eecs370.github.io

Project 1 EECS 370 (Fall 2021)

Worth: 100 points
Assigned: Tuesday, August 31, 2021
Part 1a Due: 11:55 PM EDT, Thursday, September 16, 2021
Part 1s & 1m Due: 11:55 PM EDT, Thursday, September 23, 2021

0. Starter Code

We provide starter code for the project.

1. Purpose

This project is intended to help you understand the instructions of a very simple assembly language and how to assemble programs into machine language.

2. Problem

This project has three parts. In the first part, you will write a program to take an assembly-language program and produce the corresponding machine language. In the second part, you will write a behavioral simulator for the resulting machine code. In the third part, you will write a short assembly-language program to multiply two numbers.

3. LC-2K Instruction-Set Architecture

In this and several of the following projects, you will be gradually “building” the LC-2K (Little Computer 2000). The LC-2K is very simple, but it is general enough to solve complex problems. For this project, you will only need to know the instruction set and instruction format of the LC-2K.

The LC-2K is an 8-register, 32-bit computer. All addresses are word-addresses. The LC-2K has 65536 words of memory. By assembly-language convention, register 0 will always contain 0 (i.e. the machine will not enforce this, but no assembly-language program should ever change register 0 from its initial value of 0).

3.1. Instruction Formats

There are 4 instruction formats (bit 0 is the least-significant bit). Bits 31-25 are unused for all instructions, and should always be 0.

R-type instructions (add, nor) bits 24-22: opcode
bits 21-19: reg A
bits 18-16: reg B
bits 15-3: unused (should all be 0)
bits 2-0: destReg
I-type instructions (lw, sw, beq) bits 24-22: opcode
bits 21-19: reg A
bits 18-16: reg B
bits 15-0: offsetField (a 16-bit, 2’s complement number with a range of -32768 to 32767)
J-type instructions (jalr) bits 24-22: opcode
bits 21-19: reg A
bits 18-16: reg B
bits 15-0: unused (should all be 0)
O-type instructions (halt, noop) bits 24-22: opcode
bits 21-0: unused (should all be 0)

3.2. Description of Machine Instructions

Assembly language name for instruction Opcode in binary (bits 24, 23, 22) Action
add
(R-type format)
000 Add contents of regA with contents of regB, store results in destReg.
nor
(R-type format)
001 Nor contents of regA with contents of regB, store results in destReg. This is a bitwise nor; each bit is treated independently.
lw
(I-type format)
010 Load regB from memory. Memory address is formed by adding offsetField with the contents of regA. Behavior is defined only for memory addresses in the range [0, 65535].
sw
(I-type format)
011 Store regB into memory. Memory address is formed by adding offsetField with the contents of regA. Behavior is defined only for memory addresses in the range [0, 65535].
beq
(I-type format)
100 If the contents of regA and regB are the same, then branch to the address PC+1+offsetField, where PC is the address of this beq instruction.
jalr
(J-type format)
101 First store PC+1 into regB, where PC is the address of this jalr instruction. Then branch to the address contained in regA. Note that this implies if regA and regB refer to the same register, the net effect will be jumping to PC+1.
halt
(O-type format)
110 Increment the PC (as with all instructions), then halt the machine (let the simulator notice that the machine halted).
noop
(O-type format)
111 Do nothing.

4. LC-2K Assembly Language and Assembler (40%)

The first part of this project is to write a program to take an assembly-language program and translate it into machine language. You will translate assembly-language names for instructions, such as beq, into their numeric equivalent (e.g. 100), and you will translate symbolic names for addresses into numeric values. The final output will be a series of 32-bit instructions (instruction bits 31-25 are always 0).

The format for a line of assembly code is (<white> means a series of tabs and/or spaces):

label<white>instruction<white>field0<white>field1<white>field2<white>comments

The leftmost field on a line is the label field. Valid labels contain a maximum of 6 characters and can consist of letters and numbers (but must start with a letter). The label is optional (the white space following the label field is required). Labels make it much easier to write assembly-language programs, since otherwise you would need to modify all address fields each time you added a line to your assembly-language program!

