tutorial4.md

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title: CS2100 Tutorial 4 author: Qi Ji date: Week 6 ...

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Eyeballing addresses

a. op $t1, $t2

* NA, NA
* 15000, 20000

b. op $s2, 100($zero)

* NA, 100
* 200, 1000

c. op $t4, 40($s2)

* NA, 240
* 30000, 2400

d. op $s3, 200($zero)

* NA, 200
* 240, 2000

e. op $t3, $zero($t1)

* Not a valid instruction.

f. op $s1, 140($s1)

* NA, 300
* 160, 3000

Different ISAs

Translate these 3 C statements

a0 = a1 + a2;
a1 = a0 + a2;
a2 = a0 + a1;

Stack

push @a1
push @a2
add
pop @a0

push @a0
push @a2
add
pop @a1

push @a0
push @a1
add
pop @a2

Accumulator

load @a1
add @a2
store @a0

load @a0
add @a2
store @a1

load @a0
add @a1
store @a2

Memory-Memory

add @a0, @a1, @a2
add @a1, @a0, @a2
add @a2, @a0, @a1

Register-register

load $r0, @a0
load $r1, @a1
load $r2, @a2
add $r3, $r1, $r2
store $r3, @a0
add $r3, $r0, $r2
store $r3, @a1
add $r3, $r0, $r1
store $r3, @a2

Follow-up ISA

• Opcode costs 3 bits
• 128 bytes of addressable memory (\implies) 128 distinct addresses (\implies) need 7 bits to specify memory location.

Stack : 3 bits for opcode + 7 bits for address gives 10 bits

Accumulator : Same as stack

Memory-memory : 3 bits for opcode + 3 addresses in a single instruction gives 24 bits = 3 bytes

Register-register : 5 registers available, so need 3 bits to specify register.

* add costs 3 bits for opcode + 9 bits for 3 registers giving 12 bits
* load/store costs 3 bits for opcode + 3 bits for register + 7 bits for memory giving 13 bits (3 bytes needed!)

For code size, multiply number of bytes per instruction by number of instructions.