ECE291 Computer Engineering II Lecture 6

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ECE291Computer Engineering IILecture 6
Lecture 7

• Dr. Zbigniew Kalbarczyk
• University of Illinois at Urbana- Champaign

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Outline

• Program organization
• MASM directives
• Multiplication
• Division

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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

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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...

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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

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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

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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 CSCSEG DESCSEG SSSTACK

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Example Program Structure

• TITLE ECE291MPXXX
• 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 dspmsgnear, dspoutnear, kbdinnear
• extrn rsavenear, rrestnear, binascnear
  

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Example Program Structure (cont.)

• LIBMPXXX Routines (Your code will replace
  calls to these functions)
• extrn LibKbdHandlernear
• extrn LibMouseHandlernear
• extrn LibDisplayResultnear
• extrn MPXXXXITnear
• Stack
  
• stkseg segment stack
  STACK SEGMENT
• db 64 dup ('STACK ') 648
  512 Bytes of Stack
• stkseg ends
• Begin Code/Data
  
• cseg segment public 'CODE'
  CODE SEGMENT
• assume cscseg, dscseg, ssstkseg,
  esnothing

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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

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Example Program Structure (cont.)

• Procedures
  
• KbdHandler PROC NEAR
• ltYour code heregt
• KbdHandler ENDP
• MouseHandler PROC NEAR
• ltYour code heregt
• MouseHandler ENDP
• DisplayResult PROC NEAR
• ltYour code heregt
• DisplayResult ENDP

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Example Program Structure (cont.)

• 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
• lthere comes your main proceduregt
• CALL MPXXXXIT Exit
  to DOS
• MAIN ENDP
• CSEG ENDS
• END MAIN

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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 lt (232 - 1)
• signed (-215) (-215) 230 lt (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

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Multiplication (cont.)

• Two different instructions for multiplication
• MUL Multiply unsigned
• IMUL Integer Multiply (2s 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)

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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)

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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

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Multiplication

• 8-bit multiplication
• Multiplicand AL
• Multiplier (8-bit register,
• 8-bit memory variable)
• AX PRODUCT
• 32-bit multiplication
• Multiplicand EAX
• Multiplier (32-bit register,
• 32-bit memory variable)
• EDX, EAX PRODUCT (High word in EDX Low word
  in EAX)
• 32-bit multiplication is available only on 80386
  and above

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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)
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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
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Division (cont.)

• Two different instructions for division
• DIV Division unsigned
• IDIV Integer Division (2s 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

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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

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Division (cont.)

• 16-bit/8-bit
• Dividend AX
• Divisor (8-bit register, 8-bit memory variable)
• Quotient AL
• Remainder AH
• 64-bit/32-bit
• Dividend EDX, EAX (high double word in EDX, low
  double word in EAX)
• Divisor (32-bit register, 32-bit memory
  variable)
• Quotient EAX
• Remainder EDX
• Available on 80386 and above

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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)
• DXAX xxxx xxxx xxxx xxxx snnn nnnn nnnn nnnn
  (before)
• DXAX ssss ssss ssss ssss snnn nnnn nnnn nnnn
  (after)
• CWDE (convert double to double-word extended) -
  80386 and above

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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 gt 8 bit division is performed
• Quotient will be written to AL gt 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)

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Division (Example)

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

Unsigned MOV AL, NUMB get
NUMB MOV AH, 0 zero extend DIV NUMB1 MOV
ANSQ, AL save quotient MOV ANSR, AH
save remainder
Signed MOV AL, NUMB get NUMB CBW
signed-extend IDIV NUMB1 MOV ANSQ,
AL save quotient MOV ANSR, AH
save remainder
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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 BL ADD AH, AH double remainder CMP AH,
BL test for rounding JB NEXT INC AL NEXT