After the optional label is white space. Then follows the instruction field, where the instruction can be any of the assembly-language instruction names listed in the above table. After more white space comes a series of fields. All fields are given as decimal numbers or labels. The number of fields depends on the instruction, and unused fields should be ignored (treat them like comments).

R-type instructions (add, nor) instructions require 3 fields: 
    - field0 is regA
    - field1 is regB
    - field2 is destReg

I-type instructions (lw, sw, beq) require 3 fields: 
    - field0 is regA
    - field1 is regB
    - field2 is either a numeric value for offsetField or a
        symbolic address. Numeric offsetFields can be positive
        or negative; symbolic addresses are discussed below.

J-type instructions (jalr) require 2 fields: 
    - field0 is regA
    - field1 is regB.

O-type instructions (noop and halt) require no fields.

Symbolic addresses refer to labels. For lw or sw instructions, the assembler should compute offsetField to be equal to the address of the label. This could be used with a zero base register to refer to the label, or could be used with a non-zero base register to index into an array starting at the label. For beq instructions, the assembler should translate the label into the numeric offsetField needed to branch to that label.

After the last used field comes more white space, then any comments. The comment field ends at the end of a line. Comments are vital to creating understandable assembly-language programs, because the instructions themselves are rather cryptic.

All valid LC-2K programs must have a newline character at the end of each line. Some text editors enforce this for the last line of the file, and some do not. Failing to have a newline on the last line of assembly code can lead to strange bugs that can be quite difficult to find. Thus, be sure to end your assembly language files with newlines. In many editors, you can guarantee this by putting a blank line at the end of the file, and then deleting the blank line. Leaving a blank line in place is not recommended, as this will typically cause assembly to fail with an unrecognized opcode error.

In addition to LC-2K instructions, an assembly-language program may contain directions for the assembler. The only assembler directive we will use is .fill (note the leading period). .fill tells the assembler to put a number into the place where the instruction would normally be stored. .fill instructions use one field, which can be either a numeric value or a symbolic address. For example, .fill 32 puts the value 32 where the instruction would normally be stored. .fill with a symbolic address will store the address of the label. In the example below, .fill start will store the value 2, because the label start is at address 2. The bounds of the numeric value for .fill instructions are \(-2^{31}\) to \((+2^{31}-1)\) (-2147483648 to 2147483647).

The assembler should make two passes over the assembly-language program. In the first pass, it will calculate the address for every symbolic label. Assume that the first instruction is at address 0. In the second pass, it will generate a machine-language instruction (in decimal) for each line of assembly language. For example, here is an assembly-language program (that counts down from 5, stopping when it hits 0).

        lw      0       1       five    load reg1 with 5 (symbolic address)
        lw      1       2       3       load reg2 with -1 (numeric address)
start   add     1       2       1       decrement reg1
        beq     0       1       2       goto end of program when reg1==0
        beq     0       0       start   go back to the beginning of the loop
        noop
done    halt                            end of program
five    .fill   5
neg1    .fill   -1
stAddr  .fill   start                   will contain the address of start (2)

And here is the corresponding machine language:

(address 0): 8454151 (hex 0x810007)
(address 1): 9043971 (hex 0x8a0003)
(address 2): 655361 (hex 0xa0001)
(address 3): 16842754 (hex 0x1010002)
(address 4): 16842749 (hex 0x100fffd)
(address 5): 29360128 (hex 0x1c00000)
(address 6): 25165824 (hex 0x1800000)
(address 7): 5 (hex 0x5)
(address 8): -1 (hex 0xffffffff)
(address 9): 2 (hex 0x2)

Be sure you understand how the above assembly-language program got translated to machine language.

Since your programs will always start at address 0, your program should only output the contents, not the addresses.

