ITC161 Computer Systems Assignment - MARIE and Instruction Set Architecture, Charles Sturt University, Australia
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Question 1 - MARIE Assembly
In this task you will write a program in MARIE assembly language, to work out if a given year is leap year or not. Leap years are those that are divisible by 4, except the years that are divisible by 100. But the centurial years are leap years if they are divisible by 400. You will do this task in two steps.
(a) Remainder
(a) MARIE PROGRAM:
Org 100
Input /Input A value
Store A / A = AC
Input /Input B value
Store B / B = AC
If, Load A /AC = A
Skipcond 800 /skip and continue
Jump EndIf /jump if condition is not satisfy
Then, Load A /AC = A
Subt B / AC = AC - B
Store A /A =AC
Load C /AC = C
Add One /AC=AC+1
Store C /C = AC
Jump If /Jump if condition satisfy
EndIf, Load C /AC = C
Output
Halt, Output
C, DEC 0
A, DEC 0
B, DEC 0
One, DEC 1
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(b) Leap Year
(b) If a year is divisible by 4 and 100 but not divisible by 400, it's not a leap year.
If a year is divisible by 4 but not divisible by 100, it's a leap year.
If a year is not divisible by 1, it's not a leap year.
This logic is implemented in the above program using nested if...else statement.
Program:
Org 100
Input /Input A value
Store A / A = AC
Input /Input B value
Store B / B = AC
If, Load A /AC = A
Skipcond 800 /skip and continue
Jump EndIf /jump if condition is not satisfy
Then, Load A /AC = A
Subt B / AC = AC - B
Store A /A =AC
Load C /AC = C
Add One /AC=AC+1
Store C /C = AC
Jump If /Jump if condition satisfy
Load D /AC = D
Store A /A = AC
/Skipcond 800 /skip and continue
/ Jump EndIf /jump if condition is not satisfy
Load A /AC = A
Subt B / AC = AC - B
Store A /A =AC
Load C /AC = C
Add One /AC=AC+1
Store C /C = AC
Jump If /Jump if condition satisfy
EndIf, Load C /AC = C
Halt, Output
C, DEC 0
A, DEC 0
B, DEC 0
D, DEC 100
One, DEC 1
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Question 2 - Instruction Set Architectures
The following instruction needs to be evaluated on different types of architectures.
R = K - (L + M/N) * P
Write assembly programs to evaluate the above expression for the following computer architectures:
(a) One operand instructions
Given expression: R = K - (L + M/N) * P
M[ ] is any memory location
LOAD M AC = M[M]
DIV N AC = AC / M[N]
STORE A M[A] = AC
ADD L AC = AC + M[L]
STORE B M[B] = AC
MUL P AC = AC *P
STORE C M[C] = AC
LOAD K AC = K
SUBT C AC = AC - M[C]
STORE R M[R] = AC
OUTPUT OUTPUT DISPLAY AS AC CONTENT
HALT
K, Dec 10
L, Dec 02
M, Dec 04
N, Dec 02
P, Dec 02
(b) Two operand instructions
Given expression: R = K - (L + M/N) * P
Let R1, R2, R3 are registers
M[ ] is any memory location
Given that R, L, M, N, P are memory locations
MOV R1, M R1 = M[M]
DIV R1, N R1 = R1 / M[N]
MOV R2, L R2 = M[L]
ADD R1, R2 R1 = R1+R2
MOV R3, P R3 = M[P]
MUL R1, R3 R1 = R1*R3
MOV R3, K R3 = K
SUBT R3, R1 R3 = R3 - R1
MOV R, R1 M[R] = R3
OUTPUT OUTPUT DISPLAY AS AC CONTENT
HALT
K, Dec 10
L, Dec 02
M, Dec 04
N, Dec 02
P, Dec 02
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(c) Three operand instructions
Given expression: R = K - (L + M/N) * P
Let R1, R2, R3 are registers
M[ ] is any memory location
Given that R, L, M, N, P are memory locations
DIV R1, M, N R1 = M[M] / M[N]
ADD R1, L R1 = R1+M[L]
MUL R1, P R1 = R1* M[P]
SUBT R, K, R1 R = K - R1
OUTPUT OUTPUT DISPLAY AS AC CONTENT
HALT
K, Dec 10
L, Dec 02
M, Dec 04
N, Dec 02
P, Dec 02
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(d) Zero-address instructions (stack-based)
Given expression: R = K - (L + M/N) * P
Let R1, R2, R3 are registers
M[ ] is any memory location
Given that R, L, M, N, P are memory locations
PUSH M TOP = M[M]
PUSH N TOP = M[N]
DIV TOP = M[M]/M[N]
PUSH L TOP = M[L]
ADD TOP = M[L]+M[M]/M[N]
PUSH P TOP = M[P]
MUL TOP = M[P]*(M[L]+M[M]/M[N])
POP A M[A] = TOP
PUSH K TOP = K
PUSH A TOP = M[A]
SUBT TOP = K - M[A]
POP R M[R] = TOP
OUTPUT OUTPUT DISPLAY AS AC CONTENT
HALT
K, Dec 10
L, Dec 02
M, Dec 04
N, Dec 02
P, Dec 02
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