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(1)

Intel x86 Instruction Set Architecture

Computer Organization and Assembly Languages p g z y g g

Yung-Yu Chuang

(2)

Data Transfers Instructions

(3)

MOV instruction

• Move from source to destination. Syntax:

MOV d ti ti

MOV destination, source

• Source and destination have the same size

N h d i d

• No more than one memory operand permitted

• CS, EIP, and IP cannot be the destination

• No immediate to segment moves

(4)

MOV instruction

.data

count BYTE 100 wVal WORD 2 .code

mov bl,count V l mov ax,wVal mov count,al

mov al,wVal ; error mov ax count ; error mov ax,count ; error mov eax,count ; error

4

(5)

Exercise . . .

Explain why each of the following MOV statements are invalid:

invalid:

.data

bVal BYTE 100 bVal2 BYTE ? wVal WORD 2 dVal DWORD 5 .code

mov ds,45 ; a.

mov esi,wVal ; b.

i dV l

mov eip,dVal ; c.

(6)

Memory to memory

.data

var1 WORD ? var1 WORD ? var2 WORD ?

d .code

mov ax, var1 mov var2, ax

6

(7)

Copy smaller to larger

.data

count WORD 1 count WORD 1 .code

mov ecx, 0,

mov cx, count d

.data

signedVal SWORD -16 ; FFF0h code

.code

mov ecx, 0 ; mov ecx, 0FFFFFFFFh mov cx signedVal

mov cx, signedVal

(8)

Zero extension

When you copy a smaller value into a larger destination, the MOVZX instruction fills (extends) the upper half of ( ) pp the destination with zeros.

1 0 0 0 1 1 1 1 Source 0 1 0 0 0 1 1 1 1 Source

0 movzx r32,r/m8

movzx r32,r/m16 movzx r16,r/m8

1 0 0 0 1 1 1 1 Destination 0 0 0 0 0 0 0 0

movzx r16,r/m8

mov bl,10001111b

movzx ax,bl ; zero-extension

h d b

8

The destination must be a register.

(9)

Sign extension

The MOVSX instruction fills the upper half of the destination with a copy of the source operand's sign bit

with a copy of the source operand s sign bit.

1 0 0 0 1 1 1 1 Source

1 0 0 0 1 1 1 1 Destination 1 1 1 1 1 1 1 1

mov bl,10001111b

movsx ax,bl ; sign extension

(10)

MOVZX MOVSX

From a smaller location to a larger one

mov bx, 0A69Bh

movzx eax, bx ; EAX=0000A69Bh movzx edx, bl ; EDX=0000009Bh movzx cx, bl ; EAX=009Bh

mov bx, 0A69Bh

movsx eax, bx ; EAX=FFFFA69Bh movsx edx, bl ; EDX=FFFFFF9Bh movsx cx, bl ; EAX=FF9Bh

10

(11)

LAHF/SAHF (load/store status flag from/to AH) .data

fl BYTE ? saveflags BYTE ? .code

lahf

mov saveflags, ah mov saveflags, ah ...

h fl

mov ah, saveflags

sahf

(12)

EFLAGS

12

(13)

XCHG Instruction

XCHG exchanges the values of two operands. At least one operand must be a register. No immediate operands are p g p permitted.

data .data

var1 WORD 1000h var2 WORD 2000h .code

xchg ax,bx ; exchange 16-bit regs xchg ah,al ; exchange 8-bit regs xchg var1,bx ; exchange mem, reg

xchg eax ebx ; exchange 32-bit regs xchg eax,ebx ; exchange 32-bit regs

(14)

Exchange two memory locations

.data

1 1000h

var1 WORD 1000h var2 WORD 2000h .code

mov ax, val1 xchg ax, val2 mov val1, ax

14

(15)

Arithmetic Instructions

(16)

Addition and Subtraction

• INC and DEC Instructions

d I i

• ADD and SUB Instructions

• NEG Instruction

• Implementing Arithmetic Expressions

• Flags Affected by Arithmetic

• Flags Affected by Arithmetic

– Zero Sign – Sign – Carry

Overflow – Overflow

16

(17)

INC and DEC Instructions

• Add 1, subtract 1 from destination operand

– operand may be register or memory

• INC destination

• Logic: destination  destination + 1

• DEC destination

• Logic: destination  destination – 1

(18)

INC and DEC Examples

.data

myWord WORD 1000h myWord WORD 1000h

myDword DWORD 10000000h .code

inc myWord ; 1001h dec myWord ; 1000h

inc myDword ; 10000001h mov ax 00FFh

mov ax,00FFh

inc ax ; AX = 0100h mov ax,00FFh,

inc al ; AX = 0000h

18

(19)

Exercise...