8454151
9043971
655361
16842754
16842749
29360128
25165824
5
-1
2

4.1. Running Your Assembler

Write your program to take two command-line arguments. The first argument is the file name where the assembly-language program is stored, and the second argument is the file name where the output (the machine-code) is written. For example, with a program name of assemble, an assembly-language program in program.as, the following would generate a machine-code file program.mc:

./assemble program.as program.mc

Note that the format for running the command must use command-line arguments for the file names (rather than standard input and standard output). Your program should store only the list of decimal numbers in the machine-code file, one instruction per line. The decimal numbers will range from \(-2^{31}\) to \((+2^{31}-1)\) (-2147483648 to 2147483647). Any deviation from this format (e.g. extra spaces or empty lines) will render your machine-code file ungradeable. Any other output that you want the program to generate (e.g. debugging output) can be printed to standard output.

4.2. Error Checking

Your assembler should catch the following errors in the assembly-language program:

Your assembler should exit(1) if it detects an error and exit(0) if it finishes without detecting any errors. Your assembler should NOT catch simulation-time errors, i.e. errors that would occur at the time the assembly-language program executes (e.g. branching to address -1, infinite loops, etc.).

4.3. Test Cases

An integral (and graded) part of writing your assembler will be to write a suite of test cases to validate any LC-2K assembler. This is common practice in the real–world software companies maintain a suite of test cases for their programs and use this suite to check the program’s correctness after a change. Writing a comprehensive suite of test cases will deepen your understanding of the project specification and your program, and it will help you a lot as you debug your program.

The test cases for the assembler part of this project will be short assembly-language programs that serve as input to an assembler. You will submit your suite of test cases together with your assembler, and we will grade your test suite according to how thoroughly it exercises an assembler. Each test case may be at most 50 lines long, and your test suite may contain up to 20 test cases. These limits are much larger than needed for full credit (the solution test suite is composed of 5 test cases, each < 10 lines long). See Section 7 for how your test suite will be graded.

Hints: The example assembly-language program above is a good case to include in your test suite, though you’ll need to write more test cases to get full credit. Remember to create some test cases that test the ability of an assembler to check for the errors in Section 4.2.

Note: All instructions should appear before any .fill’s. Instructions and .fill’s should not be interleaved. e.g. Your assembly programs should look like this:

  noop
  noop
  .fill 0
  .fill 1

They should NOT look like this:

  noop
  .fill 0
  noop
  .fill 1

This won’t be enforced for project 1, but assembly programs with interleaved instructions and .fill’s will not work properly in project 2. Note that many students like to reuse their project 1 assembly tests for project 2.

4.4. Assembler Hints

Since offsetField is a 2’s complement number, it can only store numbers ranging from -32768 to 32767. For symbolic addresses, your assembler will compute offsetField so that the instruction refers to the correct label.

Remember that offsetField is only a 16-bit 2’s complement number. Since Linux integers are 32 bits, you’ll have to chop off all but the lowest 16 bits for negative values of offsetField.

5. Behavioral Simulator (40%)

The second part of this assignment is to write a program that can simulate any legal LC-2K machine-code program. The input for this part will be the machine-code file that you created with your assembler. With a program name of simulate and a machine-code file of program.mc, your program should be run as follows:

./simulate program.mc > output

This directs all printfs to the file output.

The simulator should begin by initializing all registers and the program counter to 0. The simulator will then simulate the program until the program executes a halt.

The simulator should call printState (included below) before executing each instruction and once just before exiting the program. This function prints the current state of the machine (program counter, registers, memory). printState will print the memory contents for memory locations defined in the machine-code file (addresses 0-9 in the Section 4 example).

5.1 Test Cases

As with the assembler, you will write a suite of test cases to validate any LC-2K simulator.

The test cases for the simulator part of this project will be short, valid assembly-language programs that, after being assembled into machine code, serve as input to a simulator. You will submit your suite of test cases together with your simulator, and we will grade your test suite according to how thoroughly it exercises an LC-2K simulator. Each test case may be at most 50 lines and may execute at most 200 instructions on a correct simulator, and your test suite may contain up to 20 test cases. These limits are much larger than needed for full credit (the solution test suite is composed of a couple test cases, each executing less than 40 instructions). See Section 7 for how your test suite will be graded.

Warning: Behavior is defined only for accesses to memory addresses in the range [0, 65535]. In your test cases, do not access memory addresses outside of this range with LW or SW instructions. This is NOT one of the errors you are required to check for, but assembly programs with undefined behavior may execute differently on your simulator than on our reference simulator.

