microprocessor pdf - م.م.أنمار خليل

ﺤﺎﺴﺒﺎﺕ ﻤﺘﻘﺩﻤﺔ ﺍﻟﻤﺭﺤﻠﺔ ﺍﻟﺜﺎﻟﺜﺔ ﻫﻨﺩﺴﺔ ﻜﻬﺭﻭ ﻤﻴﻜﺎﻨﻴﻙ ﺍﻟﻤﺩﺭﺱ ﻤﺴﺎﻋﺩ ﺍﻨﻤﺎﺭ ﺨﻠﻴل ﺍﺒﺭﺍﻫﻴﻡ

٨٤

ARITHMETIC INSTRUCTIONS

The 8086 can perform arithmetic operations on binary numbers (signed or

unsigned) and on decimal numbers (packed or unpacked). As Table 1 shows,

there

are

instructions

for

the

four

standard

arithmetic

functions-addition,

subtraction, multiplication, and division. There are also instructions that "sign-

extend" operands; these let you combine data of different sizes (e.g., add a byte

to a word).

ﺔ!!!ﯿﺋﺎﻨﺜﻟا مﺎ!!!ﻗرﻻا ﻰ!!!ﻠﻋ تﺎ!!!ﯿﻠﻤﻌﺑ مﻮ!!!ﻘﯾ نا ﺞﻟﺎ!!!ﻌﻤﻠﻟ ﻦ!!!ﻜﻤﯾ

)

ﺎرة!!!ﺷا نوﺪ!!!ﺑ ﻢ!!!ﻗر ، ﺎرة!!!ﺷﺎﺑ ﻢ!!!ﻗر

(

ﺮﯾﺔ!!!ﺸﻌﻟا مﺎ!!!ﻗرﻻاو

(packed or unpacked)

!!!ﺬﻟﻚ اﻻﯾﻌ!!!ﻛو ﺔ!!!ﻌﺑرﻻا تﺎ!!!ﯿﻠﻤﻌﻟا لوﺪ!!!ﺠﻟا ﻦﯿ!!!ﺒﯾو

ﺎ

ﺬا!!!ھو ةرﺎ!!!ﺷﻻا ﻊ!!!ﺳﻮﯾ يﺬ!!!ﻟا تاز

ﯾﺴﻤﺢ ﺑﻀﻢ ﺑﯿﺎﻧﺎت ﺑﺎﺣﺠﺎم ﻣﺨﺘﻠﻔﺔ

٩٤

1-Addition Instructions

1. Add (ADD) and Add With Carry (ADC)

The ADD and ADC instructions can add 8- or 16-bit operands. ADD adds a source and

destination operand and puts the result in the destination. Symbolically, we can represent

this as

ﯾﻤﻜﻦ ﻻﯾﻌﺎزي

ADD and ADC

ان ﺗﺠﻤﻊ ﻣﻌﻠﻮﻣﺎت ﺑ

ﻄﻮل

8 or 16 bit

وﯾﺘﻢ اﺿﺎﻓﺔ اﻟﻤﺼﺪر اﻟﻰ اﻟﻤﺴﺘﻘﺮ وﯾﺨﺰن

اﻟﻨﺎﺗ

ﺞ ﻓﻲ اﻟﻤﺴﺘﻘﺮ

destination = destination + source

ADC does the same thing as ADD, except it includes the Carry Flag (CF) in the addition,

like this:

وﯾﺸﺎﺑﮫ اﯾﻌﺎز

ADC

اﯾﻌﺎز

ADD

ﺑﺎﺳﺘﺜﻨﺎء اﻧﮫ ﯾﺠﻤﻊ اﯾﻀﺎ

ﻗﯿﻤﺔ

Carry Flag (CF)

destination = destination + source + carry

Now you can see why the 8086 has two separate add instructions. One

instruction, ADD, can add single-byte or single-word numbers and the low-order

terms of 32-bits numbers. The other, ADC, is used to add the higher-order terms

of two 32-bits numbers.