Show the value of the destination operand after each of the following instructions executes:g

.data

myByte BYTE 0FFh, 0 .code

l B t AL FFh

mov al,myByte ; AL = mov ah,[myByte+1] ; AH =

dec ah ; AH =

FFh 00h FFh

dec ah ; AH

inc al ; AL =

dec ax ; AX =

FFh 00h FEFF

(20)

ADD and SUB Instructions

•ADD destination, source

• Logic: destination  destination + source

•SUB destination, source

• Logic: destination  destination – source

• Same operand rules as for the MOV instruction Same operand rules as for the MOV instruction

20

(21)

ADD and SUB Examples

data .data

var1 DWORD 10000h var2 DWORD 20000h

.code ; ---EAX---

mov eax,var1 ; 00010000h add eax,var2 ; 00030000h add ax,0FFFFh ; 0003FFFFh add eax 1 ; 00040000h add eax,1 ; 00040000h sub ax,1 ; 0004FFFFh

(22)

NEG (negate) Instruction

Reverses the sign of an operand. Operand can be a register or memory operand

.data

1

register or memory operand.

valB BYTE -1

valW WORD +32767 code

.code

mov al,valB ; AL = -1 neg alg ; AL = +1

neg valW ; valW = -32767

22

(23)

Implementing Arithmetic Expressions

HLL compilers translate mathematical expressions into assembly language. You can do it also. For example:

Rval DWORD ?

y g g p

Rval = -Xval + (Yval – Zval) Rval DWORD ?

Xval DWORD 26 Yval DWORD 30 Zval DWORD 40 .code

mov eax,Xval

neg eax ; EAX = -26 mov ebx Yval

mov ebx,Yval

sub ebx,Zval ; EBX = -10

(24)

Exercise ...

Translate the following expression into assembly language.

Do not permit Xval Yval or Zval to be modified:

Do not permit Xval, Yval, or Zval to be modified:

Rval = Xval - (-Yval + Zval)

mov ebx,Yval

Assume that all values are signed doublewords.

mov ebx,Yval neg ebx

add ebx,Zval mov eax,Xval sub eax,ebx

R l

mov Rval,eax

24

(25)

Flags Affected by Arithmetic

• The ALU has a number of status flags that

reflect the o tcome of arithmetic (and bit ise) reflect the outcome of arithmetic (and bitwise) operations

b d h f h d i i d

– based on the contents of the destination operand

• Essential flags:

– Zero flag – destination equals zero – Sign flag – destination is negative

– Carry flag – unsigned value out of range – Overflow flag – signed value out of range

• The MOV instruction never affects the flags.

(26)

Zero Flag (ZF)

Whenever the destination operand equals Zero, the Zero flag is set.

mov cx,1

sub cx 1 ; CX = 0 ZF = 1 g

sub cx,1 ; CX = 0, ZF = 1 mov ax,0FFFFh

inc ax c a ; AX = 0, ZF = 1; 0, inc ax ; AX = 1, ZF = 0

A flag is set when it equals 1.

A flag is clear when it equals 0.

26

(27)

Sign Flag (SF)

The Sign flag is set when the destination operand is negative The flag is clear when the destination is

0

negative. The flag is clear when the destination is positive.

mov cx,0

sub cx,1 ; CX = -1, SF = 1 add cx 2 ; CX = 1 SF = 0 add cx,2 ; CX = 1, SF = 0

h fl f h d h h b

The sign flag is a copy of the destination's highest bit:

mov al,0

sub al,1 ; AL=11111111b, SF=1

(28)

Carry Flag (CF)

• Addition and CF: copy carry out of MSB to CF

S b i d CF i d f

• Subtraction and CF: copy inverted carry out of MSB to CF

• INC/DEC do not affect CF

• Applying NEG to a nonzero operand sets CF pp y g p

28

(29)

Exercise . . .

For each of the following marked entries, show the values of the destination operand and the Sign, Zero,

mov ax 00FFh

p g , ,

and Carry flags:

mov ax,00FFh

add ax,1 ; AX= SF= ZF= CF=

sub ax,1 ; AX= SF= ZF= CF=

0100h 0 0 0 00FFh 0 0 0 add al,1 ; AL= SF= ZF= CF=

mov bh,6Ch

dd bh 95h BH SF ZF CF

00h 0 1 1

01h 0 0 1

add bh,95h ; BH= SF= ZF= CF=

mov al,2

01h 0 0 1 mov al,2

sub al,3 ; AL= SF= ZF= CF=FFh 1 0 1

(30)

Overflow Flag (OF)

The Overflow flag is set when the signed result of an operation is invalid or out of range

operation is invalid or out of range.

; Example 1 mov al +127 mov al,+127

add al,1 ; OF = 1, AL = ??

; Example 2

mov al,7Fh ; OF = 1, AL = 80h add al,1

The two examples are identical at the binary level The two examples are identical at the binary level because 7Fh equals +127. To determine the value of the destination operand, it is often easier to calculate

30

in hexadecimal.

(31)

A Rule of Thumb

• When adding two integers, remember that the Overflow flag is only set when

Overflow flag is only set when . . .

