Assembly Languages Assembly Languages

Computer Organization &

Computer Organization &

Assembly Languages Assembly Languages

Pu-Jen Cheng

Integer Arithmetic

Adapted from the slides prepared by Kip Irvine for the book, Assembly Language for Intel-Based Computers, 5th Ed.

Chapter Overview

„ Shift and Rotate Instructions

„ Shift and Rotate Applications

„ Multiplication and Division Instructions

„ Extended Addition and Subtraction

„ ASCII and Unpacked Decimal Arithmetic

„ Packed Decimal Arithmetic

Shift and Rotate Instructions

„ Logical vs Arithmetic Shifts

„ SHL Instruction

„ SHR Instruction

„ SAL and SAR Instructions

„ ROL Instruction

„ ROR Instruction

„ RCL and RCR Instructions

„ SHLD/SHRD Instructions

Logical vs Arithmetic Shifts

„ A logical shift fills the newly created bit position with zero:

A ith ti hift fill th l t d bit

CF

0

• An arithmetic shift fills the newly created bit position with a copy of the number’s sign bit:

CF

SHL Instruction

„ The SHL (shift left) instruction performs a logical left shift on the destination operand, filling the

lowest bit with 0.

0

CF

• Operand types for SHL: SHL destination,count

SHL reg,imm8 SHL mem,imm8 SHL reg,CL SHL mem,CL

(Same for all shift and rotate instructions)

Fast Multiplication

mov dl,5 shl dl,1

Shifting left 1 bit multiplies a number by 2

0 0 0 0 1 0 1 0

0 0 0 0 0 1 0 1 = 5

= 10

Before:

After:

S f f

mov dl,5

shl dl,2 ; DL = 20

Shifting left n bits multiplies the operand by 2ⁿ For example, 5 * 2² = 20

SHR Instruction

„ The SHR (shift right) instruction performs a logical right shift on the destination operand. The highest bit position is filled with a zero.

0

CF

mov dl,80

shr dl,1 ; DL = 40

shr dl,2 ; DL = 10

Shifting right n bits divides the operand by 2ⁿ

SAL and SAR Instructions

„ SAL (shift arithmetic left) is identical to SHL.

„ SAR (shift arithmetic right) performs a right arithmetic shift on the destination operand.

CF

An arithmetic shift preserves the number's sign.

mov dl,-80

sar dl,1 ; DL = -40

sar dl,2 ; DL = -10

Your turn . . .

mov al,6Bh ; 01101011

shr al,1 a.

shl al,3 b.

mov al,8Ch ; 10001100

sar al,1 c.

Indicate the hexadecimal value of AL after each shift:

35h A8h

sar al,1 c. C6h

sar al,3 d.

C6h F8h

ROL Instruction

„ ROL (rotate) shifts each bit to the left

„ The highest bit is copied into both the Carry flag and into the lowest bit

„ No bits are lost

CF

mov al,11110000b

rol al,1 ; AL = 11100001b

mov dl,3Fh

rol dl,4 ; DL = F3h

ROR Instruction

„ ROR (rotate right) shifts each bit to the right

„ The lowest bit is copied into both the Carry flag and into the highest bit

„ No bits are lost

CF

mov al,11110000b

ror al,1 ; AL = 01111000b

mov dl,3Fh

ror dl,4 ; DL = F3h

Your turn . . .

mov al,6Bh ; 01101011

ror al,1 a.

rol al,3 b.

Indicate the hexadecimal value of AL after each rotation:

B5h ADh

RCL Instruction

„ RCL (rotate carry left) shifts each bit to the left

„ Copies the Carry flag to the least significant bit

„ Copies the most significant bit to the Carry flag

CF

clc ; CF = 0

mov bl,88h ; CF,BL = 0 10001000b rcl bl,1 ; CF,BL = 1 00010000b rcl bl,1 ; CF,BL = 0 00100001b

RCR Instruction

„ RCR (rotate carry right) shifts each bit to the right

„ Copies the Carry flag to the most significant bit

„ Copies the least significant bit to the Carry flag

CF

stc ; CF = 1

mov ah,10h ; CF,AH = 1 00010000b rcr ah,1 ; CF,AH = 0 10001000b

Your turn . . .

stc

mov al,6Bh

rcr al,1 a.

rcl al,3 b.