واﻻن ﻧﺴﺘﻄﯿﻊ ان اﻟﻤﻌﺎﻟﺞ

ﻟﮫ اﯾﻌﺎزي ﺟﻤﻊ ﻣﻨﻔﺼﻠﯿﻦ ﻋﻦ ﺑﻌﻀﮭﺎ اﻟﺒﻌﺾ

اﻻول ھﻮ اﯾﻌﺎز

ADD

واﻟﺬي ﺑﺎﻣﻜﺎﻧﮫ ان

ﯾﺠﻤﻊ ﺑﺎﯾﺖ واﺣﺪ او ﻛﻠﻤﺔ واﺣﺪة او اﻟﺠﺰء اﻻدﻧﻰ ﻣﻦ اﻟﺮﻗﻢ

32-bits

وﯾﺴﺘﺨﺪم اﯾﻌﺎز

ADC

ﻟﺠﻤﻊ اﻟﺠﺰء اﻻ

ﻋﻠﻰ ﻣﻦ

اﻟﺮﻗﻢ

32-bits

For example:

ADD AX , CX

adds the 16-bit contents of AX and CX and returns the result in AX. If your operands are

longer than 16 bits, you can use this kind of sequence:

اﻻﯾﻌﺎز اﻟﺴﺎﺑﻖ ﯾﺠﻤﻊ ﻣﺤﺘﻮى

16 bit

ﻟﻤﺴﺠﻠﻲ

AX and CX

واﻟﻨﺎﺗﺞ ﯾ

ﺨﺰن ﻓﻲ ﻣﺴﺠﻞ

واذاﻛﺎن ﻃﻮل اﻟﺮﻗﻤﯿﻦ

اﻃﻮل ﻣﻦ

16-bit

ﻓﺎﻧﻨﺎ ﻧﺴﺘﺨﺪم ھﺬا اﻟﻨﻮع ﻣﻦ ﺗﺴﻠﺴﻞ اﻻﯾﻌﺎزات

٠٥

ADD AX,CX ;Add low – order 16 bits,

ADD BX,DX ;then high – order 16 bits,

which adds the 32-bit number in CX and DX to the one in AX and BX. Here, the ADC

instruction includes any carry out of (CX) + (AX) into (DX)+ (BX).

وﺗﺴﻠﺴﻞ اﻻﯾﻌﺎزات ﯾﻘﻮم ﺑﺠﻤﻊ رﻗﻤﯿﻦ ﺑﻄﻮل

32-bit

ﺣﯿﺚ ﯾﻜﻮن

اﻟﺮﻗﻢ

اﻻول ﻓﻲ اﻟﻤﺴﺠﻠﯿﻦ

CX and DX

واﻟﺮﻗﻢ

اﻟﺜﺎﻧﻲ ﻓﻲ اﻟﻤﺴﺠﻠﯿﻦ

AX and BX

وﯾﻘﻮم اﻻﯾﻌﺎز

ADC

ﺑﺎﺧﺬ ﺑﻨﻈﺮ اﻻﻋﺘﺒﺎر

carry out

ان ﺣﺪث ﻣﻦ ﻋﻤﻠﯿﺔ ا

ﻟﺠﻤﻊ

ﺑﯿﻦ اﻟﻤﺴﺠﻠﯿﻦ

(DX)+ (BX)

اﻟﻰ ﻋﻤﻠﯿﺔ اﻟﺠﻤﻊ ﺑﯿﻦ اﻟﻤﺴﺠﻠﯿﻦ

(CX) + (AX)

You may also add a memory operand to a general register (or vice versa) or add an

immediate value to a register or to memory. Some examples are:

وﯾﻤﻜﻦ ان ﯾﺠﻤﻊ ﻣﻮﻗﻊ ذاﻛﺮة ﻣﻊ ﻣﺴﺠﻞ ﻋﺎم و

اﻟﻌﻜﺲ ﺑﺎﻟﻌﻜﺲ او اﺿﺎﻓﺔ اﻟﻘﯿﻤﺔ اﻟﻤﺒﺎﺷﺮة اﻟﻰ ﻣﺴﺠﻞ او ﻣﻮﻗﻊ اﻟﺬاﻛﺮة وھﺬه

ﺑﻌﺾ اﻻﻣﺜﻠﺔ ﻋﻠﻰ ھﺬا اﻟﻤﻮﺿﻮع

ADD AX , MEM_WORD; Add a memory operand to a register

ADD MEM_WORD , AX ; or vice versa

ADD AL , 10 ; Add a constant to a register

ADD MEM_BYTE , 0FH ; or to a memory location

Note that most combinations are legal, but you may not add memory-to memory, nor may

you use an immediate value as a destination.

