digits of an address in binary show what the address number is divisible by.
3
New cards
What are the binary address requirements for different data alignments?
Alignment Binary Requirements:

* Half-word (2 byte) addresses:
* xxxxx0x \ldots xxxx0 (address in binary must end in a zero)
* Word (4 bytes) addresses:
* xxxx00x \ldots xxx00 (address in binary must end in two zeros)
* Double-word (8 bytes) addresses:
* xxx000x \ldots xx000
* Quad-word (16 bytes) addresses:
* xx0000x \ldots x0000
4
New cards
What is the difference between Little Endian and Big Endian byte ordering?
Little Endian:
* The first byte is stored at the least significant position in the word.
* Example ordering:
* 3 2 1 0 -> "DCBA"

Big Endian:
* The first byte is stored at the most significant position in the word.
* Example ordering:
* 0 1 2 3 -> "AB