Department of Electrical and Computer Engineering

The University of Texas at Austin

EE 360N, Spring 2009
Lab Assignment 1
Due: Sunday, February 8, 11:59 pm
Yale N. Patt, Instructor
Ramapriyan Chakravarthy, Khubaib, Vivekanand Venugopal, TAs

Introduction

The purpose of this lab is to reinforce the concepts of assembly language and assemblers. In this lab assignment, you will write an LC-3b Assembler, whose job will be to translate assembly language source code into the machine language (ISA) of the LC-3b. You will also write a program to solve a problem in the LC-3b Assembly Language.

In Lab Assignments 2 and 3, you will close the loop by completing the design of two types of simulators for the LC-3b, and test your assembler by having the simulators execute the program you wrote and assembled in this lab.

Part I: Write an assembler for the LC-3b Assembly Language

The general format of a line of assembly code, which will be the input to your assembler, is as follows:

        label opcode operands ; comments

The leftmost field on a line will be the label field. Valid labels consist of from one to 20 alphanumeric characters (i.e., a capital or lowercase letter of the alphabet, or a decimal digit), starting with a letter of the alphabet. A valid label cannot be the same as an opcode or a pseudo-op. A valid label must start with a character other than ‘x’ and consist solely of alphanumeric characters – A to Z, a to z, 0 to 9. The label is optional, i.e., an assembly language instruction can leave out the label. A label is necessary if the program is to branch to that instruction or if the location contains data that is to be addressed explicitly.

The opcode field can be any one of the following instructions:

ADD, AND, BR, HALT, JMP, JSR, JSRR, LDB, LDW, LEA, NOP, 
NOT, RET, LSHF, RSHFL, RSHFA, RTI, STB, STW, TRAP, XOR

The number of operands depends on the operation being performed. It can consist of register names, labels, or constants (immediates). If a hexadecimal constant is used, it must be prefixed with the ‘x’ character. Similarly, decimal constants must be prefixed with a ‘#’ character.

Optionally, an instruction can be commented, which is good style if the comment contains meaningful information. Comments follow the semicolon and are not interpreted by the Assembler. Note that the semicolon prefaces the comment, and a newline ends the comment. Other delimiters are not allowed.

In this lab assignment, the NOP instruction translates into the machine language instruction 0x0000.

Note that you should also implement the HALT instruction as TRAP x25. Other TRAP commands (GETC, IN, OUT, PUTS) need not be recognized by your assembler for this assignment.

In addition to LC-3b instructions, an assembly language also contains pseudo-ops, sometimes called macro directives. These are messages from the programmer to the assembler that assist the assembler in performing the translation process. In the case of our LC-3b Assembly Language, we will only require three pseudo-ops to make our lives easier: .ORIG, .END, and .FILL.

An assembly language program will consist of some number of assembly language instructions, delimited by .ORIG and .END. The pseudo-op .END is a message to the assembler designating the end of the assembly language source program. The .ORIG pseudo-op provides two functions: it designates the start of the source program, and it specifies the location of the first instruction in the object module to be produced. For example, .ORIG N means “the next instruction will be assigned to location N.” The pseudo-op .FILL W assigns the value W to the corresponding location in the object module. W is regarded as a word (16-bit value) by the .FILL pseudo-op.

The task of the assembler is that of line-by-line translation. The input is an assembly language file, and the output is an object (ISA) file (consisting of hexadecimal digits). To make it a little more concrete, here is a sample assembly language program:

;This program counts from 10 to 0 
        .ORIG x3000       
        LEA R0, TEN       ;This instruction will be loaded into memory location x3000
        LDW R1, R0, #0 
START   ADD R1, R1, #-1 
        BRZ DONE 
        BR START 
                          ;blank line 
DONE    TRAP x25          ;The last executable instruction 
TEN     .FILL x000A       ;This is 10 in 2's comp, hexadecimal 
        .END              ;The pseudo-op, delimiting the source program

And its corresponding ISA program:

       0x3000
       0xE005
       0x6200 
       0x127F 
       0x0401 
       0x0FFD 
       0xF025 
       0x000A

Note that each line of the output is a four digit hex number, prefixed with “0x”, representing the 16-bit machine instruction. The reason that your output should be prefixed with “0x” is because the simulator for Lab Assignment 2 that you will write in C expects the input data to be expressed in hex, and C syntax requires hex data to start with "0x". Also note that BR instruction is assembled as the unconditional branch, BRnzp.

When this program is loaded into the simulator, the instruction 0xE005 will be loaded into the memory location specified by the first line of the program, which is x3000. As instructions consist of two bytes, the second instruction, 0x6200, will be loaded into memory location x3002. Thus, memory locations x3000 to x300C will contain the program.

We have included below another example of an assembly language program, and the result of the assembly process. In this case, the .ORIG pseudo-op tells the assembler to place the program at memory address #4096.

        .ORIG #4096 
A       LEA R1, Y
        LDW R1, R1, #0
        LDW R1, R1, #0 
        ADD R1, R1, R1 
        ADD R1, R1, x-10  ;x-10 is the negative of x10 
        BRN A 
        HALT 
Y       .FILL #263 
        .FILL #13 
        .FILL #6 
        .END

would be assembled into the following:

       0x1000 
       0xE206
       0x6240
       0x6240 
       0x1241 
       0x1270 
       0x09FA 
       0xF025 
       0x0107 
       0x000D 
       0x0006

Important note: even though this program will assemble correctly, it may not do anything useful.

