Assembly 05

Arithmetic Flags and Instructions

Status Flags• Zero flag (ZF)

Indicate that the execution of the last instruction that affects the zero flag has a zero result, ZF = 1 for zero result and ZF = 0 otherwise.

• Carry flag (CF) Indicates that the result of an arithmetic operation on unsigned numbers is out of range of the destination (register or memory).

• Overflow flag (OF) Indicates whether an operation on signed numbers has produced a result that is out of range.

• Sign flag (SF) Indicates the sign of the result of an operation, 0 for positive numbers and 1 for negative numbers.

• Auxiliary flag (AF) Indicates whether an operation has produced a result that has generated a carry out of or a borrow into the low-order four bits

• Parity flag (PF).Indicates the parity of the 8-bit result produced by an operation; The parity flag is set if the byte contains an even number of 1 bits and cleared if there are an odd number of 1bits

Zero FlagSet the ZF

mov AL,0FHadd AL,0F1H

Set the ZFmov EAX,1dec EAX

Example sub EAX, EAXmov EDX,M

L1:mov ECX, N

L2:inc EAXloop L2dec EDXjnz L1

Exit:mov sum, EAX

Carry Flag• 100H require 9 bit to store

• Rang of values

• The CF is set as the result is too small for unsigned number

• Inc and Dec instructions does not affect the CF• Other Usage

– To propagate carry or borrow in multiword addition or subtraction operations.

– To detect overflow/underflow conditions.– To test a bit using the shift/rotate family of

instructions.

Overflow Flag• Range of values

• Sets the OFmov AL, 72H ; 72H = 114D

add AL, 0EH ; 0EH = 14D

• and thismov AX, 0FFFBH ; AX = -5

sub AX, 7FFDH ; subtract 32,765 from AX

• The overflow flag is also affected by – shift, multiply, and divide operations.

Usage

• The main purpose of the overflow flag is to indicate whether an arithmetic operation on signed numbers has produced an out-of-range result.

The Sign Flag• Clear SF (SF = 0 ), 112D is positive

mov EAX,15

add EAX,97

• Set SF (SF = 1), (15 – 97) is negative

mov EAX,15

sub EAX,97

Sign and Unsigned Q: How does the system know how your program is interpreting a given bit pattern? A: The answer is that the processor does not have a clue.• It is up to our program logic to interpret a given bit pattern correctly.• The processor assumes both interpretations and sets the carry and overflow flags.

Examplemov AL,72Hadd AL,0EH

• Unsigned Number (CF = 0)The result 80H (128) fits 8-bit unsigned

• Signed Number (OF = 1)The results 80H (128) is outside the 8-bit signed

• Thus both flags are set and it is up to our program logic to consider the right flag. – If unsigned numbers are represented,

disregard the OF as use the CF.

Example

mul & imul instructions

mov AL,10mov DL,26mul DL

mov DL, 25 ; DL = 25mov AL, 0F6H ; AL = -10imul DL

div & idiv instructionsmov AX,251mov CL,12div CL

xor DX, DX ; clear DXmov AX, 141BH ; AX = 5147Dmov CX, 012CH ; CX = 300Ddiv CX

Code Example; displays a signed 8-bit integer that is in AL register.;-----------------------------------------------------------PutInt8:

enter 3,0 ; reserves 3 bytes of buffer spacepush AXpush BXpush ESItest AL,80H ; negative number?jz positive

negative:PutCh ’-’ ; sign for negative numbersneg AL ; convert to magnitude

positive:mov BL,10 ; divisor = 10sub ESI,ESI ; ESI = 0 (ESI points to buffer)

repeat1:sub AH,AH ; AH = 0 (AX is the dividend)div BL

Code Example; AX/BL leaves AL = quotient & AH = remainderadd AH,’0’ ; convert remainder to ASCIImov [EBP+ESI-3], AH ; copy into the bufferinc ESIcmp AL,0 ; quotient = zero?jne repeat1 ; if so, display the number

display_digit:dec ESImov AL,[EBP+ESI-3]; display digit pointed by ESIPutCh ALjnz display_digit ; if ESI<0, done displaying

display_done:pop ESI ; restore registerspop BXpop AXleave ; clears local buffer spaceret