5.2 Simulator Hints

Be careful how you handle offsetField for lw, sw, and beq. Remember that it’s a 2’s complement 16-bit number, so you need to convert a negative offsetField to a negative 32-bit integer on the Linux workstations (by sign extending it). One way to do this is to use the following function, also given in the starter code:

int convertNum(int num)
{
    /* convert a 16-bit number into a 32-bit Linux integer */
    if (num & (1<<15) ) {
        num -= (1<<16);
    }
    return(num);
}

An example run of the simulator (not for the specified task of multiplication) is included in Section 12.

6. Assembly-Language Multiplication (20%)

The third part of this assignment is to write an assembly-language program to multiply two numbers. Input the numbers by reading memory locations called mcand and mplier. The result should be stored in register 1 when the program halts. You may assume that the two input numbers are at most 15 bits and are positive; this ensures that the (positive) result fits in an LC-2K word. Remember that shifting left by one bit is the same as adding the number to itself. Given the LC-2K instruction set, it’s easiest to modify the algorithm so that you avoid the right shift. Submit a version of the program that computes (\(1103 \times 7043\)).

Your multiplication program must be reasonably efficient — it must be at most 50 lines long and execute at most 1000 instructions for any valid input (this is several times longer and slower than the solution). To achieve this, you are strongly encouraged to consider using a loop and shift algorithm to perform the multiplication; algorithms such as successive addition (e.g. multiplying \(5 \times 6\) by adding 5 six times) will take too long.

7. Grading, Auto-Grading, and Formatting

We will grade primarily on functionality, including error handling, correctly assembling and simulating all instructions, input and output format, method of executing your program, correctly multiplying, and comprehensiveness of the test suites.

To help you validate your project, your submission will be graded automatically after submission. You may then continue to work on the project and re-submit. To deter you from using the autograder as a debugger, you will receive feedback from the autograder only for the first THREE SUBMISSIONS on any given day. That is, you will receive feedback with your score only three times on any given day. All subsequent submissions will be silently graded. Your final score will be derived from your overall best submission to the autograder.

The feedback from the autograder will not be very illuminating; it won’t tell you where your problem is or give you the test programs. The purpose of the autograder is to let you know that you should keep working on your project (rather than thinking it’s perfect and ending up with a 0). The best way to debug your program is to generate your own test cases, figure out the correct answers, and compare your program’s output to the correct answer. This is also one of the best ways to learn the concepts in the project.

The student suite of test cases for the assembler and simulator parts of this project will be graded according to how thoroughly they test an LC-2K assembler or simulator. We will judge thoroughness of the test suite by how well it exposes potential bugs in an assembler or simulator.

For the assembler test suite, the auto-grader will use each test case as input to a set of buggy assemblers. A test case exposes a buggy assembler by causing it to generate a different answer from a correct assembler. The test suite is graded based on how many of the buggy assemblers were exposed by at least one test case. This is known as mutation testing in the research literature on automated testing. Your test suite is run on 13 buggy assemblers. To receive all Mutation Testing points, your test suite must expose at least 11/13 of the buggy assemblers.

For the simulator test suite, the auto-grader will correctly assemble each test case, then use it as input to a set of buggy simulators. A test case exposes a buggy simulator by causing it to generate a different answer from a correct simulator. The test suite is graded based on how many of the buggy simulators were exposed by at least one test case. Your test suite is run on 7 buggy assemblers. To receive all Mutation Testing points, your test suite must expose at least 6/7 of the buggy assemblers. Note that the test cases for the simulator should all be valid, correct assembly language programs.

Because all programs will be auto-graded, you must be careful to follow the exact formatting rules in the project description:

  1. (assembler) Follow exactly the format for inputting the assembly-language program and outputting the machine-code file.

  2. (assembler) Call exit(1) if you detect errors in the assembly-language program. Call exit(0) if you finish without detecting errors.

  3. (simulator) Don’t modify printState or stateStruct at all. Download this code into your program electronically (don’t re-type it) to avoid typos.