Indicate the hexadecimal value of AL after each rotation:

B5h AEh

SHLD Instruction

„ Shifts a destination operand a given number of bits to the left

„ The bit positions opened up by the shift are filled by the most significant bits of the source operand

Th d i ff d

„ The source operand is not affected

„ Syntax:

SHLD destination, source, count

„ Operand types:

SHLD reg16/32, reg16/32, imm8/CL SHLD mem16/32, reg16/32, imm8/CL

SHLD Example

.data

wval WORD 9BA6h

9BA6 AC36

wval AX

Shift wval 4 bits to the left and replace its lowest 4 bits with the high 4 bits of AX:

.code B f

mov ax,0AC36h shld wval,ax,4

9BA6 AC36 BA6A AC36 Before:

After:

SHRD Instruction

„ Shifts a destination operand a given number of bits to the right

„ The bit positions opened up by the shift are filled by the least significant bits of the source operand The source operand is not affected

„ The source operand is not affected

„ Syntax:

SHRD destination, source, count

„ Operand types:

SHRD reg16/32, reg16/32, imm8/CL SHRD mem16/32, reg16/32, imm8/CL

SHRD Example

mov ax,234Bh mov dx,7654h

Shift AX 4 bits to the right and replace its highest 4 bits with the low 4 bits of DX:

Before: ^{7654 234B}

DX AX

shrd ax,dx,4

After: ^{7654 4234}

Your turn . . .

mov ax,7C36h mov dx,9FA6h

shld dx,ax,4 ; DX =

Indicate the hexadecimal values of each destination operand:

FA67h 36FAh shrd dx,ax,8 ; DX = ^36FAh

What's Next

„ Shift and Rotate Instructions

„ Shift and Rotate Applications

„ Multiplication and Division Instructions

„ Extended Addition and Subtraction

„ ASCII and Unpacked Decimal Arithmetic

„ Packed Decimal Arithmetic

Shift and Rotate Applications

„ Shifting Multiple Doublewords

„ Binary Multiplication

„ Displaying Binary Bits

„ Isolating a Bit String

Shifting Multiple Doublewords

„ Programs sometimes need to shift all bits within an array, as one might when moving a bitmapped

graphic image from one screen location to another.

„ The following shifts an array of 3 doublewords 1 bit to the rightg

.data

ArraySize = 3

array DWORD ArraySize DUP(99999999h) ; 1001 1001...

.code

mov esi,0

shr array[esi + 8],1 ; high dword

rcr array[esi + 4],1 ; middle dword, include Carry rcr array[esi],1 ; low dword, include Carry

[esi+8] [esi+4] [esi]

Binary Multiplication

„ We already know that SHL performs unsigned multiplication efficiently when the multiplier is a power of 2.

„ You can factor any binary number into powers of 2.

¾ For example to multiply EAX * 36 factor 36 into 32 + 4

¾ For example, to multiply EAX 36, factor 36 into 32 + 4 and use the distributive property of multiplication to carry out the operation:

EAX * 36

= EAX * (32 + 4)

= (EAX * 32)+(EAX * 4)

mov eax,123 mov ebx,eax

shl eax,5 ; mult by 2⁵ shl ebx,2 ; mult by 2² add eax,ebx

Your turn . . .

mov ax,2 ; test value

mov dx,ax

shl dx 4 ; AX * 16

Multiply AX by 26, using shifting and addition instructions. Hint: 26 = 16 + 8 + 2.

shl dx,4 ; AX 16

push dx ; save for later

mov dx,ax

shl dx,3 ; AX * 8

shl ax,1 ; AX * 2

add ax,dx ; AX * 10

pop dx ; recall AX * 16

add ax,dx ; AX * 26

Displaying Binary Bits

Algorithm: Shift MSB into the Carry flag; If CF = 1,

append a "1" character to a string; otherwise, append a "0" character. Repeat in a loop, 32 times.

.data

buffer BYTE 32 DUP(0),0 .code

mov ecx,32

mov esi,OFFSET buffer L1: shl eax,1

mov BYTE PTR [esi],'0' jnc L2

mov BYTE PTR [esi],'1' L2: inc esi

loop L1

Isolating a Bit String

„ The MS-DOS file date field packs the year, month, and day into 16 bits:

DH DL

0 1 0

0

0 0 1 1 0 1 1 0 1 0 1 0

Year Month Day

9-15 5-8 0-4

Field:

Bit numbers:

Isolating a Bit String (cont.)