وﯾﺠﺐ ان ﻧﻼﺣﻆ اﻧﮫ ﻻﯾﻤﻜﻦ ان ﻧﺠﻤﻊ ﻣﻮﻗﻌﯿﻦ ذاﻛﺮة ﻣﺒﺎﺷﺮة او اﺳﺘﺨﺪام اﻟﻘﯿﻤﺔ اﻟﻤﺒﺎﺷﺮة ﻛﻤﺴﺘﻘﺮ

2.(AAA and DAA) Instructions Adjust for Addition

The 8086 always adds numbers as if they were binary. What happens if they are

binary-coded decimal (BCD) numbers? Let's find out by looking at an examples.

ان ﻣﻌﺎﻟﺞ

8086

ﯾﻘﻮم ﺑﺠﻤﻊ اﻻرﻗﺎم ﻛﺎرﻗﺎم ﺛﻨﺎﺋﯿﺔ وﻟﻜﻦ

اذا ﻛﺎن اﻻرﻗﺎ

م ﻣﻦ ﻧﻮﻋﯿﺔ

BCD

ﻓﻜﯿﻒ ﯾﺘﻢ اﻟﺘﻌﺎﻣﻞ ﻣﻊ ھﺬا

اﻟﻤﻮﺿﻮع ﻓﻠﻨﺮى ﺑﻌﺾ اﻻﻣﺜﻠﺔ

١٥

1. Unpacked BCD number and AAA Instruction

If you add the unpacked BCD numbers 26 and 42, the 8086 performs this Hexadecimal

addition:

اذا ﺗﻢ اﺿﺎﻓﺔ رﻗﻤﯿﻦ

unpacked BCD

ھﻤﺎ

26 and 42

ﻓﺎن اﻟﻤ

ﻌﺎﻟﺞ ﺳﯿﻘﻮم ﺑﺎﺟﺮاء ﻋﻤﻠﯿﺔ اﻟﺠﻤﻊ ﺑﺎﻟﻨﻈﺎم اﻟﺴﺎدس

ﻋﺸﺮي وﻛﺎﻻﺗﻲ

42 +

68 hexadecimal

08 BCD

If we use an AAA instruction after the addition operation will clear the 4 bit in the left

of register AL and leave the 4 bit in the right of AL and have no effect on AH and no

effect on flags because the Low digit is less the number 9.

اذا ﺗﻢ اﺳﺘﺨﺪام اﯾﻌﺎز

AAA

ﻟﺘﺤﻮﯾﻞ ﻧﺎﺗﺞ اﻟﺠﻤﻊ ﻣﻦ اﻟﻨﻈﺎم اﻟﺴﺎدس اﻟﻌﺸﺮي اﻟﻰ ﻧﻈﺎم

BCD

ﻓﺎن اﯾﻌﺎز

AAA

ﯾﻘﻮم ﺑﺘﺼﻔﯿﺮ اﻻرﺑﻌﺔ اﻟﻰ اﻟﯿﺴﺎر

High DIGIT

ﻣﻦ اﻟﻤﺴﺠﻞ

وﯾﺘﺮك اﻻرﺑﻌﺔ اﻟﻰ اﻟ

ﯿﻤﯿﻦ

LOW DIGIT

ﺑﺪون ﺗﻐﯿﯿﺮ وﻟﯿﺲ ھﻨﺎك ﺗﺎﺛﯿﺮ ﻋﻠﻰ

وﻻ ﻋﻠﻰ

FLAGS

ﻻن اﻟﺮﻗﻢ اﻟﻨﺎﺗﺞ ﻓﻲ

LOW DIGIT

اﻗﻞ ﻣﻦ اﻟﺘﺴﻌﺔ

This instruction is deal with only 4 bit in the right of AL and only it deal with AL

clear the digit in the left and add 1 to the AH and set 1 to AF and CF and check the

value of PF and SF.

