ECE291 Computer Engineering II Lecture 6 & Lecture 7

ECE291Computer Engineering II

Lecture 6 & Lecture 7

Dr. Zbigniew Kalbarczyk

University of Illinois at Urbana- Champaign

Z. Kalbarczyk ECE291

Outline

• Program organization

• MASM directives

• Multiplication

• Division


Program Organization

• Create block structure and/or pseudocode on paper to get a clear concept of program control flow and data structures

• Break the total program into logical procedures/macros

• Use jumps, loops, etc. where appropriate

• Use descriptive names for variables

– noun_type for types

– nouns for variables

– verbs for procedures/functions


Debugging Hints

• Good program organization helps

• Programs do not work the first time

• Strategy to find problems

– Use DEBUG breakpoints to check program progress

– Use COMMENT to temporarily remove sections of code

– "print" statements announce milestones in program

• Test values/cases

– Try forcing registers/variables to test output of a procedure

– Use "print" statements to display critical data

• Double-check your own logic (Did you miss a special case?)

• Try a different algorithm, if all else fails...


MASM Directives

• General

– TITLE my_program.asm

• Includes drive:\path\filename

• Definitions

– DB define byte (8bits)

– DW define word (16 bits)

– DD define doubleword (32 bits)

– EQU names a constant

• Labels


MASM Directives

• Macros

– Instead of using procedures, which require both stack and time resources, macros are fast and flexible

– Advantages:

• speed; no call instruction

• readability - easier to understand program function

– Drawbacks -

• space using the MACRO multiple times duplicates the code

• tricky to debug (in particular when you have nested MACROs)

• Procedures

– “name” PROC NEAR/FAR

– ENDP


MASM Directives (cont.)

• References to procedures

– EXTERN “name” NEAR/FAR/BYTE/WORD

– PUBLIC “ name”

• Segment definition

– SEGMENT “name” PUBLIC/STACK

– ENDS

• Segment register “Hooks”

– ASSUME CS:CSEG; DES:CSEG; SS:STACK


Example Program Structure

TITLE ECE291:MPXXX

COMMENT * In this MP you will develop program which take input from the keyboard…………

*

;====== Constants=================================================

;ASCII values for common characters

CR EQU 13

LF EQU 10

ESCKEY EQU 27

;====== Externals =================================================

; -- LIB291 Routines

extrn dspmsg:near, dspout:near, kbdin:near

extrn rsave:near, rrest:near, binasc:near


Example Program Structure (cont.);==== LIBMPXXX Routines (Your code will replace calls to these

functions)

extrn LibKbdHandler:near

extrn LibMouseHandler:near

extrn LibDisplayResult:near

extrn MPXXXXIT:near

;====== Stack =====================================================

stkseg segment stack ; *** STACK SEGMENT ***

db 64 dup ('STACK ') ; 64*8 = 512 Bytes of Stack

stkseg ends

;====== Begin Code/Data ============================================

cseg segment public 'CODE' ; *** CODE SEGMENT ***

assume cs:cseg, ds:cseg, ss:stkseg, es:nothing


Example Program Structure (cont.)

;====== Variables =================================================

inputValid db 0 ; 0: InputBuffer is not ready

; 1: InputBuffer is ready

;-1: Esc key pressed

operandsStr db 'Operands: ','$'

OutputBuffer db 16 dup(?),'$' ; Contains formatted output

; (Should be terminated with '$')

MAXBUFLENGTH EQU 24

InputBuffer db MAXBUFLENGTH dup(?),'$' ; Contains one line of ; user input

include graphData.dat ; data

PUBLIC OutputBuffer, inputValid, operandsStr

PUBLIC graphData



;====== Procedures ===========================================

KbdHandler PROC NEAR

<Your code here>

KbdHandler ENDP

MouseHandler PROC NEAR

<Your code here>

MouseHandler ENDP

DisplayResult PROC NEAR

<Your code here>

DisplayResult ENDP



;====== Main Procedure ========================================

MAIN PROC FAR

MOV AX, CSEG ;Use common code and data segment

MOV DS, AX

MOV AX, 0B800h ;Use extra segment to access video screen

MOV ES, AX

<here comes your main procedure>

CALL MPXXXXIT ; Exit to DOS

MAIN ENDP

CSEG ENDS

END MAIN


Multiplication

• The product after a multiplication is always a double-width product, e.g,

– if we multiply two 16-bit numbers , they generate a 32-bit product

– unsigned: (216 - 1) * (216 - 1) = (232 - 2 * 216 + 1 < (232 - 1)

– signed: (-215) * (-215) = 230 < (231 - 1)

– overflow cannot occur

• Modification of Flags

– Most flags are undefined after multiplication

– O and C flags clear to 0 if the result fit into half-size register

– e.g., if the most significant 16 bits of the product are 0, both flags C and O clear to 0


Multiplication (cont.)