mov ax,dx ; make a copy of DX shr ax,5 ; shift right 5 bits and al 00001111b ; clear bits 4 7

mov al,dl ; make a copy of DL and al,00011111b ; clear bits 5-7

mov day,al ; save in day variable

mov al,dh ; make a copy of DH shr al,1 ; shift right 1 bit mov ah,0 ; clear AH to 0

add ax,1980 ; year is relative to 1980 mov year,ax ; save in year

and al,00001111b ; clear bits 4-7

mov month,al ; save in month variable

What's Next

„ Shift and Rotate Instructions

„ Shift and Rotate Applications

„ Multiplication and Division Instructions

„ Extended Addition and Subtraction

„ ASCII and Unpacked Decimal Arithmetic

„ Packed Decimal Arithmetic

Multiplication and Division Instructions

„ MUL Instruction

„ IMUL Instruction

„ DIV Instruction

„ Signed Integer Division

„ CBW, CWD, CDQ Instructions

„ IDIV Instruction

„ Implementing Arithmetic Expressions

MUL Instruction

„ The MUL (unsigned multiply) instruction multiplies an 8-, 16-, or 32-bit operand by either AL, AX, or EAX.

„ The instruction formats are:

MUL r/m8 MUL r/m16 MUL r/m32

Implied operands:

MUL Examples

100h * 2000h, using 16-bit operands:

.data

val1 WORD 2000h val2 WORD 100h .code

mov ax,val1

The Carry flag indicates whether or not the upper half of the product

mul val2 ; DX:AX = 00200000h, CF=1 contains t i

significant digits.

mov eax,12345h mov ebx,1000h

mul ebx ; EDX:EAX = 0000000012345000h, CF=0

12345h * 1000h, using 32-bit operands:

Your turn . . .

mov ax,1234h mov bx,100h mul bx

What will be the hexadecimal values of DX, AX, and the Carry flag after the following instructions execute?

DX = 0012h, AX = 3400h, CF = 1

Your turn . . .

mov eax,00128765h mov ecx,10000h mul ecx

What will be the hexadecimal values of EDX, EAX, and the Carry flag after the following instructions execute?

EDX = 00000012h, EAX = 87650000h, CF = 1

IMUL Instruction

„ IMUL (signed integer multiply ) multiplies an 8-, 16-, or 32-bit signed operand by either AL, AX, or EAX

„ Preserves the sign of the product by sign-

extending it into the upper half of the destination registerg

Example: multiply 48 * 4, using 8-bit operands:

mov al,48 ; 110000 mov bl,4

imul bl ; AX = 00C0h, OF=1

OF=1 because AH is not a sign extension of AL.

IMUL Examples

Multiply 4,823,424 * −423:

mov eax,4823424 mov ebx,-423

imul ebx ; EDX:EAX = FFFFFFFF86635D80h, OF=0

OF=0 because EDX is a sign extension of EAX.

Your turn . . .

mov ax,8760h mov bx,100h imul bx

What will be the hexadecimal values of DX, AX, and the Carry flag after the following instructions execute?

DX = FF87h, AX = 6000h, OF = 1

DIV Instruction

„ The DIV (unsigned divide) instruction performs 8- bit, 16-bit, and 32-bit division on unsigned integers

„ A single operand is supplied (register or memory operand), which is assumed to be the divisor

I i f

„ Instruction formats:

DIV r/m8 DIV r/m16 DIV r/m32

Default Operands:

DIV Examples

Divide 8003h by 100h, using 16-bit operands:

mov dx,0 ; clear dividend, high mov ax,8003h ; dividend, low

mov cx,100h ; divisor

div cx ; AX = 0080h, DX = 3

Same division, using 32-bit operands:

mov edx,0 ; clear dividend, high mov eax,8003h ; dividend, low

mov ecx,100h ; divisor

div ecx ; EAX = 00000080h, DX = 3

Your turn . . .

mov dx,0087h mov ax,6000h

What will be the hexadecimal values of DX and AX after the following instructions execute? Or, if divide overflow occurs, you can indicate that as your answer:

,

mov bx,100h div bx

DX = 0000h, AX = 8760h

Your turn . . .

mov dx,0087h

What will be the hexadecimal values of DX and AX after the following instructions execute? Or, if divide overflow occurs, you can indicate that as your answer:

mov ax,6002h mov bx,10h div bx

Divide Overflow

Signed Integer Division

„ Signed integers must be sign-extended before division takes place

¾ fill high byte/word/doubleword with a copy of the low byte/word/doubleword's sign bit

„ For example, the high byte contains a copy of the sign bit from the low byte:

