Assignment Solution

GRADING:

Q1 and Q2 are mandatory questions. In Q1 you will have to make an assembler. In Q2 you have to make a simulator for which detailed information is mentioned in the respective questions.Q3 is a bonus question.

For Q1 and Q2:

Grading will be based on the number of test cases that your program passes.

    1. Assembler: The test cases are divided into 3 sets:

        a. ErrorGen: These tests are supposed to generate errors

        b. simpleBin: These are simple test cases which are supposed to generate a binary.

        c. hardGen: These are hard test cases which are supposed to generate a binary.

    2. Simulator: The test cases are divided into 2 sets:

        a. simpleBin: These are simple test cases which are supposed to generate a trace.

        b. hardGen: These are hard test cases which are supposed to generate a trace.

The TA will grade the errorGen cases manually.

For Q3:

For the bonus question, you need to generate some graphs which you must show to the TA during the final evaluation. The test cases for the bonus question will be provided to you separately well before the date of the deadline. Till then you can make some new test cases to try this on your own.

EVALUATION PROCEDURE:

    1. Date for the demo of the mid and final evaluation will be announced in due time.

    2. On the day of your demo, a compressed archive of all tests will be shared with you on the google classroom. This archive will include other test cases as well which will not be provided to you beforehand.
    3. On the day of evaluation, you must

        a. Prove that you are not running code written after the deadline by running “git log HEAD” which prints the date and time of the commit pointed to by the HEAD. You must also run “git status” to show that you don’t have any uncommitted changes.
        b. Prove the integrity of the tests archive by computing the sha256sum of the archive. To compute the checksum, you can run “sha256sum <path/to/the/archive>”. The TA will then match the checksum to verify the integrity.
    4. Then you can extract the archive and replace the “automatedTesting/tests” directory.

    5. Then you need to execute the automated testing infrastructure, which will run all the tests and finally print your score.
    6. The TA will verify the correctness for the test cases which are supposed to generate errors. You do not need to run these tests by yourself. The testing infrastructure will do this automatically for you.

PLAGIARISM

    1. Any copying of code from your peers or from the internet will invoke institute Plagiarism policy.
    2. Provide proper references if you're taking your code from some other resource. Needless to say, if the said code is the main part of the assignment, You will be awarded 0 marks.
    3. If you are found indulging in any bad practise to circumvent the above mentioned evaluation procedure, you will be awarded 0 marks and institute plagiarism policy will be applied.






THIS IS INTENTIONALLY LEFT BLANK
Assignment Description


ISA description:

Consider a 16 bit ISA with the following instructions and opcodes, along with the syntax of an assembly language which supports this ISA.

The ISA has 6 encoding types of instructions. The description of the types is given later.

Opcode
Instruction
Semantics
Syntax
Type





00000
Addition
Performs reg1 = reg2
add reg1 reg2 reg3
A


+ reg3. If the




computation




overflows, then the




overflow flag is set







00001
Subtraction
Performs reg1 = reg2
sub reg1 reg2 reg3
A


- reg3. In case reg3




> reg2, 0 is written




to reg1 and overflow




flag is set.







00010
Move
Performs reg1 = $Imm
mov reg1 $Imm
B

Immediate
where Imm is a 8 bit




value.







00011
Move Register
Performs reg1 =
mov reg1 reg2
C


reg2.







00100
Load
Loads data from
ld reg1 mem_addr
D


mem_addr into reg1.







00101
Store
Stores data from
st reg1 mem_addr
D


reg1 to mem_addr.







00110
Multiply
Performs reg1 = reg2
mul reg1 reg2 reg3
A


x reg3. If the




computation




overflows, then the




overflow flag is




set.







00111
Divide
Performs reg3/reg4.
div reg3 reg4
C


Stores the quotient




in R0 and the




remainder in R1.







01000
Right Shift
Right shifts reg1 by
rs reg1 $Imm
B


$Imm, where $Imm is




an 8 bit
value.







01001
Left Shift
Left shifts reg1 by
ls reg1 $Imm
B


$Imm, where $Imm is




an 8 bit
value.








01010
Exclusive OR
Performs
bitwise XOR
xor reg1 reg2 reg3
A


of reg2 and reg3.




Stores the result in




reg1.









01011
Or
Performs
bitwise OR
or reg1 reg2 reg3
A


of reg2 and reg3.




Stores the result in




reg1.









01100
And
Performs
bitwise AND
and reg1 reg2 reg3
A


of reg2 and reg3.




Stores the result in




reg1.









01101
Invert
Performs
bitwise NOT
not reg1 reg2
C


of reg2. Stores the




result in reg1.








01110
Compare
Compares
reg1 and
cmp reg1 reg2
C


reg2 and
sets up the




FLAGS register.








