Integer Arithmetic-computer organization & assembly languages

Integer Arithmetic

Computer Organization and Assembly

Languages

Yung-Yu Chuang

2005/11/17

Announcements

• Assignment #2 is due next week.

• Assignment #3 box filter is online now, due on

12/4.

What is an image

• We can think of an image as a function, f: R

2

R:

– f(r, c) gives the intensity at position (r, c) – defined over a rectangle, with a finite range:

• f: [0,h-1]x[0,w-1]  [0,255]

c r

A digital image

• The image can be represented as a matrix of

integer values

110 110 100 100 100 100 100 100 100 100 120 130 100 100 100 100 100 100 100 100 110 100 100 100 130 110 120 110 100 100 100 100 100 110 90 100 90 100 100 110 130 100 100 130 100 90 130 110 120 100 100 100 100 120 100 130 110 120 110 100 100 100 100 90 110 80 120 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100

c

r

Assignment #3 Box Filter

unsigned int c_blur(unsigned char *img_in,

unsigned char *img_out, int knl_size, int w, int h) {

for each row r

for each column c calculate k(r, c) save it

}

• Memory layout

• Only integer arithmetic operations

• MD5/CRC32 checksum

Assignment #3 Box Filter

for (int i = 0; i < height; i++) { for (int j = 0; j < width; j++) { pixel = 0; pixel_num = 0;

for (int y=-knl_size/2; y<=knl_size/2; y++) { for (int x=-knl_size/2; x<=knl_size/2; x++) { int x_s = j + x;

int y_s = i + y;

/* make sure that it's in the image */

if (x_s>=0 && x_s<w && y_s>=0 && y_s<h) { pixel += img_in[y_s * w + x_s];

pixel_num++; }

} }

pixel = pixel / pixel_num;

img_out[i * width + j] = (unsigned char)pixel; }

Chapter 7 Integer Arithmetic Overview

• Shift and Rotate Instructions

• Shift and Rotate Applications

• Multiplication and Division Instructions

• Extended Addition and Subtraction

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:

• An arithmetic shift fills the newly created bit

position with a copy of the number’s sign bit:

SHL instruction

• The SHL (shift left) instruction performs a

logical left shift on the destination operand,

filling the lowest bit with 0.

• Operand types: SHL destination,count

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

Fast multiplication

mov dl,5 shl dl,1

Shifting left 1 bit multiplies a number by 2

mov dl,5

shl dl,2 ; DL = 20

Shifting left n bits multiplies the operand by 2n

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.

mov dl,80

shr dl,1 ; DL = 40

shr dl,2 ; DL = 10

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.

An arithmetic shift preserves the number's

sign.

mov dl,-80

sar dl,1 ; DL = -40

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

mov al,11110000b

rol al,1 ; AL = 11100001b

mov dl,3Fh

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

mov al,11110000b

ror al,1 ; AL = 01111000b

mov dl,3Fh

Your turn . . .

mov al,6Bh shr al,1 a. shl al,3 b. mov al,8Ch sar al,1 c. sar al,3 d.

Indicate the hexadecimal value of AL after

each shift:

35h A8h C6h F8h

Your turn . . .

mov al,6Bh

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

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

stc ; CF = 1

mov ah,10h ; CF,AH = 00010000 1

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

• Syntax:

SHLD destination, source, count

• 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

SHLD example

.data

wval WORD 9BA6h .code

mov ax,0AC36h shld wval,ax,4

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

Before: After:

SHRD instruction

• Syntax:

SHRD destination, source, count

• 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

SHRD example

mov ax,234Bh mov dx,7654h shrd ax,dx,4

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

Before: After:

Your turn . . .

mov ax,7C36h mov dx,9FA6h

shld dx,ax,4 ; DX =

shrd dx,ax,8 ; DX =

Indicate the hexadecimal values of each

destination operand:

FA67h 36FAh

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 a

n array, as one might when moving a bitmapped gr

aphic image from one screen location to another.

• The following shifts an array of 3 doublewords 1 bi

t to the right:

mov esi,0

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

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

Binary multiplication

• We already know that SHL performs unsigned

multiplication efficiently when the multiplier is

a power of 2.

• Factor any binary number into powers of 2.

– 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 shl ebx,2 add eax,ebx

Your turn . . .

mov ax,2 ; test value

mov dx,ax

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

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

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.

