Reference no: EM132312193
MARIE and Instruction Set Architecture
Task
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. For example 1980 is a leap year because it is divisible by 4 only, 2100 is not a leap year because it is divisible by both 4 and 100, and 2000 is a leap year because it is divisible by 400.
You will do this task in two steps.
(a) Remainder
To work out if a number is exactly divisible by another number, the obvious method is to perform division and look at the remainder. A zero remainder means dividend is exactly divisible by divisor. In this part, you will write a MARIE code for calculating remainder of a division. The following pseudo code implements the algorithm for remainder.
get X
get Y
repeat until X < Y:
X ← X - Y
display X
Your program should implement the above algorithm in MARIE. It will get X (dividend) and Y (divisor) as input from the user, and display the final X value which would be the remainder of division.
(b) Leap Year
In this task you will use the code written in part (a) to find out if a given year is leap year or not. The program will ask user for an input year (e.g. 1996) and work out and display a ‘1' or ‘0' in the output to indicate if given year is leap or not.
An algorithm to determine leap year is given below (with comments). Implement this algorithm in MARIE.
if (year modulo 4 ≠ 0)
|
|
|
/
|
if year
|
is
|
not divisible
|
by
|
4
|
then output 0 and exit
|
|
|
/
|
then it
|
is
|
a common year
|
|
|
else if (year modulo 100
|
≠
|
0)
|
/
|
if year
|
is
|
not divisible
|
by
|
100
|
then output 1 and exit
|
|
|
/
|
then it
|
is
|
a leap year
|
|
|
else if (year modulo 100
|
≠
|
0)
|
/
|
if year
|
is
|
not divisible
|
by
|
400
|
then output 0 and exit
|
|
|
/
|
then it
|
is
|
a common year
|
|
|
else output 1 and exit
|
|
|
/ |
otherwise it
|
is |
a leap year |
|
|
As you can see from the pseudo code, the modulo calculation is applied three times to get the final result. You will need to copy the modulo code from part (a) and paste it (and adapt it) 3 times in your program.
Assume that user will always provide valid numbers as input, that is, do not worry about dealing with invalid input data.
Submit two source files for this question as described later.
Question 2 - Instruction Set Architectures
The following instruction needs to be evaluated on different types of architectures.
R = K - (L + M/N) * P
Where R, L, M, N, P are all memory locations (or addresses).
Write assembly programs to evaluate the above expression for the following computer architectures:
(a) One operand instructions
(b) Two operand instructions
(c) Three operand instructions
(d) Zero-address instructions (stack-based)
In addition to usual memory access instructions, assume the following arithmetic instructions are available: ADD, SUBT, DIV, MULT. For some programs you may need to use temporary memory locations. Label those locations as A, B, C... .
Comment every single instruction of your program, explaining what is happening at that step.
Do not attempt to simplify the expression or rearrange its terms. But of course take care of operator precedence (multiplication and division first, followed by addition and division).
Attachment:- MARIE and Instruction Set Architecture.rar