1 0 0 0 1 1 1 1

1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1

CBW, CWD, CDQ Instructions

„ The CBW, CWD, and CDQ instructions provide important sign-extension operations:

¾ CBW (convert byte to word) extends AL into AH

¾ CWD (convert word to doubleword) extends AX into DX

¾ CDQ (convert doubleword to quadword) extends EAX into EDX

EDX

„ Example:

mov eax,0FFFFFF9Bh ; (-101)

cdq ; EDX:EAX = FFFFFFFFFFFFFF9Bh

IDIV Instruction

„ IDIV (signed divide) performs signed integer division

„ Same syntax and operands as DIV instruction

Example: 8-bit division of –48 by 5

mov al -48 mov al,-48

cbw ; extend AL into AH mov bl,5

idiv bl ; AL = -9, AH = -3

IDIV Examples

Example: 16-bit division of –48 by 5

mov ax,-48

cwd ; extend AX into DX mov bx,5

idiv bx ; AX = -9, DX = -3

Example: 32-bit division of –48 by 5

mov eax,-48

cdq ; extend EAX into EDX mov ebx,5

idiv ebx ; EAX = -9, EDX = -3

Your turn . . .

mov ax,0FDFFh ; -513 cwd

What will be the hexadecimal values of DX and AX after the following instructions execute? Or, if divide overflow occurs, you can indicate that as your answer:

mov bx,100h ; 256 idiv bx

DX = FFFFh (−1), AX = FFFEh (−2)

Divide Overflow

„

Divide overflow

mov ax, 1000h mov bl, 10h div bl

div bl

It causes a CPU interrupt and halts the program.

(divided by zero cause similar results)

Unsigned Arithmetic Expressions

„ Some good reasons to learn how to implement integer expressions:

¾ Learn how do compilers do it

¾ Test your understanding of MUL, IMUL, DIV, IDIV

¾ Check for overflow (Carry and Overflow flags)( y g ) Example: var4 = (var1 + var2) * var3

; Assume unsigned operands mov eax,var1

add eax,var2 ; EAX = var1 + var2 mul var3 ; EAX = EAX * var3 jc TooBig ; check for carry mov var4,eax ; save product

Signed Arithmetic Expressions

Example: eax = (-var1 * var2) + var3

mov eax,var1 neg eax

imul var2

jo TooBig ; check for overflow add eax,var3

jo TooBig ; check for overflow

Example: var4 = (var1 * 5) / (var2 – 3)

mov eax,var1 ; left side mov ebx,5

imul ebx ; EDX:EAX = product mov ebx,var2 ; right side

sub ebx,3

idiv ebx ; EAX = quotient mov var4,eax

Signed Arithmetic Expressions

Example: var4 = (var1 * -5) / (-var2 % var3);

mov eax,var2 ; begin right side neg eax

cdq ; sign-extend dividend

idiv var3 ; EDX = remainder mov ebx,edx ; EBX = right side mov eax,-5 ; begin left side

imul var1 ; EDX:EAX = left side idiv ebx ; final division

mov var4,eax ; quotient

Sometimes it's easiest to calculate the right-hand term of an expression first.

Your turn . . .

mov eax,20 i l b

Implement the following expression using signed 32-bit integers:

eax = (ebx * 20) / ecx

imul ebx idiv ecx

Your turn . . .

push edx

push eax ; EAX needed later

Implement the following expression using signed 32-bit integers. Save and restore ECX and EDX:

eax = (ecx * edx) / eax

mov eax,ecx

imul edx ; left side: EDX:EAX pop ebx ; saved value of EAX idiv ebx ; EAX = quotient

pop edx ; restore EDX, ECX

Your turn . . .

mov eax,var1 mov edx,var2

Implement the following expression using signed 32-bit integers. Do not modify any variables other than var3:

var3 = (var1 * -var2) / (var3 – ebx)

neg edx

imul edx ; left side: EDX:EAX mov ecx,var3

sub ecx,ebx

idiv ecx ; EAX = quotient mov var3,eax

What's Next

„ Shift and Rotate Instructions

„ Shift and Rotate Applications

„ Multiplication and Division Instructions

„ Extended Addition and Subtraction

„ ASCII and UnPacked Decimal Arithmetic

„ Packed Decimal Arithmetic

Extended Addition and Subtraction

„ ADC Instruction

„ Extended Precision Addition

„ SBB Instruction

„ Extended Precision Subtraction

Extended Precision Addition

„ Adding two operands that are longer than the computer's word size (32 bits).