The Assembly Process

Your assembler should make two passes of the input file. In the first pass, all the labels should be bound to specific memory addresses. You create a symbol table to contain those bindings. Whenever a new instruction label is encountered in the input file, it is assigned to the current memory address.

The second pass performs the translation from assembly language to machine language, one line at a time. It is during this pass that the output file should be generated.

You should write your program to take two command-line arguments. The first argument is the name of a file that contains a program written in LC-3b assembly language, which will be the input to your program. The second argument is the name of the file to which your program will write its output. In other words, this is the name of the file which will contain the LC-3b machine code corresponding to the input assembly language file. For example, we should be able to run your assembler with the following command-line input:

assemble <source.asm> <output.obj>

where assemble is the name of the executable file corresponding to your compiled and linked program; source.asm is the input assembly language file, and output.obj; is the output file that will contain the assembled code.

You will need to include some basic error checking within your assembler to handle improperly constructed assembly language programs. Your assembler must detect three types of errors and must return three different error codes. The errors to be detected are undefined labels (error code 1), invalid opcodes (error code 2), and invalid constants (error code 3). An invalid constant is a constant that is too large to be assembled into an LC-3b instruction. If the .ORIG pseudo-op contains an address that is greater than an address that can be represented in the 16-bit address space, your program should return error code 3. Also, if the .ORIG statement specifies an address that is not word-aligned, your program should return error code 3. Your program must return the error codes via the exit(n) function, where n denotes the error code number. If the assembly language program does not contain any errors, you must exit with error code 0. Exiting with the correct codes is very important since they will be used in the grading process. On Linux, you can determine the exit code of your assembler by executing echo $? right after running the assembler.

This error checking is the bare minimum that we expect. You can return error code 4 for any other errors you find. Just be sure that the errors don't fall within the first three categories specified above.

Examples of error codes

Error code 1: undefined label

A label is used by an instruction but the label is not in the symbol table, e.g.

Error code 2: invalid opcode

An invalid opcode is one that is not defined in the LC-3b ISA, e.g.

Error code 3: invalid constant

An invalid constant is a constant that is too large to be assembled into an LC-3b instruction. An odd constant that follows .ORIG is also an invalid constant.

Examples:

Error code 4: other error

These errors which do not belong to any of the above categories.

Examples:

If a label and an instruction that uses it are too far apart and the offset cannot be specified properly in the machine code, you should produce error code 4.

Your assembler should accept an “empty” program, i.e. one with just a valid .ORIG and a .END. E.g. the following assembly program would be assembled to only one line containing the starting address (0x3000).

	.ORIG x3000
	.END

Note: your assembler needs to recognize only labels as operands for LEA, BR, and JSR instructions. For example, if the following line is in an input assembly language program, your assembler can exit with error code 4:

	LEA R1, x100

Part II: Write a program to solve the following problem

Write an LC-3b assembly language program that multiplies two 8-bit unsigned numbers.

The following example illustrates the multiplication of two positive 4-bit numbers. The first number (multiplicand) is 0110 while the second number (multiplier) is 0101.

         0110
       x 0101
        ------
         0110
        0000	
       0110
    + 0000
    ----------
      0011110

You may notice that the result of the multiplication is obtained by adding four partial products which correspond to the four multiplier bits. Each partial product is either the multiplicand left shifted by an appropriate amount or zero depending upon whether the corresponding multiplier bit is a 1 or a 0.

Your assembly language program must begin at memory location x3000. You may assume that before your program is loaded into memory and run, addresses x3100 and x3101 contain the two 8-bit unsigned numbers that have to be multiplied. Your program should store the 16-bit product in memory location x3102.

You will have no way of determining if your assembly language code works (yet!), but you can use it to determine if your assembler works!

Hint: For shifting a number, use the appropriate SHF instruction encoding.

Requirements

Important note: because we will be evaluating your code in Unix, please be sure your code compiles using gcc with the -ansi flag. This means that you need to write your code in C such that it conforms to the ANSI C standard.

You can use the following command to compile your code:

gcc -ansi -o assemble assembler.c

To complete Lab Assignment 1, you will need to turn in the following:

  1. A C file called "assembler.c" containing an adequately documented listing of your LC-3b Assembler.
  2. A source listing (LC-3b Assembly Language) of the program described above called "mul.asm".

Submit your code electronically following the posted instructions.

Things to watch for:

Be sure that your assembler can handle comments on any line, including lines that contain pseudo-ops and lines that contain only comments. Be careful with comments that follow a HALT, NOP or RET instructions – these instructions take no operand.

Your assembler should allow hexadecimal and decimal constants after both ISA instructions, like ADD, and pseudo-ops, like .FILL.

The whole assembly process is case insensitive. That is, the labels, opcodes, operands, and pseudo-ops can be in upper case, lower case, or both, and are still interpreted the same. The parser function given in the useful code page converts every line into lower case before parsing it.

You can assume that there will be at most 255 labels in an assembly program. You can also assume that the number of characters on a line will not exceed 255.

Your assembler needs to support all 8 variations of BR:

        BRn LABEL                       BRz LABEL
        BRp LABEL                       BRnz LABEL
        BRnp LABEL                      BRzp LABEL
        BR LABEL                        BRnzp LABEL