CSCI 2500 — Computer Organization
Homework 05 (document version 1.0) — Due 9 April 2021
MIPS Circuits in C
• This homework is due by the Midnight EST on the above date via a Submitty gradeable.
• This homework is to be completed individually. Do not share your code with anyone else.
• Homeworks are available approximately two weeks before they’re due. Plan to start each
homework early. You can ask questions during office hours, in the Submitty forum, and
during your lab session.
1 Implementing an ALU, Multiplier, and Shifter
For this assignment, you’ll be building upon the 1-bit ALU circuit you developed for Lab 5. Specif-
ically, you’ll be using your logic gates to implement circuits for a 32-bit ALU, a 16-bit multiplier,
and a 32-bit left shifter. Your circuit will be supporting 8 different MIPS instructions given in the
table below.
Instruction Input symbol OP bits Operation
and & 000 Logical AND
or | 001 Logical OR
nor n 010 Logical NOR
add + 011 Integer addition
sub - 100 Integer subtraction
slt < 101 Set if less than
sll s 110 Shift left logical
mul * 111 Integer multiplication
The OP bits are an array of three BITS that will be sent to your control circuit. These bits will
determine which circuit output gets sent back to main. See the hw05.c template file for function
prototypes and more detail into the design and implementation specifications.
1.1 Data Format
The input/output as well as intermediate format for storing integers will be an extension of the
BIT type we used for Lab 5. 32-bit 2’s complement numbers will be stored as binary in an array
of 32 BITs. For these arrays, the most significant bit is at index 31 and the least significant bit is
at index 0. The first function you’ll need to implement is probably convert to binary, to convert
the input decimal integers to their BIT representations.
1.2 32-bit ALU
See slides 30 and 31 of Chapter-3d-ALU.pdf for logical diagrams of the 1-bit ALU and 32-bit ripple
ALU that you’ll be implementing. The 1-bit ALU will extend the Lab 5 ALU to also include
functionality for set-less-than (slt). You can use a for loop here to simplify the 32 ’ripples’. You’ll
also need to consider logical NOR. You can implement it into your ALU with an additional control
bit to flip the OR output, or you can implement the equivalent within your Control function.
1.3 16-bit Multiplier
See slides 49 and 50 for logic diagrams for your multiplier. You should use your 32-bit ALU circuit
as well as your shifter circuit for this implementation. Note that since we’re just implementing a
32-bit ALU, we can effectively only multiply two 16-bit numbers without overflow. You are allowed
to use a for loop to simulate the ’Control’ of your circuit.
1.4 32-bit Left Shifter
You’ll implement two circuits related to shifting. First, the shifter function will shift the input by
1-bit either left or right, depending on the control bit. One possible design is given here: http://
www.mathcs.emory.edu/~jallen/Courses/355/Syllabus/1-circuits/shifter.html. This cir-
cuit will be useful for your multiplier. The second function is left shifter32, which will shift
input A to the left by the amount specified in input B. There’s multiple ways to implement such
a circuit. I give you pseudocode one possible approach in the template, which doesn’t require a
considerable number of gates.
1.5 ALU Control
The control circuit will use the OP bits to determine which operation to perform, or from which
operation to return results to main. This circuit can also be used to determine which control bits
to set for your ALU. See the template file for details on my approach.
1.6 Assignment Rules
We don’t usually have much in the way of rules for our assignments, but this one is a special case.
The primary purpose of this assignment is to demonstrate how basic logic gates can be combined
together into complex circuits. As a rule, whatever you implement in your solution file should be
a loose representation of an implementable circuit. The below are guidelines to this end:
1. No modifications to main().
2. You are allowed to implement additional basic gates – e.g., a three input AND gate – to
simplify your other circuits.
3. If desired, you can modify the function prototypes given in the template. For example, you
can pass an additional control bit to your ALU for the sake of outputting NOR instead of
OR.
2
4. No if() or if()-else() logical control statements, or equivalent.
5. You can use basic for loops only (for int i = 0; i < N; ++i). Different loop structures are not
allowed, as they can easily be used as a logical equivalent to if-else statements.
6. No calls to any external or library functions.
7. The above rules are to basically ensure that you’re sticking to using gates for all logical control
and operations. No tricky workarounds are allowed.
2 Input and output
Below are examples of inputs and expected outputs for the assignment. Generally, you should
expect positive or negative numbers for all inputs except for the shift amount for shift left logical
(note: if you implement the circuits as given in the Chapter-3d pdf, you should have to do no
additional work to account for negative inputs).
bash$ ./hw05
12 & 5
00000000000000000000000000001100
&
00000000000000000000000000000101
=
00000000000000000000000000000100
bash$ ./hw05
12 | 5
00000000000000000000000000001100
|
00000000000000000000000000000101
=
00000000000000000000000000001101
bash$ ./hw05
12 n 5
00000000000000000000000000001100
n
00000000000000000000000000000101
=
11111111111111111111111111110010
bash$ ./hw05
12 + 5
00000000000000000000000000001100
+
00000000000000000000000000000101
=
3
00000000000000000000000000010001
bash$ ./hw05
12 + -5
00000000000000000000000000001100
+
11111111111111111111111111111011
=
00000000000000000000000000000111
bash$ ./hw05
12 - 5
00000000000000000000000000001100
-
00000000000000000000000000000101
=
00000000000000000000000000000111
bash$ ./hw05
12 < 5
00000000000000000000000000001100
<
00000000000000000000000000000101
=
00000000000000000000000000000000
bash$ ./hw05
-12 < 5
11111111111111111111111111110100
<
00000000000000000000000000000101
=
00000000000000000000000000000001
bash$ ./hw05
12 s 5
00000000000000000000000000001100
s
00000000000000000000000000000101
=
00000000000000000000000110000000
bash$ ./hw05
12 * 5
00000000000000000000000000001100
*
00000000000000000000000000000101
=
4
00000000000000000000000000111100
bash$ ./hw05
12 * -5
00000000000000000000000000001100
*
11111111111111111111111111111011
=
11111111111111111111111111000100
3 Submission and Grading Criteria
For this assignment, you will submit your code to the Submitty gradeable. A code template hw05.c
is provided for your convenience. Do your best to stick to the above rules, as the TA grade is mostly
based on your adherence to them. The below will be the grading criteria for the assignment.
1. Autograding: 60%
• Standard visible and hidden test cases
2. TA grading: 40%
• Solution is representative of an implementable circuit
• Solution otherwise adheres to the rules given above
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