¾ Virtually no limit to the size of the operands

„ The arithmetic must be performed in steps

The Carry value from each step is passed on to the next

¾ The Carry value from each step is passed on to the next step.

ADC Instruction

„ ADC (add with carry) instruction adds both a source operand and the contents of the Carry flag to a destination operand.

„ Operands are binary values

¾ Same syntax as ADD, SUB, etc.

„ Example

¾ Add two 32-bit integers (FFFFFFFFh +

FFFFFFFFh), producing a 64-bit sum in EDX:EAX:

mov edx,0

mov eax,0FFFFFFFFh add eax,0FFFFFFFFh

adc edx,0 ;EDX:EAX = 00000001FFFFFFFEh

Extended Addition Example

„ Task: Add 1 to EDX:EAX

¾ Starting value of EDX:EAX: 00000000FFFFFFFFh

¾ Add the lower 32 bits first, setting the Carry flag.

¾ Add the upper 32 bits, and include the Carry flag.

d 0 t h lf

mov edx,0 ; set upper half mov eax,0FFFFFFFFh ; set lower half add eax,1 ; add lower half adc edx,0 ; add upper half

EDX:EAX = 00000001 00000000

SBB Instruction

„ The SBB (subtract with borrow) instruction

subtracts both a source operand and the value of the Carry flag from a destination operand.

„ Operand syntax:

¾ Same as for the ADC instruction

Extended Subtraction Example

„ Task: Subtract 1 from EDX:EAX

¾ Starting value of EDX:EAX: 0000000100000000h

¾ Subtract the lower 32 bits first, setting the Carry flag.

¾ Subtract the upper 32 bits, and include the Carry flag.

mov edx 1 ; set upper half mov edx,1 ; set upper half mov eax,0 ; set lower half

sub eax,1 ; subtract lower half sbb edx,0 ; subtract upper half EDX:EAX = 00000000 FFFFFFFF

What's Next

„ Shift and Rotate Instructions

„ Shift and Rotate Applications

„ Multiplication and Division Instructions

„ Extended Addition and Subtraction

„ ASCII and UnPacked Decimal Arithmetic

„ Packed Decimal Arithmetic

ASCII and Packed Decimal Arithmetic

„ Binary Coded Decimal

„ ASCII Decimal

„ AAA Instruction

„ AAS Instruction

„ AAM Instruction

„ AAD Instruction

„ Packed Decimal Integers

„ DAA Instruction

„ DAS Instruction

Representation of Numbers

„ Numbers are in ASCII form

¾ when received from keyboard

¾ when sending to the display

„ Binary form is efficient to process numbers internally

„ Requires conversion between these two number representations

Advantages of ASCII Arithmetic

„ Avoid conversion overheads between two formats

„ Avoid danger of the round-off errors that occur with floating-point numbers

Representations of Decimal Numbers

„ ASCII representation

„ BCD representation

¾ Unpacked BCD

¾ Packed BCD

ASCII Decimal

„ A number using ASCII Decimal representation stores a single ASCII digit in each byte

¾ For example, 5,678 is stored as the following sequence of hexadecimal bytes:

35 36 37 38

Binary-Coded Decimal

„ Binary-coded decimal (BCD) integers use 4 binary bits to represent each decimal digit

„ A number using unpacked BCD

representation stores a decimal digit in the lower four bits of each byte

lower four bits of each byte

¾ For example, 5,678 is stored as the following sequence of hexadecimal bytes:

05 06 07 08

AAA Instruction

„ The AAA (ASCII adjust after addition)

instruction adjusts the binary result of an ADD or ADC instruction. It makes the result in AL consistent with ASCII decimal representation.

¾ The Carry value, if any ends up in AHy , y p

„ Example: Add '8' and '2'

mov ah,0

mov al,'8' ; AX = 0038h add al,'2' ; AX = 006Ah

aaa ; AX = 0100h (adjust result) or ax,3030h ; AX = 3130h = '10'

Processing ASCII Numbers

ASCII addition

Example 1 Example 2

34H = 00110100B 36H = 00110110B 35H = 00110101B 37H = 00110111B

11 1 1 11 11 1

69H = 01101001B 6DH = 01101101B

Should be 09H Should be 13H

Ignore 6 Ignore 6 and add 9 to D

„ The AAA instruction performs these adjustments to the byte in AL register

AAS Instruction

„ The AAS (ASCII adjust after subtraction)

instruction adjusts the binary result of an SUB or SBB instruction. It makes the result in AL consistent with ASCII decimal representation.