– Two positive operands are added and their sum is negative

negative

– Two negative operands are added and their sum is positive

positive

What will be the values of OF flag?

l 80h mov al,80h

add al,92h ; OF = mov al,-2

(32)

Signed/Unsigned Integers: Hardware Viewpoint

• All CPU instructions operate exactly the same on signed and unsigned integers

on signed and unsigned integers

• The CPU cannot distinguish between signed and i d i t

unsigned integers

• YOU, the programmer, are solely responsible for using the correct data type with each

instruction

32

(33)

Overflow/Carry Flags: Hardware Viewpoint

• How the ADD instruction modifies OF and CF:

CF ( t f th MSB) – CF = (carry out of the MSB)

– OF = (carry out of the MSB) XOR (carry into the MSB)

• How the SUB instruction modifies OF and CF:

NEG th d ADD it t th d ti ti – NEG the source and ADD it to the destination – CF = INVERT (carry out of the MSB)

OF ( f h MSB) XOR ( i h MSB) – OF = (carry out of the MSB) XOR (carry into the MSB)

(34)

Auxiliary Carry (AC) flag

• AC indicates a carry or borrow of bit 3 in the destination operand

destination operand.

• It is primarily used in binary coded decimal (BCD) ith ti

(BCD) arithmetic.

mov al, oFh

add al, 1 ; AC = 1 add al, 1 ; AC 1

34

(35)

Parity (PF) flag

• PF is set when LSB of the destination has an even number of 1 bits

even number of 1 bits.

mov al, 10001100b

add al, 00000010b ; AL=10001110, PF=1 , ,

sub al, 10000000b ; AL=00001110, PF=0

(36)

Jump and Loop

(37)

JMP and LOOP Instructions

• Transfer of control or branch instructions

– unconditional – conditional

• JMP Instruction

• LOOP LOOP Instruction Instruction

• LOOP Example

S i I t A

• Summing an Integer Array

• Copying a String

(38)

JMP Instruction

• JMP is an unconditional jump to a label that is usually within the same procedure.y p

• Syntax: JMP target

• Logic: EIP  target

• Example:

top:

. .

jmp top

38

(39)

LOOP Instruction

• The LOOP instruction creates a counting loop

• Syntax: LOOP target

• Syntax: LOOP target

• Logic:

ECX ECX 1

• ECX  ECX – 1

• if ECX != 0, jump to target

• Implementation:

• The assembler calculates the distance, in bytes, , y , between the current location and the offset of the target label. It is called the relative offset.

• The relative offset is added to EIP.

(40)

LOOP Example

The following loop calculates the sum of the integers 5 + 4 + 3 +2 + 1:tege s 5 3 :

00000000 66 B8 0000 mov ax,0

00000004 B9 00000005 5

offset machine code source code 00000004 B9 00000005 mov ecx,5 00000009 66 03 C1 L1:add ax,cx 00000009 66 03 C1 L1:add ax,cx 0000000C E2 FB loop L1

0000000E

When LOOP is assembled, the current location = 0000000E.

Looking at the LOOP machine code, we see that –5 (FBh) g , ( ) is added to the current location, causing a jump to

location 00000009:

00000009  0000000E + FB 40

(41)

Exercise . . .

If the relative offset is encoded in a single byte,

(a) what is the largest possible backward jump?

(b) what is the largest possible forward jump?

(a) 128 ( )

(b) +127

Average sizes of machine instructions are about 3 Average sizes of machine instructions are about 3 bytes, so a loop might contain, on average, a

(42)

Exercise . . .

mov ax,6 What will be the final value of AX? mov ecx,44

L1:

inc ax

10 inc ax

loop L1

How many times will the loop

execute? mov ecx,0

execute? X2:

inc ax loop X2 4,294,967,296

loop X2

42

(43)

Nested Loop

If you need to code a loop within a loop, you must save the outer loop counter's ECX value. In the following example, oute loop cou te s C value. t e ollow g e a ple, the outer loop executes 100 times, and the inner loop 20 times.

.data

count DWORD ? d

.code

mov ecx,100 ; set outer loop count L1:

mov count,ecx ; save outer loop count mov ecx,20 ; set inner loop count L2:

L2:...

loop L2 ; repeat the inner loop

(44)

Summing an Integer Array

The following code calculates the sum of an array of 16 bit integers

.data

16-bit integers.

intarray WORD 100h,200h,300h,400h .code

di OFFSET i t dd

mov edi,OFFSET intarray ; address

mov ecx,LENGTHOF intarray ; loop counter

mov ax 0 ; zero the sum

mov ax,0 ; zero the sum

L1:

add ax [edi] ; add an integer add ax,[edi] ; add an integer add edi,TYPE intarray ; point to next loop L1 ; repeat until ECX = 0

44

p ; p

(45)

Copying a String

good use of g SIZEOF

The following code copies a string from source to target.

.data

source BYTE "This is the source string",0 target BYTE SIZEOF source DUP(0) 0

target BYTE SIZEOF source DUP(0),0 .code

mov esi,0 ; index register mov ecx,SIZEOF source ; loop counter L1:

L1:

mov al,source[esi] ; get char from source mov target[esi],al ; store in the target inc esi ; move to next char

(46)

Conditional Processing

(47)

Status flags - review

• The Zero flag is set when the result of an operation equals zero.q

• The Carry flag is set when an instruction generates a result that is too large (or too small) for the

destination operand.

