Chapter 7 Introduction to LC-3 Assembly Language

Chapter 7Introduction to LC-3Assembly Language • Assembly Language • Assembly Process • Using the Editor & Simulator for Assembly Language Programming

LC-3 Assembly Language Syntax • Each line of a program is one of the following: • an instruction • an assember directive (or pseudo-op) • a comment • Whitespace (between symbols) and case are ignored. • Comments (beginning with “;”) are also ignored. • An instruction has the following format: LABEL OPCODE OPERANDS ;COMMENTS optional mandatory Example:Loop1 ADD R3, R3, #-1 ; Decrement R3

Assembler Directives • Pseudo-operations • do not refer to operations executed by program • used by assembler • look like instruction, but “opcode” starts with dot

Compute the Sum of 12 Integers • Program begins at location x3000. • Integers begin at location x3100. R1  x3100R3  0 (Sum)R2  12(count) R2=0? R4  M[R1] R3  R3+R4R1  R1+1 R2  R2-1 NO YES R1: “Array” index pointer (Begin with location 3100) R3: Accumulator for the sum of integers R2: Loop counter (Count down from 12) R4: Temporary register to store next integer

Compute the Sum of 12 Integers R1: “Array” index pointer (Begin with location 3100) R3: Accumulator for the sum of integers R2: Loop counter (Count down from 12) R4: Temporary register to store next integer

Compute the Sum of 12 Integers - Program R1  x3100R3  0 (Sum)R2  12(count) .ORIG x3000 ; Add 12 integers ; R1: Pointer to integer ; R2: Loop counter ; R3: Accumulator ; R4: Temporary register LD R1 DATAADDR ; Load pointer ; to integers AND R3, R3, #0 ; Accumulator = 0 AND R2, R2, #0 ; Counter = 12 ADD R2, R2, #12 ; Add integers LOOP BRZ STOP ; Stop when done LDR R4, R1, #0 ; Add next integer ADD R3, R3, R4 ADD R1, R1, #1 ; Inc pointer ADD R2, R2, #-1 ; Dec counter BRNZP LOOP STOP BRNZP STOP ; Stop DATAADDR .FILL x3100 .END Note: Used DATAADDR to hold address of DATA. Why? R2=0? R4  M[R1] R3  R3+R4R1  R1+1 R2  R2-1 NO YES

Compute the Sum of 12 Integers - Data .ORIG x3100 ; Data section DATA .FILL x0001 ; 12 integers .FILL x0002 .FILL x0004 .FILL x0008 .FILL xFFFF .FILL xFFFE .FILL xFFFC .FILL xFFF8 .FILL x0007 .FILL x0004 .FILL x0002 .FILL x0003 .END

Compute the Sum of 12 Integers • Use the LC3 Editor to enter the program and data and store them as add1.asm data1.asm • Use the LC3 Editor to assemble them: add1.asm  add1.obj data1.asm  data1.obj • Then use the LC3 Simulator to test them: load add1.obj and data1.obj • Set the PC, appropriate breakpoints, and execute the program (single step or run)

Sum of 12 Integers in One Package .ORIG x3000 ; Add 12 integers ; R1: Pointer to integer ; R2: Loop counter ; R3: Accumulator ; R4: Temporary register LEA R1, DATA ; Load pointer to integers AND R3, R3, #0 ; Accumulator = 0 AND R2, R2, #0 ; Counter = 12 ADD R2, R2, #12 LOOP BRZ STOP ; Stop when done LDR R4, R1, #0 ; Add next integer ADD R3, R3, R4 ADD R1, R1, #1 ; Inc pointer ADD R2, R2, #-1 ; Dec counter BRNZP LOOP STOP BRNZP STOP ; Stop ; Data section DATA .FILL x0001 ; 12 integers .FILL x0002 .FILL x0004 .FILL x0008 .FILL xFFFF .FILL xFFFE .FILL xFFFC .FILL xFFF8 .FILL x0007 .FILL x0004 .FILL x0002 .FILL x0003 .END Note: Used LEA to load pointer to data. Why?

One Pass vs Two Pass Assemblers • What does the assembler need to do? • Check for syntax errors • Build a symbol table • Assemble statements • Use pseudo-op instructions • Resolve Addresses • Create a load module • Two Pass – Checks for syntax errors and builds the Symbol Table during first pass, resolves operand addresses during second pass. • One Pass – Checks for syntax errors, builds the Symbol Table, and resolves operand addresses during the first pass. So why have a two pass?

clear R3 add R3 to R2 decrement R1 R1 = 0? No Yes HALT An Assembly Language Program Example ; ; Program to multiply a number by the constant 6 ; .ORIG x3050 LD R1, SIX ; Constant 6 LD R2, NUMBER ; Number AND R3, R3, #0 ; Clear R3 ; The product. ; The multiply loop AGAIN ADD R3, R3, R2 ; Accumulate product ADD R1, R1, #-1 ; Dec counter BRp AGAIN HALT NUMBER .BLKW 3 ; Value of Number SIX .FILL x0006 .END Symbol Table:Symbol Address AGAIN x3053 NUMBER x3057 SIX x305A

What if there were More than One Object (Load) File • Example Symbol Table: Symbols Externals Exports Addresses Start x3000 Number x300A Data ? Value x30E0 • The “Linker/Loader” would generate another “global symbol table” to resolve Externals & Exports at Load time. It would only address the Externals (Imports) and Exports (Internals). • An Export is a value make available to other modules • An External is a value expected to be defined in another module

Trap Codes • LC-3 assembler provides “pseudo-instructions” foreach trap code, so you don’t have to remember their TRAP #’s.