¾ It places the Carry value, if any, in AH

„ Example: Subtract '9' from '8'

mov ah,0

mov al,'8' ; AX = 0038h sub al,'9' ; AX = 00FFh

aas ; AX = FF09h, CF=1

pushf ; save CF or al,30h ; AL = '9‘

popf ; restore CF

68 -29 39

AAM Instruction

„ The AAM (ASCII adjust after multiplication) instruction adjusts the binary result of a MUL instruction. The multiplication must have been performed on unpacked BCD numbers.

mov bl,05h ; first operand mov al,06h ; second operand

mul bl ; AX = 001Eh

aam ; AX = 0300h

AAD Instruction

„ The AAD (ASCII adjust before division) instruction adjusts the unpacked BCD dividend in AX before a division operation

.data

quotient BYTE ? remainder BYTE ? .code

mov ax,0307h ; dividend

aad ; AX = 0025h

mov bl,5 ; divisor

div bl ; AX = 0207h

mov quotient,al mov remainder,ah

What's Next

„ Shift and Rotate Instructions

„ Shift and Rotate Applications

„ Multiplication and Division Instructions

„ Extended Addition and Subtraction

„ ASCII and UnPacked Decimal Arithmetic

„ Packed Decimal Arithmetic

Packed Decimal Arithmetic

„ Packed decimal integers store two decimal digits per byte

¾ For example, 12,345,678 can be stored as the following sequence of hexadecimal bytes:

12 34 56 78

Packed decimal is also known as packed BCD.

Good for financial values – extended precision possible, without rounding errors.

DAA Instruction

„ The DAA (decimal adjust after addition) instruction converts the binary result of an ADD or ADC

operation to packed decimal format.

¾ The value to be adjusted must be in AL

¾ If the lower digit is adjusted the Auxiliary Carry flag is

¾ If the lower digit is adjusted, the Auxiliary Carry flag is set.

¾ If the upper digit is adjusted, the Carry flag is set.

DAA Logic

If (AL(lo) > 9) or (AuxCarry = 1) AL = AL + 6

If (AL(hi) > 9) or Carry = 1 AL = AL + 60h

DAA Examples

„ Example: calculate BCD 35 + 48

mov al,35h

add al,48h ; AL = 7Dh

daa ; AL = 83h, CF = 0

• Example: calculate BCD 35 + 65Example: calculate BCD 35 + 65

mov al,35h

add al,65h ; AL = 9Ah

daa ; AL = 00h, CF = 1

• Example: calculate BCD 69 + 29

mov al,69h

add al,29h ; AL = 92h

daa ; AL = 98h, CF = 0

DAS Instruction

„ The DAS (decimal adjust after subtraction)

instruction converts the binary result of a SUB or SBB operation to packed decimal format.

„ The value must be in AL

DAS Logic

If (AL(lo) > 9) OR (AuxCarry = 1) AL = AL − 6;

If (AL(hi) > 9) or (Carry = 1) AL = AL − 60h;

DAS Examples

„ Example: subtract BCD 48 – 35

mov al,48h

sub al,35h ; AL = 13h

das ; AL = 13h CF = 0

• Example: subtract BCD 62 – 35Example: subtract BCD 62 35

mov al,62h

sub al,35h ; AL = 2Dh, CF = 0

das ; AL = 27h, CF = 0

• Example: subtract BCD 32 – 29

mov al,32h

sub al,29h ; AL = 09h, CF = 0

daa ; AL = 03h, CF = 0

DAS Examples

„ Example: subtract BCD 32 – 39

mov al,32h

sub al,39h ; AL = F9h, CF = 1

das ; AL = 93h, CF = 1

Steps:

AL = F9h

ACF = 1, so subtract 6 from F9h AL = F3h

F > 9, so subtract 60h from F3h AL = 93h, CF = 1

432 -139 293

Summary

„ Shift and rotate instructions are some of the best tools of assembly language

¾ finer control than in high-level languages

¾ SHL, SHR, SAR, ROL, ROR, RCL, RCR

MUL d DIV i t ti

„ MUL and DIV – integer operations

¾ close relatives of SHL and SHR

¾ CBW, CDQ, CWD: preparation for division

„ Extended precision arithmetic: ADC, SBB

„ ASCII decimal operations (AAA, AAS, AAM, AAD)

„ Packed decimal operations (DAA, DAS)