MIPS Memory Addressing

Byte vs word addressing, alignment, and the “address of the last byte/word” calculation (Q3(c) most semesters).

Byte addressing

Memory is one large single-dimension array; an address is an index into it. MIPS uses byte addressing: each address points to 1 byte (8 bits), and a load/store’s unit is a byte. MIPS uses a 32-bit byte address.

Words

Most data items are 4 bytes wide. For MIPS a word = 32 bits = 4 bytes.
Registers hold 32 bits, matching a word.
Word address = the address of the first byte in the word.
Words are aligned: word addresses are 0, 4, 8, 12, … so the low 2 bits of a word address are always 00.

word 0

word 4

word 8

…

Byte addresses grouped into words — the word address is its first byte.

MIPS-32 address space

Quantity	Value
Total bytes	$2^{32}$ bytes
Byte addresses	$0 \dots 2^{32} - 1$
Address of last byte	$2^{32} - 1$
Total words	$2^{30}$ words
Word addresses	$0, 4, 8, \dots, 2^{32} - 4$
Address of last word	$2^{32} - 4$

Why

2^{30}

words?

2^{32}\ \text{bytes} \div 4\ \text{bytes/word} = 2^{30}\ \text{words}

. The last word starts 4 bytes before the end of memory, so its (first-byte) address is

2^{32} - 4

MIPS-64 (Autumn-24 Q3b)

“Calculate the address of the last word in MIPS-64, assuming everything is similar to MIPS-32.”

Scale the widths up by analogy. In MIPS-32 the address width (32) equals the word width (32 bits = 4 bytes). Keeping the same relationship for MIPS-64: 64-bit addresses, and a word = 64 bits = 8 bytes.

\begin{aligned} \text{Total bytes} &= 2^{64} \\ \text{Address of last byte} &= 2^{64} - 1 \\ \text{Word} = 8\ \text{bytes} \;&\Rightarrow\; \text{last word address} = 2^{64} - 8 \end{aligned}

Primary answer: last byte =

2^{64} - 1

, last word =

2^{64} - 8

(8-byte words). If your instructor keeps the word at 4 bytes (“similar” = same word size), then last word =

2^{64} - 4

— state your assumption either way.

Max offset of a load (Spring-26 Q3c)

lw $s1, offset($s2): the offset lives in the 16-bit immediate field and is signed.

\text{Max positive offset} = 2^{15} - 1 = 32767 = 0111\,1111\,1111\,1111_2

Machine instruction with the max offset (lw op = 100011, $s2 = 18 = 10010, $s1 = 17 = 10001):

lw $s1, 32767($s2)