• The Sign flag is set if the destination operand is

negative, and it is clear if the destination operand is positive.

Th O fl fl i t h i t ti t

• The Overflow flag is set when an instruction generates an invalid signed result.

Less important:

• Less important:

– The Parity flag is set when an instruction generates an even number

(48)

NOT instruction

• Performs a bitwise Boolean NOT operation on a single destination operand

single destination operand

• Syntax: (no flag affected)

NOT destination NOT destination

• Example:

mov al 11110000b

NOT mov al, 11110000b

not al

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

NOT

inverted

1 1 0 0 0 1 0 0 inverted

48

(49)

AND instruction

• Performs a bitwise Boolean AND operation between each pair of matching bits in two between each pair of matching bits in two operands

S t (O 0 C 0 SZP)

• Syntax: (O=0,C=0,SZP)

AND destination, source AND

• Example:

mov al, 00111011b and al, 00001111b

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

AND

(50)

OR instruction

• Performs a bitwise Boolean OR operation between each pair of matching bits in two between each pair of matching bits in two operands

S t (O 0 C 0 SZP)

• Syntax: (O=0,C=0,SZP)

OR destination, source

OR

• Example:

mov dl, 00111011b or dl, 00001111b

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

OR

set unchanged

50

0 0 1 1 1 1 1 1 set

unchanged

(51)

XOR instruction

• Performs a bitwise Boolean exclusive-OR

operation between each pair of matching bits operation between each pair of matching bits in two operands

S t (O 0 C 0 SZP)

• Syntax: (O=0,C=0,SZP)

XOR destination, source

• Example:

XOR

mov dl, 00111011b

XOR

xor dl, 00001111b

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

XOR

(52)

Applications

(1 of 4)

• Task: Convert the character in AL to upper case.

• Solution: Use the AND instruction to clear bit 5.

mov al,'a' ; AL = 01100001b and al,11011111b ; AL = 01000001b

52

(53)

Applications

(2 of 4)

• Task: Convert a binary decimal byte into its equivalent ASCII decimal digit

equivalent ASCII decimal digit.

• Solution: Use the OR instruction to set bits 4 and 5.

mov al,6 ; AL = 00000110b or al,00110000b ; AL = 00110110b, ;

The ASCII digit '6' = 00110110b The ASCII digit 6 = 00110110b

(54)

Applications

(3 of 4)

• Task: Jump to a label if an integer is even.

• Solution: AND the lowest bit with a 1. If the result is Zero, the number was even.

mov ax,wordVal

and ax,1 ; low bit set?, ;

jz EvenValue ; jump if Zero flag set

54

(55)

Applications

(4 of 4)

• Task: Jump to a label if the value in AL is not zero.

• Solution: OR the byte with itself, then use the JNZ (jump if not zero) instruction.

or al,al

jnz IsNotZero ; jump if not zero

j ; j p

ORi g b ith it lf d t h g it l

ORing any number with itself does not change its value.

(56)

TEST instruction

• Performs a nondestructive AND operation between each pair of matching bits in two operands

pair of matching bits in two operands

• No operands are modified, but the flags are affected.

• Example: jump to a label if either bit 0 or bit 1 in AL is

• Example: jump to a label if either bit 0 or bit 1 in AL is set. test al,00000011b

jnz ValueFound jnz ValueFound

• Example: jump to a label if neither bit 0 nor bit 1 in

• Example: jump to a label if neither bit 0 nor bit 1 in AL is set.

test al 00000011b test al,00000011b jz ValueNotFound

56

(57)

CMP instruction

(1 of 3)

• Compares the destination operand to the source operandp

– Nondestructive subtraction of source from destination (destination operand is not changed)

• Syntax: (OSZCAP)

• Syntax: (OSZCAP)

CMP destination, source

• Example: destination == source

• Example: destination == source mov al,5

cmp al,5 ; Zero flag set

• Example: destination < source mov al,4

(58)

CMP instruction

(2 of 3)

• Example: destination > source mov al,6

cmp al,5 ; ZF = 0, CF = 0 (both the Zero and Carry flags are clear)

The comparisons shown so far were unsigned The comparisons shown so far were unsigned.

58

(59)

CMP instruction

(3 of 3)

The comparisons shown here are performed with

• Example: destination > source signed integers.