  4. (simulator) Call printState exactly once before each instruction executes and once just before the simulator exits. Do not call printState at any other time.

  5. (simulator) Don’t print the sequence “@@@” anywhere (except where the provided printState function prints it).

  6. (simulator) state.numMemory must be equal to the number of lines in the machine-code file.

  7. (simulator) Initialize all registers to 0.

  8. (multiplication) Store the result in register 1.

  9. (multiplication) The two input numbers must be in locations labeled mcand and mplier (lower-case).

8. Turning in the Project

Use autograder.io to submit your files. You have been added as a student to the class, so you should see EECS 370 listed as a class.

Here are the files you should submit for each project part:

1) assembler (part 1a)
    a. Your assembler, a C program named "assembler.c"
    b. Suite of test cases (each test case is an assembly-language program
        in a separate file, ending in ".as")

2) simulator (part 1s)
    a. Your simulator, a C program named "simulator.c"
    b. suite of test cases (each test case is an assembly-language program
        in a separate file, ending in ".as")

3) multiplication (part 1m)
    a. Your assembly program for multiplication, named "mult.as"

Your code will be compiled with the GCC compiler using the C99 standard. The following bash command compiles program.c and writes the executable into program. We will use this command (with the appropriate file name subsituted for “program”) to compile your submission. You are allowed to use any standard C libraries which compile with the specified flags below.

gcc program.c -lm -std=c99 -o program

The official time of submission for your project will be the time the last file is sent. If you send in anything after the due date, your project will be considered late (and will use up your late days). If you have already used up all of your late days, additional late submissions will not be scored for your project grade.

9. Code Fragment for Assembler

The focus of this class is machine organization, not C programming skills. To “build” your computer, however, you will be doing a lot of C programming. To help you, use the given fragment of the C program for the assembler. This shows how to specify command-line arguments to the program (via argc and argv), how to parse the assembly-language file, etc. This fragment is provided strictly to help you, though it may take a bit for you to understand and use the file. You may also choose to not use this fragment.

10. Code Fragment for Simulator

We have also given you some C code that may help you write the simulator. Again, you should take this merely as a hint. You may have to re-code this to make it do exactly what you want, but this should help you get started. Remember not to change stateStruct or printState.

11. C Programming Tips

Here are a few programming tips for writing C programs to manipulate bits:

1) To indicate a hexadecimal constant in C, precede the number by 0x. For example, 27 decimal is 0x1b in hexadecimal.

2) The value of the expression (a >> b) is the number “a” shifted right by “b” bits. Neither a nor b are changed. E.g. (25 >> 2) is 6. Note that 25 is 11001 in binary, and 6 is 110 in binary.

3) The value of the expression (a << b) is the number “a” shifted left by “b” bits. Neither a nor b are changed. E.g. (25 << 2) is 100. Note that 25 is 11001 in binary, and 100 is 1100100 in binary.

4) To find the value of the expression (a & b), perform a logical AND on each bit of a and b (i.e. bit 31 of a ANDed with bit 31 of b, bit 30 of a ANDed with bit 30 of b, etc.). E.g. (25 & 11) is 9, since:

    11001 (binary)
  & 01011 (binary)
---------------------
 =  01001 (binary), which is 9 decimal.

5) To find the value of the expression (a | b), perform a logical OR on each bit of a and b (i.e. bit 31 of a ORed with bit 31 of b, bit 30 of a ORed with bit 30 of b, etc.). E.g. (25 | 11) is 27, since:

    11001 (binary)
  | 01011 (binary)
---------------------
 =  11011 (binary), which is 27 decimal.

6) ~a is the bit-wise complement of a (a is not changed).

Use these operations to create and manipulate machine-code. For example:

If you’re not sure what an operation is doing, print some intermediate results to help you debug.