01111
Unconditional
Jumps to
mem_addr,
jmp mem_addr
E

Jump
where mem_addr is a




memory address.







10000
Jump If Less
Jump to mem_addr if
jlt mem_addr
E

Than
the less
than flag




is set (less than




flag = 1), where




mem_addr
is a memory




address.









10001
Jump If
Jump to mem_addr if
jgt mem_addr
E

Greater Than
the greater than




flag is set (greater




than flag = 1),




where mem_addr is a




memory address.







10010
Jump If Equal
Jump to mem_addr if
je mem_addr
E


the equal flag is




set (equal flag =




1), where mem_addr




is a memory address.







10011
Halt
Stops the machine
hlt
F


from executing until




reset








where reg(x) denotes register, mem_addr is a memory address (must be an 8-bit binary number), and Imm denotes a constant value (must be an 8-bit binary number).
The ISA has 7 general purpose registers and 1 flag register. The ISA supports an address size of 8 bits, which is double byte addressable. Therefore, each address fetches two bytes of data. This results in a total address space of 512 bytes. This ISA only supports whole number arithmetic. If the subtraction results in a negative number; for example “3 - 4”, the reg value will be set to 0 and overflow bit will be set. All the representations of the number are hence unsigned.
The registers in assembly are named as R0, R1, R2, ... , R6 and FLAGS. Each register is 16 bits.

Note: “mov reg $Imm”: This instruction copies the Imm(8bit) value in the register’s lower 8 bits. The upper 8 bits are zeroed out.

Example:

Suppose R0 has 1110_1010_1000_1110 stored, and mov R0 $13 is executed.

The final value of R0 will be 0000_0000_0000_1101.

FLAGS semantics

The semantics of the flags register are:

    • Overflow (V): This flag is set by add, sub and mul, when the result of the operation overflows. This shows the overflow status for the last executed instruction.

    • Less than (L): This flag is set by the “cmp reg1 reg2” instruction if reg1 < reg2

    • Greater than (G): This flag is set by the “cmp reg1 reg2” instruction if the value of reg1 > reg2
    • Equal (E): This flag is set by the “cmp reg1 reg2” instruction if reg1 = reg2

The default state of the FLAGS register is all zeros. If an instruction does not affect the FLAGS register, then the state of the FLAGS register is reset to 0 upon the execution.

The structure of the FLAGS register is as follows:

Unused 12 bits
V
L
G
E





15
3
2
1
0
The only operation allowed in the FLAGS register is “mov reg1 FLAGS”, where reg1 can be any of the registers from R0 to R6. This instruction reads FLAGS register and writes the data into reg1. All other operations on the FLAGS register are prohibited.

The cmp instruction can implicitly write to the FLAGS register. Similarly, conditional jump instructions can implicitly read the FLAGS register.

Example:

R0 has 5, R1 has 10

Implicit write: cmp R0 R1 will set the L (less than) flag in the FLAGS register.

Implicit read: jlt 10001001 will read the FLAGS register and figure out that the L flag was set, and then jump to address 10001001.

Binary Encoding

The ISA has 6 types of instructions with distinct encoding styles. However, each instruction is of 16 bits, regardless of the type.

    • Type A: 3 register type


opcode

unused

reg1

reg2

reg3

(5 bits)

(2 bits)

(3 bits)

(3 bits)

(3 bits)










15

10

8

5

2
0

    • Type B: register and immediate type


opcode

reg1

Immediate Value

(5 bits)

(3 bits)

(8 bits)






15

10

7
0

    • Type C: 2 registers type

opcode
unused
reg1

reg2
(5 bits)
(5 bits)
(3 bits)

(3 bits)





15
10
5
2
0

    • Type D: register and memory address type

opcode

reg1

Memory Address
(5 bits)

(3 bits)

(8 bits)





15
10

7
0

    • Type E: memory address type


opcode

unused

Memory Address

(5 bits)

(3 bits)

(8 bits)






15

10

7
0

    • Type F: halt


opcode

unused


(5 bits)

(11 bits)






15

10
0

Binary representation for the register are given as follows:-

Register
Address


R0
000


R1
001


R2
010


R3
011


R4
100


R5
101


R6
110


FLAGS
111



Executable binary syntax

The machine exposed by the ISA starts executing the code provided to it in the following format, until it reaches hlt instruction. There can only be one hlt instruction in the whole program, and it must be the last instruction. The execution starts from the 0th address. The ISA follows von-neumann architecture with a unified code and data memory. The variables must be allocated in the binary in the program order.


code




(last instruction) halt

variables
Questions:

Q1: Assembler:

Program an assembler for the aforementioned ISA and assembly. The input to the assembler is a text file containing the assembly instructions. Each line of the text file may be of one of 3 types:

    • Empty line: Ignore these lines

    • A label followed by an instruction

    • An instruction

    • A variable definition

Each of these entities have the following grammar:

    • The syntax of all the supported instructions is given above. The fields of an instruction are whitespace separated. The instruction itself might also have whitespace before it. An instruction can be one of the following:
        ◦ The opcode must be one of the supported mnemonic.