Example: destination source mov al,5

cmp al -2 ; Sign flag == Overflow flag cmp al, 2 ; Sign flag == Overflow flag

• Example: destination < source

• Example: destination < source mov al,-1

cmp al,5 ; Sign flag != Overflow flag c p a ,5 ; S g ag ! O e o ag

(60)

Conditions

unsigned ZF CF

destination<source 0 1 destination>source 0 0 destination>source 0 0

destination=source 1 0

signed flags

destination<source SF != OF

d ti ti SF OF

destination>source SF == OF destination=source ZF=1

60

(61)

Setting and clearing individual flags

and al, 0 ; set Zero

or al 1 ; clear Zero or al, 1 ; clear Zero or al, 80h ; set Sign

d l 7 h l i

and al, 7Fh ; clear Sign

stc ; set Carry

clc ; clear Carry

mov al, 7Fh

inc al ; set Overflow ;

(62)

Conditional jumps

Conditional jumps

(63)

Conditional structures

• There are no high-level logic structures such as if then else in the IA 32 instruction set But if-then-else, in the IA-32 instruction set. But, you can use combinations of comparisons and jumps to implement any logic structure

jumps to implement any logic structure.

• First, an operation such as CMP, AND or SUB is t d t difi d th CPU fl S d executed to modified the CPU flags. Second, a conditional jump instruction tests the flags and

h th ti fl di l

changes the execution flow accordingly.

CMP AL, 0 JZ L1

(64)

J cond instruction

• A conditional jump instruction branches to a label when specific register or flag conditions label when specific register or flag conditions are met

J d d ti ti Jcond destination

• Four groups: (some are the same) 1. based on specific flag values

2 based on equality between operands 2. based on equality between operands

3. based on comparisons of unsigned operands

4 b d i f i d d

4. based on comparisons of signed operands

64

(65)

Jumps based on specific flags

(66)

Jumps based on equality

66

(67)

Jumps based on unsigned comparisons

>≧<≦

(68)

Jumps based on signed comparisons

68

(69)

Examples

• Compare unsigned AX to BX, and copy the larger of the two into a variable named Large

mov Large,bx cmp ax,bx

t e two to a va able a ed a ge

p

jna Next

mov Large,ax N t

Next:

• Compare signed AX to BX, and copy the smaller of th t i t i bl d S ll

mov Small,ax b

the two into a variable named Small

cmp bx,ax

(70)

Examples

• Find the first even number in an array of unsigned integers

.date

i tA DWORD 7 9 3 4 6 1 integers

intArray DWORD 7,9,3,4,6,1 .code

...

mov ebx, OFFSET intArray mov ecx, LENGTHOF intArray L1: test DWORD PTR [ebx], 1

jz found

dd b 4

add ebx, 4 loop L1

...

70

...

(71)

BT (Bit Test) instruction

• Copies bit n from an operand into the Carry flag S t BT bitB

• Syntax: BT bitBase, n

– bitBase may be r/m16 or r/m32 – n may be r16, r32, or imm8

• Example: jump to label L1 if bit 9 is set in the Example: jump to label L1 if bit 9 is set in the AX register:

bt AX 9 ; CF = bit 9 bt AX,9 ; CF = bit 9

jc L1 ; jump if Carry

bit t t d l t

• BTC bitBase, n: bit test and complement

(72)

Conditional loops

Conditional loops

(73)

LOOPZ and LOOPE

• Syntax:

LOOPE destination LOOPZ destination

• Logic:

– ECX ECX – 1

– if ECX != 0 and ZF=1, jump to destination

• The destination label must be between -128 and +127 bytes from the location of the

following instruction

• Useful when scanning an array for the first

(74)

LOOPNZ and LOOPNE

• Syntax:

d i i

LOOPNZ destination LOOPNE destination

• Logic:

– ECX  ECX – 1; ECX  ECX 1;

– if ECX != 0 and ZF=0, jump to destination

74

(75)

LOOPNZ example

data

The following code finds the first positive value in an array:

.data

array SWORD -3,-6,-1,-10,10,30,40,4 sentinel SWORD 0

d .code

mov esi,OFFSET array mov ecx,LENGTHOF arrayy next:

test WORD PTR [esi],8000h ; test sign bit pushfd ; push flags on stack pushfd ; push flags on stack add esi,TYPE array

popfd ; pop flags from stack

l t ti l

loopnz next ; continue loop

jnz quit ; none found

(76)

Exercise ...

Locate the first nonzero value in the array. If none is found, let ESI point to the sentinel value:

.data

array SWORD 50 DUP(?)

, p

array SWORD 50 DUP(?) sentinel SWORD 0FFFFh .code

mov esi,OFFSET array mov ecx,LENGTHOF array

L1: cmp WORD PTR [esi],0 ; check for zero

76

quit:

(77)

Solution

.data

0

array SWORD 50 DUP(?) sentinel SWORD 0FFFFh

code .code

mov esi,OFFSET array mov ecx,LENGTHOF array

L1:cmp WORD PTR [esi],0 ; check for zero

pushfd ; push flags on stack add esi,TYPE array

popfd ; pop flags from stack loope L1 ; continue loop

loope L1 ; continue loop

jz quit ; none found

(78)

Conditional structures

Conditional structures

(79)

If statements

if then elseC T E

C

JNE else

C

T

JMP endif else:

E

E

(80)

Block-structured IF statements

Assembly language programmers can easily translate logical statements written in C++/Java into assembly logical statements written in C++/Java into assembly language. For example:

mov eax,op1 cmp eax,op2 if( op1 == op2 )