12. Example Run of Simulator

memory[0]=8454151
memory[1]=9043971
memory[2]=655361
memory[3]=16842754
memory[4]=16842749
memory[5]=29360128
memory[6]=25165824
memory[7]=5
memory[8]=-1
memory[9]=2


@@@
state:
	pc 0
	memory:
		mem[ 0 ] 8454151
		mem[ 1 ] 9043971
		mem[ 2 ] 655361
		mem[ 3 ] 16842754
		mem[ 4 ] 16842749
		mem[ 5 ] 29360128
		mem[ 6 ] 25165824
		mem[ 7 ] 5
		mem[ 8 ] -1
		mem[ 9 ] 2
	registers:
		reg[ 0 ] 0
		reg[ 1 ] 0
		reg[ 2 ] 0
		reg[ 3 ] 0
		reg[ 4 ] 0
		reg[ 5 ] 0
		reg[ 6 ] 0
		reg[ 7 ] 0
end state

@@@
state:
	pc 1
	memory:
		mem[ 0 ] 8454151
		mem[ 1 ] 9043971
		mem[ 2 ] 655361
		mem[ 3 ] 16842754
		mem[ 4 ] 16842749
		mem[ 5 ] 29360128
		mem[ 6 ] 25165824
		mem[ 7 ] 5
		mem[ 8 ] -1
		mem[ 9 ] 2
	registers:
		reg[ 0 ] 0
		reg[ 1 ] 5
		reg[ 2 ] 0
		reg[ 3 ] 0
		reg[ 4 ] 0
		reg[ 5 ] 0
		reg[ 6 ] 0
		reg[ 7 ] 0
end state

@@@
state:
	pc 2
	memory:
		mem[ 0 ] 8454151
		mem[ 1 ] 9043971
		mem[ 2 ] 655361
		mem[ 3 ] 16842754
		mem[ 4 ] 16842749
		mem[ 5 ] 29360128
		mem[ 6 ] 25165824
		mem[ 7 ] 5
		mem[ 8 ] -1
		mem[ 9 ] 2
	registers:
		reg[ 0 ] 0
		reg[ 1 ] 5
		reg[ 2 ] -1
		reg[ 3 ] 0
		reg[ 4 ] 0
		reg[ 5 ] 0
		reg[ 6 ] 0
		reg[ 7 ] 0
end state

@@@
state:
	pc 3
	memory:
		mem[ 0 ] 8454151
		mem[ 1 ] 9043971
		mem[ 2 ] 655361
		mem[ 3 ] 16842754
		mem[ 4 ] 16842749
		mem[ 5 ] 29360128
		mem[ 6 ] 25165824
		mem[ 7 ] 5
		mem[ 8 ] -1
		mem[ 9 ] 2
	registers:
		reg[ 0 ] 0
		reg[ 1 ] 4
		reg[ 2 ] -1
		reg[ 3 ] 0
		reg[ 4 ] 0
		reg[ 5 ] 0
		reg[ 6 ] 0
		reg[ 7 ] 0
end state

@@@
state:
	pc 4
	memory:
		mem[ 0 ] 8454151
		mem[ 1 ] 9043971
		mem[ 2 ] 655361
		mem[ 3 ] 16842754
		mem[ 4 ] 16842749
		mem[ 5 ] 29360128
		mem[ 6 ] 25165824
		mem[ 7 ] 5
		mem[ 8 ] -1
		mem[ 9 ] 2
	registers:
		reg[ 0 ] 0
		reg[ 1 ] 4
		reg[ 2 ] -1
		reg[ 3 ] 0
		reg[ 4 ] 0
		reg[ 5 ] 0
		reg[ 6 ] 0
		reg[ 7 ] 0
end state

@@@
state:
	pc 2
	memory:
		mem[ 0 ] 8454151
		mem[ 1 ] 9043971
		mem[ 2 ] 655361
		mem[ 3 ] 16842754
		mem[ 4 ] 16842749
		mem[ 5 ] 29360128
		mem[ 6 ] 25165824
		mem[ 7 ] 5
		mem[ 8 ] -1
		mem[ 9 ] 2
	registers:
		reg[ 0 ] 0
		reg[ 1 ] 4
		reg[ 2 ] -1
		reg[ 3 ] 0
		reg[ 4 ] 0
		reg[ 5 ] 0
		reg[ 6 ] 0
		reg[ 7 ] 0
end state