وھﺬا ﯾﻌﻨﻲ ان اﻻﯾﻌﺎز ﻓﻘﻂ ﯾﺘﻌﺎﻣﻞ ﻣﻊ اﻻرﺑﻌﺔ ﺑﺘﺎت ﻓﻲ اﻟﯿﻤﯿﻦ ﻣﻦ اﻟﻤﺴﺠﻞ

وﻓﻘﻂ اﻟﻤﺴﺠﻞ

وﻋﻨﺪﻣﺎ ﯾﺘﺠﺎوز

اﻟﻨﺎﺗﺞ ﻓﻲ

LOW DIGIT

اﻟ

ﺮﻗﻢ

ﻓﺎﻧﮫ ﯾﺼﺤﺢ ﺑﺎﺿﺎﻓﺔ

اﻟﻰ اﻟﺮﻗﻢ وﯾﻘﻮم ﺑﺘﺼﻔﯿﺮ

High Digit

اﻟﻰ اﻟﯿﺴﺎر وﻛﺬﻟﻚ

ﯾﻀﯿﻒ واﺣﺪ اﻟﻰ اﻟﻤﺴﺠﻞ

وﻧﺠﻌﻞ ﻗﯿﻢ ال

flag

ﻛﺎﻻﺗﻲ

AF=1

CF=1

وﻧﺪﻗﻖ ﻗﯿﻤﺘﻲ

PF and SF

٢٥

48 +

وﯾﻀﯿﻒ

اﻟﻰ ﻗﯿﻤﺔ ﻣﺴﺠﻞ

وﺗﻜﻮن ﻗﯿﻢ

Flags

ﻛﺎﻻﺗﻲ

AF=1, CF=1, PF=0,SF=0

2. Packed BCD number and DAA Instruction

If you add the packed BCD numbers 46 and 43, the 8086 performs this Hexadecimal

addition:

اذا ﺗﻢ اﺿﺎﻓﺔ رﻗﻤﯿﻦ

packed BCD

ھﻤﺎ

46 and 43

ﻓﺎن اﻟﻤﻌﺎﻟﺞ ﺳﯿﻘﻮم ﺑﺎﺟﺮاء ﻋﻤﻠﯿﺔ اﻟﺠﻤﻊ ﺑﺎﻟﻨﻈﺎم اﻟﺴﺎدس

ﻋﺸﺮي وﻛﺎﻻﺗﻲ

43 +

89 hexadecimal

89 BCD

If we use an DAA instruction after the addition operation will have no effect on the

two digits in the register AL and have no effect on AH and no effect on flags because

the Low digit and High digit is less the number 9.

اذا ﺗﻢ اﺳﺘ

ﺨﺪام اﯾﻌﺎز

DAA

ﻟﺘﺤﻮﯾﻞ ﻧﺎﺗﺞ اﻟﺠﻤﻊ ﻣﻦ اﻟﻨﻈﺎم اﻟﺴﺎدس اﻟﻌﺸﺮي اﻟﻰ ﻧﻈﺎم

BCD

ﻓﺎن اﯾﻌﺎز

DAA

ﻟﻦ

ﯾﻜﻮن ﻟﮫ ﺗﺎﺛﯿﺮ ﻋﻠﻰ اﻟﺮﻗﻤﯿﻦ ﻓﻲ ﻣﺴﺠﻞ

وﻻ ﻋﻠﻰ اﻟﻤﺴﺠﻞ

وﻻ ﻋﻠﻰ

Flags

وذﻟﻚ ﻻﻧﮫ ﻛﻞ ﻣﻦ اﻟﺮﻗﻤﯿﻦ

Low digit and High digit

ھﻤﺎ اﻗﻞ ﻣﻦ ﺗﺴﻌﺔ

٣٥

This DAA instruction is deal with two digits the register AL and only it deal with AL

(High digit) exceed number 9 or both exceed number 9 it will add six to correct the digit

which will exceed number 9 and have no effect on the AH and set 1 to AF and CF and

check the value of PF and SF.

