Flag Control instructions

CLC clear carry flag CF = 0STC set carry flag CF= 1CMC complement carry flag [CF] CF

ADD action

Add ADD D,S [S] + [D]] [D] [carry] [CF] flags affected: OF, SF, ZF, AF, PF, CF

ADC action

Add with carry ADC D,S [S] + [D] + [CF] [D] [carry] [CF] flags affected: OF, SF, ZF, AF, PF, CF

Destination Source Register Register Register Memory Memory Register Register Immediate Memory Immediate Accumulator Immediate

DH

Format INC D Action [D+1] → [D]

DR16

R8

M not in the debugger environment !

Flags affectedSF, ZF ,AF ,PF

ADD immediate to

accumulator0000010W Data Data if W=1

00000100 00100101

04 25

ADD AL,25H

ADD AL, BL

ADD Reg/Mem with Reg

000000DW mod reg r/m Displ-lo

00000010 11000011

02 C3

Displ-hi

00000000 1101100000 D8

Only 1 binary coded decimal digit is placed in 1 byte

2 binary coded decimal digits are packed in 1 byte

Packed BCD

5610 0101 0110

Unpacked BCD

5610 0000 0101 00000110

Assume that AL contains 32H which is the ASCII code for 210 and BL contains 34H which is the ASCII code for 410 and AH has been cleared.

What is the result of executing the following instructions?

ADD AL, BL

AAA

AAA = ASCII adjust after addition 00110110

36

Note that the CF and AF remain cleared

AX = 66

AX = 6

machine code

Assume that AL contains 35H which is the ASCII code for 510 and BL contains 37H which is the ASCII code for 710 and AH has been cleared.

What is the result of executing the following instructions?

ADD AL, BL

AAA

AX = 6C

AX = 01 02Note that CF and AF are set

(Unpacked BCD)

Repeat assuming that AL contains 06H and BL contains 04H and AH has been cleared.

What is the result of executing the following instructions?

ADD AL, BL

DAA

DAA = decimal adjust after addition 00100111

27

Assume that AL contains 08H and BL contains 04H and AH has been cleared.

What is the result of executing the following instructions?

ADD AL, BL

DAA

0C0012 (packed BCD)

0A

0010 (packed BCD)

SUB register/memory with register to either

001010DW

00101011 00000101

2B 05

SUB AX, [DI]

MOV register/memory to/from register

100010DW mod reg r/m Displ-lo

10001001 00000100

89 04

Displ-hi

MOV [SI], AX

mod reg r/m Displ-lo Displ-hi

MOV immediate to

register1011W reg Data Data if W=1

10111110 00000000

BE 00

MOV SI, 200H

MOV DI, 100H

MOV register/memory to/from register

100010DW mod reg r/m Displ-lo

10001011 00000100

8B 04

Displ-hi

00000010

02

10111111 00000000 00000001

BF 00 01

MOV AX,[SI]

MOV AX, [DI] 10001011 000001018B 05

INC register 01000 reg

INC SI

01000110

46

INC DI

01000111

47

SBB register/memory and register to either

000110DW mod reg r/m Displ-lo

00011011 00000101

1B 05

Displ-hi

SBB AX,[DI]

MOV DS,AXMOV AX, 1000

BE 00 02

BF 00 01

8B 04

2B 05

89 04

46

46

8B 04

47

47

1B 05

89 04

EDS:200 78 56 34 12

EDS:100 89 67 45 23

DDS:200 EF EE EE EE

Mov si,0200

Mov di,0100

Mov ax,[si]

Sub ax,[di]

Mov [si],ax

Inc si

Inc si

Mov ax,[si]

Inc di

Inc di

Sbb ax,[di]

Mov [si],ax

Negate instruction:

MOV AX,0060

NEG AX

0000 0000 0110 0000

1111 1111 1010 0000 2’complement

F F A 0

B328 B083

001010DWSubtract m/reg and reg to either Mod reg r/m Disp-lo Disp-hi

00101010 11000011

2A C3

DAS decimal adjust after subtraction 00101111

2F

B83800

Subtract immediate from accumulator

0010110W data data if W=1

2C 39

AAS ASCII adjust for subtraction

00111111

3F

If AL=-1 and CL=-2

MUL CL

If AL=-1 and CL=-2

IMUL CL

AL=FF CL= FE

Unsigned multiplication

Result AX= FD02

Result AX= 2

Signed multiplication

CF is set because there is a carry from the LSByteVF is set because the result cannot fit in 1 Byte

MOV BX,0009MOV AX,0007MUL BLAAM

BCD Arithmetics

AX=003F

MOV AX, 0603AAD AX=003F

Signed COMPARE example

MOV AX, 99 AX=1001 1001 (-10310)MOV BX, 1B BX=0001 1011 (+2710)CMP AX, BX 2’scomplement [1001 1001] = 0110 01112 = 6716 = - 10310

1B16 = 2710

-103 – 27 = -13010 = 2’s complement [ 2816] = 2’s complement [ 0010 1000] = 1101 10002 = D816 1001 1001 - 0001 1011 0111 1110 AF=1 CF=0 OF=0