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

Isolating a bit string

• The MS-DOS file date field packs the year

Isolating a bit string

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

add ax,1980 ; year is relative to 198 0

mov year,ax ; save in year

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

mov month,al ; save in month variable mov al,dl ; make a copy of DL

and al,00011111b ; clear bits 5-7

Multiplication and division instructions

• MUL Instruction

• IMUL Instruction

• DIV Instruction

• Signed Integer Division

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

MUL examples

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

.data

val1 WORD 2000h val2 WORD 100h .code

mov ax,val1

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

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

contains significant digits.

mov eax,12345h mov ebx,1000h

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

Your turn . . .

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

What will be the hexadecimal values of (E)DX, (E)AX, and the Carry flag after the following instructions

execute? DX = 0012h, AX = 3400h, CF = 1 mov eax,00128765h mov ecx,10000h mul ecx 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 register

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

mov al,48 mov bl,4

imul bl ; AX = 00C0h, 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

• 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

Your turn . . .

mov dx,0087h mov ax,6000h mov bx,100h div bx

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

Signed integer division

• Signed integers must be sign-extended before

division takes place

– fill high byte/word/doubleword with a copy of the l ow byte/word/doubleword's sign bit

• For example, the high byte contains a copy of

the sign bit from the low byte:

CBW, CWD, CDQ instructions

• The CBW, CWD, and CDQ instructions pr

ovide important sign-extension operatio

ns:

– 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 in

to EDX

• For example:

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

IDIV instruction

• IDIV (signed divide) performs signed integer division • Uses same operands as DIV

Example: 8-bit division of –48 by 5

mov al,-48

cbw ; extend AL into AH mov bl,5

IDIV examples

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

Example: 16-bit division of –48 by 5

mov ax,-48

cwd ; extend AX into DX

mov bx,5

Your turn . . .

mov ax,0FDFFh ; -513

cwd

mov bx,100h idiv bx

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

Divide overflow

• Divide overflow happens when the quotient is t

oo large to fit into the destination.

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

It causes a CPU interrupt and halts the program.

(divided by zero cause similar results)

Implementing arithmetic expressions

• Some good reasons to learn how to implement

expressions:

– Learn how compilers do it

– 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

Implementing arithmetic expressions

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

mov eax,var1 neg eax

mul var2

jo TooBig ; check for overflow add eax,var3

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

mov eax,var1 ; left side mov ebx,5

mul ebx ; EDX:EAX = product

mov ebx,var2 ; right side sub ebx,3

div ebx ; final division

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 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 mul ebx

div ecx

Implement the following expression using sign

ed 32-bit integers:

Your turn . . .

push ecx push edx

push eax ; EAX needed later

mov eax,ecx

mul edx ; left side: EDX:EAX

pop ecx ; saved value of EAX

div ecx ; EAX = quotient

pop edx ; restore EDX, ECX

pop ecx

Implement the following expression using signed

32-bit integers. Save and restore ECX and EDX:

Your turn . . .

mov eax,var1 mov edx,var2 neg edx

mul edx ; left side: edx:eax

mov ecx,var3 sub ecx,ebx

div ecx ; eax = quotient

mov var3,eax

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

Extended addition and subtraction

• ADC Instruction

• Extended Addition Example

• SBB Instruction

ADC instruction

• ADC (add with carry) instruction adds both a

source operand and the contents of the Carry

flag to a destination operand.

• Example: Add two 32-bit integers (FFFFFFFFh

+ FFFFFFFFh), producing a 64-bit sum:

mov edx,0

mov eax,0FFFFFFFFh add eax,0FFFFFFFFh

Extended addition example

• Add two integers of any size

• Pass pointers to the addends and sum • ECX indicates the number of words

L1:

mov eax,[esi] ; get the first integer adc eax,[edi] ; add the second integer pushfd ; save the Carry flag

mov [ebx],eax ; store partial sum

add esi,4 ; advance all 3 pointers add edi,4

add ebx,4

popfd ; restore the Carry flag loop L1 ; repeat the loop

Extended addition example

.data

op1 QWORD 0A2B2A40674981234h op2 QWORD 08010870000234502h sum DWORD 3 dup(?)

; = 0000000122C32B0674BB5736 .code

...

mov esi,OFFSET op1 ; first operand mov edi,OFFSET op2 ; second operand mov ebx,OFFSET sum ; sum operand

mov ecx,2 ; number of doublewords call Extended_Add

SBB instruction

• The SBB (subtract with borrow) instruction subtracts bo th a source operand and the value of the Carry flag fro m a destination operand.

• The following example code performs 64-bit subtraction . It sets EDX:EAX to 0000000100000000h and subtracts 1 from this value. The lower 32 bits are subtracted first, s etting the Carry flag. Then the upper 32 bits are subtra cted, including the Carry flag:

mov edx,1 ; upper half

mov eax,0 ; lower half

sub eax,1 ; subtract 1