وھﺬا ﯾﻌﻨﻲ ان اﻻﯾﻌﺎز

DAA

ﯾﺘﻌﺎﻣﻞ ﻣﻊ

اﻟﺮﻗﻤﯿﻦ ﻓﻲ

اﻟﻤﺴﺠﻞ

وﻋﻨﺪﻣﺎ ﯾﺘﺠﺎوز اﻟﻨﺎﺗﺞ ﻓﻲ

LOW DIGIT

اﻟﺮﻗﻢ

اوﻋﻨﺪﻣﺎ ﯾﺘﺠﺎوز اﻟﻨﺎﺗﺞ ﻓﻲ

High DIGIT

اﻟﺮﻗﻢ

اوﻋﻨﺪﻣﺎ ﯾﺘﺠﺎوز اﻟﻨﺎﺗﺞ ﻓﻲ اﻟﺮﻗﻤﯿﻦ اﻟﺮﻗﻢ

ﻓﺎﻧﮫ ﯾﺼﺤﺢ

ﺑﺎﺿﺎﻓﺔ

اﻟﻰ اﻟﺮﻗﻢ

اﻟﺬي ﺗﺠﺎوز اﻟﺮﻗﻢ ﺗﺴﻌﺔ وﻟﯿﺲ ﻟﮫ ﺗﺎﺛﯿﺮ ﻋﻠﻰ

اﻟﻤ

ﺴﺠﻞ

وﻧﺠﻌﻞ ﻗﯿﻢ ال

flag

ﻛﺎﻻﺗﻲ

AF=1

CF=1

وﻧﺪﻗﻖ ﻗﯿﻤﺘﻲ

PF and SF

89 +

ﻻﯾﺆﺛﺮ ﻋﻠﻰ

ﻗﯿﻤﺔ ﻣﺴﺠﻞ

وﺗﻜﻮن ﻗﯿﻢ

Flags

ﻛﺎﻻﺗﻲ

AF=1, CF=1, PF=1,SF=0

3. Increment Destination by One (INC)

The INC instruction adds 1 to a register or memory operand but, unlike ADD,

does not affect the Carry Flag (CF). INC is convenient for increasing loop

counters .It can also be used to increase an index register or pointer when you

are accessing consecutive locations in memory. Examples are:

ان اﯾﻌﺎز

INC

ﯾﻀﯿﻒ واﺣﺪ

اﻟﻰ ﻣﺤﺘﻮى ﻣﺴﺠﻞ او ﻣﻮﻗﻊ اﻟﺬاﻛﺮة وﻟﻜﻨﮫ ﻟﯿﺲ ﻣﺜﻞ اﯾﻌﺎز

ADD

ﻓﺎﻧﮫ ﻻﯾﺆﺛﺮ ﻋﻠﻰ

وھﻮ ﯾﺴﺘﺨﺪم ﻓﻲ زﯾﺎدة ﻋﺪادات اﻟﺤﻠﻘﺎت اﻟﺘﻜﺮارﯾﺔ وﯾﻤﻜﻦ ان ﯾﺰﯾﺪ ﻗﯿﻤﺔ ﻣﺴﺠﻞ ﻓﮭﺮﺳﺔ او ﻣﺆﺷﺮ وذﻟﻚ ﻋﻨﺪ ﻣﻌﺎﻟﺠﺔ ﻣﻮاﻗﻊ

ﻣﺘﺴﻠﺴﻠﺔ ﻓﻲ اﻟﺬاﻛﺮة واﻻﻣﺜﻠﺔ ﻟﻠﻤﻮﺿﻮع ھﻲ

٤٥

INC CX ;Increment a word register

INC AL ;Increment a byte register

INC MEM_BYTE ; Increment a byte in memory

INC MEM_WORD ; Increment a word in memory

2-Subtraction Instructions

1. How the Microprocessor Subtracts

Like every other general-purpose microprocessor, the 8086, has no internal subtraction

unit. However, it does have an addition unit an adder and can subtract numbers by adding

them.