X = 1;

jne L1 mov X,1 j L2 X 1;

else

X = 2;

jmp L2 L1: mov X,2 L2:

;

L2:

80

(81)

Example

Implement the following pseudocode in assembly language. All values are unsigned:

cmp ebx ecx cmp ebx,ecx ja next

mov eax,5 if( ebx <= ecx )

{

5 ,

mov edx,6 next:

eax = 5;

edx = 6;

} }

(82)

Example

Implement the following pseudocode in assembly language. All values are 32-bit signed integers:

mov eax var1 if( var1 <= var2 ) mov eax,var1 cmp eax,var2 jle L1

if( var1 <= var2 ) var3 = 10;

else j

mov var3,6 mov var4,7 else

{

var3 = 6;

jmp L2

L1: mov var3,10 L2:

var3 6;

var4 = 7;

} L2:

}

82

(83)

Compound expression with AND

• When implementing the logical AND operator, consider that HLLs use short-circuit evaluation

• In the following example, if the first expression is false, the second expression is skipped:

if (al > bl) AND (bl > cl) X = 1;

(84)

Compound expression with AND

if (al > bl) AND (bl > cl) X = 1;

This is one possible implementation . . .

cmp al,bl ; first expression...

ja L1

p p

j

jmp next L1:

cmp bl,cl ; second expression...

ja L2 jmp next jmp next

L2: ; both are true

mov X,1 ; set X to 1

84

next:

(85)

Compound expression with AND

if (al > bl) AND (bl > cl)

But the following implementation uses 29% less code X = 1;

by reversing the first relational operator. We allow the program to "fall through" to the second expression:

cmp al,bl ; first expression...

jbe next ; quit if false

cmp bl,cl ; second expression...

jbe next ; quit if false mov X 1 both are true mov X,1 ; both are true next:

(86)

Exercise . . .

Implement the following pseudocode in assembly language. All values are unsigned:

cmp ebx ecx cmp ebx,ecx ja next

cmp ecx,edx if( ebx <= ecx

&& ecx > edx )

{ p ,

jbe next mov eax,5 {

eax = 5;

edx = 6; mov edx,6

next:

edx = 6;

}

(There are multiple correct solutions to this problem )

86

(There are multiple correct solutions to this problem.)

(87)

Compound Expression with OR

• In the following example, if the first expression is true, the second expression is skipped:

if (al > bl) OR (bl > cl) X = 1;

(88)

Compound Expression with OR

if (al > bl) OR (bl > cl) X = 1;

We can use "fall-through" logic to keep the code as short as possible:

cmp al,bl ; is AL > BL?

ja L1 ; yes

cmp bl,cl ; no: is BL > CL?

jbe next ; no: skip next statement

L1 X 1 t X t 1

L1:mov X,1 ; set X to 1 next:

88

(89)

WHILE Loops

A WHILE loop is really an IF statement followed by the body of the loop followed by an unconditional jump to body of the loop, followed by an unconditional jump to the top of the loop. Consider the following example:

while( eax < ebx) eax = eax + 1;

_while:

cmp eax,ebx ; check loop condition cmp eax,ebx ; check loop condition jae _endwhile ; false? exit loop

inc eax ; body of loop

jmp _while ; repeat the loop

(90)

Exercise . . .

while( ebx <= val1)

Implement the following loop, using unsigned 32-bit integers: while( ebx <= val1)

{

ebx = ebx + 5;

g

; val1 = val1 - 1 }

_while:

cmp ebx,val1 ; check loop condition cmp ebx,val1 ; check loop condition ja _endwhile ; false? exit loop

add ebx,5 ; body of loop dec val1

jmp while ; repeat the loop d hil

90

_endwhile:

(91)

Example: IF statement nested in a loop

while(eax < ebx) {

_while: cmp eax, ebx

j d hil

{

eax++;

if (ebx==ecx)

jae _endwhile inc eax

cmp ebx, ecx X=2;

else X=3;

jne _else mov X, 2 jmp while X=3;

}

jmp _while _else: mov X, 3

jmp _while _endwhile:

(92)

Table-driven selection

• Table-driven selection uses a table lookup to

l lti l ti t t

replace a multiway selection structure (switch-case statements in C)

• Create a table containing lookup values and the offsets of labels or procedures

• Use a loop to search the table

• Suited to a large number of comparisons Suited to a large number of comparisons

92

(93)

Table-driven selection

Step 1: create a table containing lookup values and procedure offsets:

.data

CaseTable BYTE 'A' ; lookup value

procedure offsets:

CaseTable BYTE 'A' ; lookup value

DWORD Process_A ; address of procedure EntrySize = ($ - CaseTable)

BYTE 'B' BYTE 'B'

DWORD Process_B BYTE 'C'

DWORD Process_C BYTE 'D'

DWORD Process D_

(94)