@@@
state:
	pc 3
	memory:
		mem[ 0 ] 8454151
		mem[ 1 ] 9043971
		mem[ 2 ] 655361
		mem[ 3 ] 16842754
		mem[ 4 ] 16842749
		mem[ 5 ] 29360128
		mem[ 6 ] 25165824
		mem[ 7 ] 5
		mem[ 8 ] -1
		mem[ 9 ] 2
	registers:
		reg[ 0 ] 0
		reg[ 1 ] 3
		reg[ 2 ] -1
		reg[ 3 ] 0
		reg[ 4 ] 0
		reg[ 5 ] 0
		reg[ 6 ] 0
		reg[ 7 ] 0
end state

@@@
state:
	pc 4
	memory:
		mem[ 0 ] 8454151
		mem[ 1 ] 9043971
		mem[ 2 ] 655361
		mem[ 3 ] 16842754
		mem[ 4 ] 16842749
		mem[ 5 ] 29360128
		mem[ 6 ] 25165824
		mem[ 7 ] 5
		mem[ 8 ] -1
		mem[ 9 ] 2
	registers:
		reg[ 0 ] 0
		reg[ 1 ] 3
		reg[ 2 ] -1
		reg[ 3 ] 0
		reg[ 4 ] 0
		reg[ 5 ] 0
		reg[ 6 ] 0
		reg[ 7 ] 0
end state

@@@
state:
	pc 2
	memory:
		mem[ 0 ] 8454151
		mem[ 1 ] 9043971
		mem[ 2 ] 655361
		mem[ 3 ] 16842754
		mem[ 4 ] 16842749
		mem[ 5 ] 29360128
		mem[ 6 ] 25165824
		mem[ 7 ] 5
		mem[ 8 ] -1
		mem[ 9 ] 2
	registers:
		reg[ 0 ] 0
		reg[ 1 ] 3
		reg[ 2 ] -1
		reg[ 3 ] 0
		reg[ 4 ] 0
		reg[ 5 ] 0
		reg[ 6 ] 0
		reg[ 7 ] 0
end state

@@@
state:
	pc 3
	memory:
		mem[ 0 ] 8454151
		mem[ 1 ] 9043971
		mem[ 2 ] 655361
		mem[ 3 ] 16842754
		mem[ 4 ] 16842749
		mem[ 5 ] 29360128
		mem[ 6 ] 25165824
		mem[ 7 ] 5
		mem[ 8 ] -1
		mem[ 9 ] 2
	registers:
		reg[ 0 ] 0
		reg[ 1 ] 2
		reg[ 2 ] -1
		reg[ 3 ] 0
		reg[ 4 ] 0
		reg[ 5 ] 0
		reg[ 6 ] 0
		reg[ 7 ] 0
end state

@@@
state:
	pc 4
	memory:
		mem[ 0 ] 8454151
		mem[ 1 ] 9043971
		mem[ 2 ] 655361
		mem[ 3 ] 16842754
		mem[ 4 ] 16842749
		mem[ 5 ] 29360128
		mem[ 6 ] 25165824
		mem[ 7 ] 5
		mem[ 8 ] -1
		mem[ 9 ] 2
	registers:
		reg[ 0 ] 0
		reg[ 1 ] 2
		reg[ 2 ] -1
		reg[ 3 ] 0
		reg[ 4 ] 0
		reg[ 5 ] 0
		reg[ 6 ] 0
		reg[ 7 ] 0
end state

@@@
state:
	pc 2
	memory:
		mem[ 0 ] 8454151
		mem[ 1 ] 9043971
		mem[ 2 ] 655361
		mem[ 3 ] 16842754
		mem[ 4 ] 16842749
		mem[ 5 ] 29360128
		mem[ 6 ] 25165824
		mem[ 7 ] 5
		mem[ 8 ] -1
		mem[ 9 ] 2
	registers:
		reg[ 0 ] 0
		reg[ 1 ] 2
		reg[ 2 ] -1
		reg[ 3 ] 0
		reg[ 4 ] 0
		reg[ 5 ] 0
		reg[ 6 ] 0
		reg[ 7 ] 0
end state