Program to add two single digit integers .ORIG x3000 ; begin at x3000 ; input two numbers IN ;input an integer character (ascii) {TRAP 23} LD R3, HEXN30 ;subtract x30 to get integer ADD R0, R0, R3 ADD R1, R0, #0 ;move the first integer to register 1 IN ;input another integer {TRAP 23} ADD R0, R0, R3 ;convert it to an integer ; add the numbers ADD R2, R0, R1 ;add the two integers ; print the results LEA R0, MESG ;load the address of the message string PUTS ;"PUTS" outputs a string {TRAP 22} ADD R0, R2, #0 ;move the sum to R0, to be output LD R3, HEX30 ;add 30 to integer to get integer character ADD R0, R0, R3 OUT ;display the sum {TRAP 21} ; stop HALT ;{TRAP 25} ; data MESG .STRINGZ "The sum of those two numbers is: “ HEXN30 .FILL xFFD0 ; -30 HEX HEX30 .FILL x0030 ; 30 HEX .END

Program to add two single digit integers • Enter, • Assemble, • Load • Execute, and Test the program.

Write a program to count the 1’s in register R0 ; Program to count 1's in Register R0 ; R3 is a working copy of R0 ; R1 contains the count ; R2 is a loop counter .orig x3100 ADD R3, R0, #0 ;copy R0 into R3 AND R1, R1, #0 ;clear count ADD R3, R3, #0 ;test highest bit BRZP NEXT ;count if neg ADD R1, R1, #1 NEXT AND R2, R2, #0 ;check remaining 15 bits ADD R2, R2, #-15 ; R2 = -15 (count) LOOP ADD R3, R3, R3 ;shift R3 left BRZP AGAIN ADD R1, R1, #1 ;count if neg AGAIN ADD R2, R2, #1 ;inc count BRN LOOP HALT .END

Program to Check for overflow ; Add R3=R0+R1, R2=0 indicates no overflow ; .ORIG x3000 AND R2, R2, #0 ;Initially R2=0 (no Overflow assumed) ADD R3, R0, R1 ;R3=R0+R1 ; test for overflow ADD R0, R0, #0 ;test R0 BRN NEG ;Branch if RO negative ADD R1, R1, #0 ;R0 pos, test R1 BRN DONE ;R0 pos, R1 neg -> No overflow ADD R3, R3, #0 ;R0 pos, R1 pos, maybe, test R3 BRZP DONE ;R3 also pos -> no overflow ADD R2, R2, #1 ;Set R2=1 indicating overflow BRNZP DONE R0NEG ADD R1, R1, #0 ;R0 neg, test R1 BRZP DONE ;R0 neg, R1 pos -> No overflow ADD R3, R3, #0 ;R0 neg, R1 neg, maybe, test R3 BRN DONE ;R3 also neg -> no overflow ADD R2, R2, #1 ;Set R2=1 indicating overflow DONE HALT .END

Count the occurrences of a character in a file.

; Program to count occurrences of a character in a file. ; Character to be input from the keyboard. ; Result to be displayed on the monitor. ; Program only works if no more than 9 occurrences are found. ; ; ; Initialization ; .ORIG x3000 AND R2, R2, #0 ; R2 is counter, initially 0 LD R3, PTR ; R3 is pointer to character file GETC ; R0 gets input character LDR R1, R3, #0 ; R1 gets first character from file ; ; Test character for end of file ; TEST ADD R4, R1, #-4 ; Test for EOT (ASCII x04) BRz OUTPUT ; If done, prepare the output ; ; Test character for match. If a match, increment count. ; NOT R1, R1 ADD R1, R1, R0 ; If match, R1 = xFFFF NOT R1, R1 ; If match, R1 = x0000 BRnp GETCHAR ; If no match, do not increment ADD R2, R2, #1 ; ; Get next character from file. ; GETCHAR ADD R3, R3, #1 ; Point to next character. LDR R1, R3, #0 ; R1 gets next char to test BRnzp TEST ; ; Output the count. ; OUTPUT LD R0, ASCII ; Load the ASCII template ADD R0, R0, R2 ; Covert binary count to ASCII OUT ; ASCII code in R0 is displayed. HALT ; Halt machine ; ; Storage for pointer and ASCII template ; ASCII .FILL x0030 ; ASCII offset PTR .FILL x4000 ; PTR to character file .END

Chapter 7 Introduction to LC-3 Assembly Language