• Two different instructions for multiplication

– MUL Multiply unsigned

– IMUL Integer Multiply (2’s complement)

• Multiplication is performed on bytes, words, or double words

• Which operation to perform depends on the size of the multiplier

• The multiplier can be any register or any memory location

mul cx ; AX * CX (unsigned result in DX--AX);imul BYTE PTR [si] ; AX * [word contents of memory location

; addressed by SI] (signed product in DX--AX)


Multiplication(16 bit)

The use of the AX (and DX) registers is implied!!!!!

Multiplicand AX

Multiplier (16-bit register,

16-bit memory variable)

DX, AX = PRODUCT

(High word in DX : Low word in AX)


Multiplication (cont.)

• 8086/8088 microprocessors do not allow to perform immediate multiplication

• 80286, 80386, and 80486 allow the immediate multiplication by using a special version of the multiply instruction

• Immediate multiplication must be signed multiplication and contains three operands

– 16-bit destination register

– register or memory location that contains 16-bit multiplicand

– 8-bit or 16-bit immediate data used as a multiplier

mul cx, dx, 12h ;multiplies 12h * DX and leaves ;16-bit signed product

in CX


Multiplication

• 8-bit multiplication

Multiplicand AL



AX PRODUCT

• 32-bit multiplication

Multiplicand EAX



EDX, EAX PRODUCT (High word in EDX : Low word in EAX)

– 32-bit multiplication is available only on 80386 and above


Binary Multiplication

• Long Multiplication is done through shifts and additions

• This works if both numbers are positive

• To multiply a negative numbers, the CPU will store the sign bits of the numbers, make both numbers positive, compute the result, then negate the result if necessary

0 1 1 0 0 0 1 0 (98) x 0 0 1 0 0 1 0 1 (37)

------------------------- 0 1 1 0 0 0 1 0 0 1 1 0 0 0 1 0 - - 0 1 1 0 0 0 1 0 - - - - - (3626)


Division

• X / Y = Q; R

X Dividend

Y Divisor

Q Quotient

R Remainder

Note: Remainder has the same

sign as X (Dividend)

Examples (Signed Integers)

X / Y Q R

9 / 4 2 1-9 / 4 -2 -1 9 / -4 -2 1-9 / -4 2 1


Division (cont.)

• Two different instructions for division

– DIV Division unsigned

– IDIV Integer Division (2’s complement)

• Division is performed on bytes, words, or double words

• Which operation to perform depends on the size of the divisor

• The dividend is always a double-width dividend that is divided by the operand (divisor)

• The divisor can be any register or any memory location


Division(32-bit/16-bit)

The use of the AX (and DX) registers is implied!!!!!

Dividend DX, AX (high word in DX, low word in AX)

Divisor (16-bit register, 16-bit memory variable)

Quotient AX

Remainder DX


Division (cont.)

• 16-bit/8-bit

Dividend AX


Quotient AL

Remainder AH

• 64-bit/32-bit

Dividend EDX, EAX (high double word in EDX, low double word in EAX)


Quotient EAX

Remainder EDX

• Available on 80386 and above


Division (cont.)

• Division of two equally sized words

• Prepare the dividend

– Unsigned numbers: move zero into high order-word

– Signed numbers: use signed extension (implicitly uses AL, AX, DX registers) to fill high-word with once or zeros

– CBW (convert byte to word)

AX = xxxx xxxx snnn nnnn (before)

AX = ssss ssss snnn nnnn (after)

– CWD (convert word to double)

DX:AX = xxxx xxxx xxxx xxxx snnn nnnn nnnn nnnn (before)

DX:AX = ssss ssss ssss ssss snnn nnnn nnnn nnnn (after)

– CWDE (convert double to double-word extended) - 80386 and above


Division (cont.)

• Flag settings

– none of the flag bits change predictably for a division

• A division can result in two types of errors

– divide by zero

– divide overflow (a small number divides into a large number), e.g., 3000 / 2

• AX = 3000;

• Devisor is 2 => 8 bit division is performed

• Quotient will be written to AL => but 1500 does not fit into AL

• consequently we have divide overflow

• in both cases microprocessor generates interrupt (interrupts are covered later in this course)


Division (Example)

Division of the byte contents of memory NUMB by the contents of NUMB1

Unsigned

MOV AL, NUMB ;get NUMBMOV AH, 0 ;zero extendDIV NUMB1MOV ANSQ, AL ;save quotientMOV ANSR, AH ;save remainder

Signed

MOV AL, NUMB ;get NUMBCBW ;signed-extendIDIV NUMB1MOV ANSQ, AL ;save quotientMOV ANSR, AH ;save remainder


Division (cont.)

• What do we do with remainder after division?

– use the remainder to round the result

– drop the remainder to truncate the result

– if the division is unsigned, rounding requires that remainder is compared with half the divisor to decide whether to round up the quotient

– e.g., sequence of instructions that divide AX by BL and round the result

DIV BLADD AH, AH ;double remainderCMP AH, BL ;test for roundingJB NEXTINC AL

NEXT:

Documents

ECE291 Computer Engineering II Lecture 6 & Lecture 7