@@@
state:
	pc 3
	memory:
		mem[ 0 ] 8454151
		mem[ 1 ] 9043971
		mem[ 2 ] 655361
		mem[ 3 ] 16842754
		mem[ 4 ] 16842749
		mem[ 5 ] 29360128
		mem[ 6 ] 25165824
		mem[ 7 ] 5
		mem[ 8 ] -1
		mem[ 9 ] 2
	registers:
		reg[ 0 ] 0
		reg[ 1 ] 1
		reg[ 2 ] -1
		reg[ 3 ] 0
		reg[ 4 ] 0
		reg[ 5 ] 0
		reg[ 6 ] 0
		reg[ 7 ] 0
end state

@@@
state:
	pc 4
	memory:
		mem[ 0 ] 8454151
		mem[ 1 ] 9043971
		mem[ 2 ] 655361
		mem[ 3 ] 16842754
		mem[ 4 ] 16842749
		mem[ 5 ] 29360128
		mem[ 6 ] 25165824
		mem[ 7 ] 5
		mem[ 8 ] -1
		mem[ 9 ] 2
	registers:
		reg[ 0 ] 0
		reg[ 1 ] 1
		reg[ 2 ] -1
		reg[ 3 ] 0
		reg[ 4 ] 0
		reg[ 5 ] 0
		reg[ 6 ] 0
		reg[ 7 ] 0
end state

@@@
state:
	pc 2
	memory:
		mem[ 0 ] 8454151
		mem[ 1 ] 9043971
		mem[ 2 ] 655361
		mem[ 3 ] 16842754
		mem[ 4 ] 16842749
		mem[ 5 ] 29360128
		mem[ 6 ] 25165824
		mem[ 7 ] 5
		mem[ 8 ] -1
		mem[ 9 ] 2
	registers:
		reg[ 0 ] 0
		reg[ 1 ] 1
		reg[ 2 ] -1
		reg[ 3 ] 0
		reg[ 4 ] 0
		reg[ 5 ] 0
		reg[ 6 ] 0
		reg[ 7 ] 0
end state

@@@
state:
	pc 3
	memory:
		mem[ 0 ] 8454151
		mem[ 1 ] 9043971
		mem[ 2 ] 655361
		mem[ 3 ] 16842754
		mem[ 4 ] 16842749
		mem[ 5 ] 29360128
		mem[ 6 ] 25165824
		mem[ 7 ] 5
		mem[ 8 ] -1
		mem[ 9 ] 2
	registers:
		reg[ 0 ] 0
		reg[ 1 ] 0
		reg[ 2 ] -1
		reg[ 3 ] 0
		reg[ 4 ] 0
		reg[ 5 ] 0
		reg[ 6 ] 0
		reg[ 7 ] 0
end state

@@@
state:
	pc 6
	memory:
		mem[ 0 ] 8454151
		mem[ 1 ] 9043971
		mem[ 2 ] 655361
		mem[ 3 ] 16842754
		mem[ 4 ] 16842749
		mem[ 5 ] 29360128
		mem[ 6 ] 25165824
		mem[ 7 ] 5
		mem[ 8 ] -1
		mem[ 9 ] 2
	registers:
		reg[ 0 ] 0
		reg[ 1 ] 0
		reg[ 2 ] -1
		reg[ 3 ] 0
		reg[ 4 ] 0
		reg[ 5 ] 0
		reg[ 6 ] 0
		reg[ 7 ] 0
end state
machine halted
total of 17 instructions executed
final state of machine:

@@@
state:
	pc 7
	memory:
		mem[ 0 ] 8454151
		mem[ 1 ] 9043971
		mem[ 2 ] 655361
		mem[ 3 ] 16842754
		mem[ 4 ] 16842749
		mem[ 5 ] 29360128
		mem[ 6 ] 25165824
		mem[ 7 ] 5
		mem[ 8 ] -1
		mem[ 9 ] 2
	registers:
		reg[ 0 ] 0
		reg[ 1 ] 0
		reg[ 2 ] -1
		reg[ 3 ] 0
		reg[ 4 ] 0
		reg[ 5 ] 0
		reg[ 6 ] 0
		reg[ 7 ] 0
end state