Table-driven selection

Step 2: Use a loop to search the table. When a match is found, we call the procedure offset stored in the current

mov ebx,OFFSET CaseTable ; point EBX to the table

, p

table entry:

mov ecx,NumberOfEntries ; loop counter

L1:cmp al,[ebx] ; match found?

jne L2 ; no: continue

call NEAR PTR [ebx + 1] ; yes: call the procedure

jmp L3 ; and exit the loop

j p ; p

L2:add ebx,EntrySize ; point to next entry

loop L1 ; repeat until ECX = 0

L3:

required for procedure

94

required for procedure pointers

(95)

Shift and rotate

Shift and rotate

(96)

Shift and Rotate Instructions

• Logical vs Arithmetic Shifts

• SHL Instruction

• SHR Instruction

• SAL and SAR Instructions

• ROL Instruction

• ROL Instruction

• ROR Instruction

• RCL and RCR Instructions

• SHLD/SHRD Instructions

96

(97)

Logical vs arithmetic shifts

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

position with zero:

0

CF

0

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

position with a copy of the number s sign bit:

(98)

SHL instruction

• The SHL (shift left) instruction performs a logical left shift on the destination operand logical left shift on the destination operand, filling the lowest bit with 0.

CF

0

CF

• Operand types:

SHL destination,count SHL reg,imm8

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

98

(99)

Fast multiplication

Shifting left 1 bit multiplies a number by 2 mov dl,5

shl dl,1

0 0 0 0 1 0 1 0

0 0 0 0 0 1 0 1 = 5

= 10

Before:

After:

, After: 0 0 0 0 1 0 1 0 = 10

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

mov dl,5

hl dl 2 20

For example, 5 * 22 = 20

shl dl,2 ; DL = 20

(100)

SHR instruction

• The SHR (shift right) instruction performs a logical right shift on the destination operand logical right shift on the destination operand.

The highest bit position is filled with a zero.

CF

0

CF

Shifting right n bits divides the operand by 2

n

mov dl,80

Shifting right n bits divides the operand by 2

shr dl,1 ; DL = 40

shr dl,2 ; DL = 10

100

(101)

SAL and SAR instructions

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

SAR ( hif i h i i h ) f i h

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

CF

• An arithmetic shift preserves the number's sign

• An arithmetic shift preserves the number s sign.

mov dl,-80

dl 1 40

sar dl,1 ; DL = -40

(102)

ROL instruction

• ROL (rotate) shifts each bit to the left

The highest bit is copied into both the Carry

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

No bits are lost

• No bits are lost

CF

mov al,11110000b

rol al,1 ; AL = 11100001b mov dl,3Fh

rol dl 4 ; DL = F3h

102

rol dl,4 ; DL = F3h

(103)

ROR instruction

• ROR (rotate right) shifts each bit to the right Th l bi i i d i b h h C fl

• 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

(104)

RCL instruction

• RCL (rotate carry left) shifts each bit to the left C i h C fl h l i ifi bi

• 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

104

(105)

RCR instruction

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

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 = 00010000 1

(106)

SHLD instruction

• Syntax: (shift left double)

SHLD d ti ti t

SHLD destination, source, count

• Shifts a destination operand a given number of bi h l f

bits to the left

• The bit positions opened up by the shift are

filled by the most significant bits of the source operand

• The source operand is not affected

106

(107)

SHLD example

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

bits with the high 4 bits of AX:

.data

wval WORD 9BA6h

d 9BA6 AC36

wval AX

Before:

.code

mov ax,0AC36h shld wval ax 4

BA6A AC36

After:

shld wval,ax,4

(108)

SHRD instruction

• Syntax:

SHRD d ti ti t

SHRD destination, source, count

• Shifts a destination operand a given number of bi h i h

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

108

(109)

SHRD example

Shift AX 4 bits to the right and replace its highest 4 bi i h h l 4 bi f DX

bits with the low 4 bits of DX:

DX AX

mov ax,234Bh mov dx,7654h

Before: 7654 234B

DX AX

,

shrd ax,dx,4 After: 7654 4234

(110)

Shift and rotate applications

• Shifting Multiple Doublewords

• Binary Multiplication

• Displaying Binary Bits p y g y

• Isolating a Bit String

110

(111)

Shifting multiple doublewords

• Programs sometimes need to shift all bits

within an array as one might when moving a within an array, as one might when moving a bitmapped graphic image from one screen location to another

location to another.

• The following shifts an array of 3 doublewords 1 bit to the right:

bit to the right:

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

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

(112)

Binary multiplication

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

F t bi b i t f 2

• Factor any binary number into powers of 2.