        ◦ A register can be one of R0, R1, … R6, and FLAGS.

        ◦ A mem_addr in jump instructions must be a label.

        ◦ A Imm must be a whole number <= 255 and >= 0.

        ◦ A mem_addr in load and store must be a variable.

    • A label marks a location in the code and must be followed by a colon (:). No spaces are allowed between label name and colon(:). A label name consists of alphanumeric characters and underscores.
A label followed by the instruction may looks like:

mylabel: add R1 R2 R3

    • A variable definition is of the following format: var xyz
which declares a 16 bit variable called xyz. This variable name can be used in place of mem_addr fields in load and store instructions. All variables must be defined at the very beginning of the assembly program. A variable name consists of alphanumeric characters and underscores.
    • Each line may be preceded by whitespace.



The assembler should be capable of:

    1. Handling all supported instructions

    2. Handling labels

    3. Handling variables

    4. Making sure that any illegal instruction (any instruction (or instruction usage) which is not supported) results in a syntax error. In particular you must handle:
        a. Typos in instruction name or register name

        b. Use of undefined variables

        c. Use of undefined labels

        d. Illegal use of FLAGS register

        e. Illegal Immediate values (less than 0 or more than 255)

        f. Misuse of labels as variables or vice-versa

        g. Variables not declared at the beginning
    h. Missing hlt instruction

    i. hlt not being used as the last instruction

    j. Wrong syntax used for instructions (For example, add instruction being used as a

type B instruction )

You need to generate distinct readable errors for all these conditions. If you find any other illegal usage, you are required to generate a “General Syntax Error”. The assembler must print out all these errors.

If the code is error free, then the corresponding binary is generated. The binary file is a text file in which each line is a 16bit binary number written using 0s and 1s in ASCII. The assembler can write less than or equal to 256 lines.

Input/Output format:

    • The assembler must read the assembly program as an input text file (stdin).

    • The assembler must generate the binary (if there are no errors) as an output text file (stdout).
    • The assembler must generate the error notifications along with line number on which the error was encountered (if there are errors) as the output text file (stdout). In case of multiple errors, the assembler may print any one of the errors.

Example of an assembly program

var X

mov R1 $10

mov R2 $100

mul R3 R1 R2

st R3 X

hlt

The above program will be converted into the following machine code

0001000100001010

0001001001100100

0011000011001010

0010101100000101

1001100000000000
Q2: Simulator:

You need to write a simulator for the given ISA. The input to the simulator is a binary file (the format is the same as the format of the binary file generated by the assembler in Q1). The simulator should load the binary in the system memory at the beginning, and then start executing the code at address 0. The code is executed until hlt is reached. After execution of each instruction, the simulator should output one line containing an 8 bit number denoting the program counter. This should be followed by 8 space separated 16 bit binary numbers denoting the values of the registers (R0, R1, … R6 and FLAGS).

<PC (8 bits)><space><R0 (16 bits)><space>...<R6 (16 bits)><space><FLAGS (16 bits)>. The output must be written to stdout. Similarly, the input must be read from stdin. After the program is halted, print the memory dump of the whole memory. This should be 256 lines, each having a 16 bit value

<16 bit data>

<16 bit data>

…..

<16 bit data>

Your simulator must have the following distinct components:

    1. Memory (MEM): MEM takes in an 8 bit address and returns a 16 bit value as the data. The MEM stores 512bytes, initialized to 0s.
    2. Program Counter (PC): The PC is an 8 bit register which points to the current instruction.

    3. Register File (RF): The RF takes in the register name (R0, R1, … R6 or FLAGS) and returns the value stored at that register.
    4. Execution Engine (EE): The EE takes the address of the instruction from the PC, uses it to get the stored instruction from MEM, and executes the instruction by updating the RF and PC.

The simulator should follow roughly the following pseudocode:

initialize(MEM);    // Load memory from stdin

PC = 0;    // Start from the first instruction

halted = false;

white(not halted)

{

Instruction = MEM.getData(PC);    // Get current instruction

halted, new_PC = EE.execute(Instruction);    // Update RF compute new_PC

PC.dump();    // Print PC

RF.dump();    // Print RF state

PC.update(new_PC);    // Update PC

}

MEM.dump()    // Print memory state

Q3: (Bonus) Memory Access Trace:

In Q2, generate a scatter plot with the cycle number on the x-axis and the memory address on the y-axis. You need to plot which memory address is accessed at what time.