Assembly Language - Machine Organization - Lecture Slides, Slides of Electric Machines

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Chapter 7

Introduction to LC-

Assembly 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 R

Compute the Sum of 12 Integers

• Program begins at location x3000.
• Integers begin at location x3100.

R1 ← x R3 ← 0 (Sum) R2 ← 12(count)

R2=0?

R4 ← M[R1]

R3 ← R3+R

R1 ← R1+

R2 ← R2-

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 Docsity.com

Compute the Sum of 12 Integers

Address Instruction Comments

x3000 1 1 1 0 0 0 1 0 1 1 1 1 1 1 1 1 R1 ← x

x3001 0 1 0 1 0 1 1 0 1 1 1 0 0 0 0 0 R3 ← 0

x3002 0 1 0 1 0 1 0 0 1 0 1 0 0 0 0 0 R2 ← 0

x3003 0 0 0 1 0 1 0 0 1 0 1 0 1 1 0 0 R2 ← 12

x3004 (^0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 1) If Z, goto x300A

x3005 (^0 1 1 0 1 0 0 0 0 1 0 0 0 0 0 0) Load next value to R

x3006 (^0 0 0 1 0 1 1 0 1 1 0 0 0 1 0 0) Add to R

x3007 (^0 0 0 1 0 0 1 0 0 1 1 0 0 0 0 1) Increment R1 (pointer)

X3008 0 0 0 1 0 1 0 0 1 0 1 1 1 1 1 1 Decrement R (counter) x3009 (^0 0 0 0 1 1 1 1 1 1 1 1 1 0 1 0) Goto x

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 Docsity.com

Compute the Sum of 12 Integers - Data

.ORIG x

; Data section

DATA .FILL x0001 ; 12 integers

.FILL x

.FILL x

.FILL x

.FILL xFFFF

.FILL xFFFE

.FILL xFFFC

.FILL xFFF

.FILL x

.FILL x

.FILL x

.FILL x

.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) Docsity.com

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?

An Assembly Language Program Example

; Program to multiply a number by the constant 6 ; .ORIG x

LD R1, SIX ; Constant 6 LD R2, NUMBER ; Number AND R3, R3, #0 ; Clear R ; 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 x

.END

Symbol Table: Symbol Address AGAIN x NUMBER x SIX x305A

clear R

add R3 to R

decrement R

R1 = 0?

HALT

No

Yes

Trap Codes

• LC-3 assembler provides “pseudo-instructions” for

each trap code, so you don’t have to remember their TRAP #’s.

Code Equivalent Description

HALT TRAP x25 Halt execution and print message to console.

IN TRAP x23 Print prompt on console,

read (and echo) one character from keyboard.

Character stored in R0[7:0].

OUT TRAP x21 Write one character (in R0[7:0]) to console.

GETC TRAP x20 Read one character from keyboard.

Character stored in R0[7:0].

PUTS TRAP x22 Write null-terminated string to console.

Address of string is in R0.

Program to add two single digit integers

.ORIG x3000 ; begin at x ; input two numbers IN ;input an integer character (ascii) {TRAP 23} LD R3, HEXN30 ;subtract x30 to get integer ADD R0, R0, R 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, R 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

Write a program to count the 1’s in register R

; Program to count 1's in Register R

; R3 is a working copy of R

; R1 contains the count

; R2 is a loop counter

.orig x

ADD R3, R0, #0 ;copy R0 into R

AND R1, R1, #0 ;clear count

ADD R3, R3, #0 ;test highest bit

BRZP NEXT ;count if neg

ADD R1, R1,

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 Docsity.com

Program to Check for overflow

; Add R3=R0+R1, R2=0 indicates no overflow ; .ORIG x

AND R2, R2, #0 ;Initially R2=0 (no Overflow assumed) ADD R3, R0, R1 ;R3=R0+R

; test for overflow ADD R0, R0, #0 ;test R BRN NEG ;Branch if RO negative ADD R1, R1, #0 ;R0 pos, test R BRN DONE ;R0 pos, R1 neg -> No overflow

ADD R3, R3, #0 ;R0 pos, R1 pos, maybe, test R 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 R BRZP DONE ;R0 neg, R1 pos -> No overflow

ADD R3, R3, #0 ;R0 neg, R1 neg, maybe, test R BRN DONE ;R3 also neg -> no overflow ADD R2, R2, #1 ;Set R2=1 indicating overflow

DONE HALT .END

; 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 x 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, R ADD R1, R1, R0 ; If match, R1 = xFFFF NOT R1, R1 ; If match, R1 = x BRnp GETCHAR ; If no match, do not increment ADD R2, R2, # ; ; 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