– For example, to multiply EAX * 36, factor 36 into 32 4 d th di t ib ti t f

+ 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

= (EAX * 32)+(EAX * 4) shl eax,5 shl ebx,2 add eax,ebx

112

add ea ,eb

(113)

Displaying binary bits

Algorithm: Shift MSB into the Carry flag; If CF = 1, append a "1" character to a string; otherwise

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

times.

mov ecx,32,

mov esi,offset buffer L1: shl eax,1

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

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

(114)

Isolating a bit string

• The MS-DOS file date field packs the year

(relative to 1980) month and day into 16 bits:

(relative to 1980), month, and day into 16 bits:

DH DL

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: 9 15 5 8 0 4

Bit numbers:

114

(115)

Isolating a bit string

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

and al,00011111b ; clear bits 5-7

mov day,al ; save in day variable mov ax,dx ; make a copy of DX

shr ax,5 ; shift right 5 bits and al 00001111b ; clear bits 4 7

and al,00001111b ; clear bits 4-7

mov month,al ; save in month variable mov al,dh ; make a copy of DX

shr al,1 ; shift right 1 bit mov ah,0 ; clear AH to 0

(116)

Multiplication and division

Multiplication and division

(117)

MUL instruction

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

Th i i f

• The instruction formats are:

MUL r/m8 MUL r/m16

MUL r/m32

Implied operands:

MUL r/m32

(118)

MUL examples

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

.data

val1 WORD 2000h al2 WORD 100h

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

val2 WORD 100h .code

mov ax,val1

half of the product

contains significant digits.

mov ax,val1

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

mov eax,12345h

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

mov eax,12345h mov ebx,1000h

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

118

(119)

IMUL instruction

• IMUL (signed integer multiply) multiplies an 8-, 16- or 32-bit signed operand by either AL AX 16 , or 32 bit signed operand by either AL, AX, or EAX (there are one/two/three operand

format))

• Preserves the sign of the product by sign- extending it into the upper half of the g pp destination register

Example: multiply 48 * 4, using 8-bit operands:p p y , g p mov al,48

mov bl,4,

imul bl ; AX = 00C0h, OF=1

(120)

DIV instruction

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

A i l d i li d ( i t

• A single operand is supplied (register or

memory operand), which is assumed to be the di i

divisor

• Instruction formats:

DIV r/m8 DIV r/m16

Default Operands:

DIV r/m32

120

(121)

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

(122)

Signed integer division

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

division takes place

– fill high byte/word/doubleword with a copy of the low byte/word/doubleword's sign bity g

• For example, the high byte contains a copy of the sign bit from the low byte: g y

1 0 0 0 1 1 1 1

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

122

(123)

CBW, CWD, CDQ instructions

• The CBW, CWD, and CDQ instructions

id i t t i t i

provide important sign-extension operations:

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

– CWD (convert word to doubleword) extends AX into DX CDQ (convert doubleword to quadword) extends EAX – CDQ (convert doubleword to quadword) extends EAX

into EDX

• For example: o e a ple:

mov eax,0FFFFFF9Bh ; -101 (32 bits) cdq ; EDX:EAX = FFFFFFFFFFFFFF9Bh

101 (64 bi )

; -101 (64 bits)

(124)

IDIV instruction

• IDIV (signed divide) performs signed integer di i i

division

• Uses same operands as DIV

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

124

(125)

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 b 5

mov ebx,5

(126)

Divide overflow

• Divide overflow happens when the quotient is too large to fit into the destination

too large to fit into the destination.

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

It causes a CPU interrupt and halts the

program. (divided by zero cause similar results)

126

(127)

Arithmetic expressions

Arithmetic expressions

(128)

Implementing arithmetic expressions

• Some good reasons to learn how to implement expressions:

expressions:

– Learn how compilers do it

T t d t di f MUL IMUL DIV d IDIV – Test your understanding of MUL, IMUL, DIV, and IDIV – Check for 32-bit overflow

Example: var4 = (var1 + var2) * var3 mov eax,var1,

add eax,var2 mul var3

jo TooBig ; check for overflow mov var4,eax ; save product

128

(129)

Implementing arithmetic expressions

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

mov eax,var1 neg eax

mul var2

jo TooBig ; check for overflow add eax,var3

Example: var4 = (var1 * 5) / (var2 – 3) mov eax,var1 o ea , a ; left side; e t s de

mov ebx,5

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

(130)

Implementing 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 ebx,edx ; EBX = right side mov eax,-5 ; begin left side

imul var1 u a ; EDX:EAX = left side; : e t s de idiv ebx ; final division

mov var4,eax ; quotient

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

130

expression first.

(131)

Exercise . . .

Implement the following expression using signed 32-bit integers:

eax = (ebx * 20) / ecx

mov eax,20 mul ebx

div ecx

(132)

Exercise . . .

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

eax = (ecx * edx) / eax

push ecx h d push edx

push eax ; EAX needed later mov eax,ecx

mov eax,ecx

mul edx ; left side: EDX:EAX pop ecx ; saved value of EAX div ecx ; EAX = quotient

pop edx ; restore EDX, ECX

132

pop ecx

(133)

Exercise . . .

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

integers. Do not modify any variables other than var3:

var3 = (var1 * -var2) / (var3 – ebx) mov eax,var1

mov edx var2 mov edx,var2 neg edx

mul edx ; left side: edx:eax mov ecx,var3

sub ecx,ebx

di i

div ecx ; eax = quotient

(134)

Extended addition and subtraction

Extended addition and subtraction

參考文獻

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