ﻣﺜﻞ ﻛﻞ اﻟﻤﻌﺎﻟﺠﺎت ذات اﻻﺳﺘﺨﺪام اﻟﻌﺎم ﻓﺎن اﻟﻤﻌﺎﻟﺞ

8086

ﻻﯾﻤﺘﻠﻚ وﺣﺪة ﻃﺮح داﺧﻠﯿﺔ ﺑﻞ ﯾﻤﺘﻠﻚ وﺣﺪة ﺟﻤﻊ وﯾﻤﻜﻨﮫ ﻣﻦ

ﻃﺮح اﻻﻋﺪاد ﻣﻦ ﺧﻼل ﺟﻤﻌﮭﻢ

To see how to subtract by adding, consider how you subtract 7 from 10.

وﻟﻨﺮى ﻛﯿﻒ ﯾﺘﻢ اﻟﻄﺮح ﻣﻦ ﺧﻼل اﻟﺠﻤﻊ ﻣﻦ ﺧﻼل ﻃﺮح اﻟﺮﻗﻢ

ﻣﻦ اﻟﺮﻗﻢ

The 8086 subtracts in two steps. First, it changes the sign of, or complements, the second

number (the subtrahend). Then it adds the minuend and complemented subtrahend to

produce the result. Because the 8086 works with base 2 (binary) numbers, the

complement is a two's-complement. To obtain the two's-complement of a binary number,

take its positive form and reverse each bit (change each 1 to 0 and each 0 to 1), then add

1 to the result.

اﻻوﻟﻰ ﯾﻘﻮم ﺑﻌﻤﻞ ﻣﻜﻤ، ﻦﯿﺗﻮﻄﺨﺑ حﺮﻄﻟﺎﺑ مﻮﻘﯾ ﺞﻟﺎﻌﻤﻟا نا

ﻠﺔ ﻟﻠﺮﻗﻢ اﻟﻤﻄﺮوح وﻣﻦ ﺛﻢ ﯾﺘﻢ اﺿﺎﻓﺔ اﻟﺮﻗﻢ اﻟﻤﻄﺮوح ﻣﻨﮫ

اﻟﻰ ﻣﻜﻤﻠﺔ اﻟﺮﻗﻢ اﻟﻤﻄﺮوح

وﺑﻤﺎ ان اﻟﻤﻌﺎﻟﺞ

8086

ﺔــﻘـﯾﺮـﻄﺑ نﻮﻜـﺘﺳ ﺔﻠـﻤـﻜﻤﻟا نﺎﻓ ﻲﺋﺎﻨﺜﻟا مﺎـﻈــﻨﻟا ﻊﻣ ﻞﻣﺎﻌﺘﯾ

two's-complement

وﻟﻠﺤﺼﻮل ﻋﻠﻰ ھﺬه اﻟﻤﻜﻤﻠﺔ ﻧﻘﻠﺐ ﻛﻞ ﺻﻔﺮ اﻟﻰ واﺣﺪ واﻟﻌﻜﺲ ﺑﺎﻟﻌﻜﺲ وﻧﻀﯿﻒ واﺣﺪ اﻟﻰ

اﻟﻨﺎﺗﺞ

٥٥

Applying this to our "10 - 7" example, the 8-bit binary representations of 10 and 7 are

00001010B and 00000111B, respectively. Take the two's complement of 7 as follows:

وﺑﺘﻄﺒﯿﻖ ھﺬه اﻟﻄﺮﯾﻘﺔ ﻋﻠﻰ ﻣﺜﺎﻟﻨﺎ

"10 - 7"

ﻓﺎن اﻟﺘﻤﺜﯿﻞ

ﻟﻠﺮﻗﻤﯿﻦ

ﺑﻄﻮل ﺛﻤﺎﻧﯿﺔ ﺑ

ﺘﺎت ﺳﯿﻜﻮن ﺑﺎﻟﺸﻜﻞ اﻟﺘﺎﻟﻲ

00001010B and 00000111B

وﻋﻤﻠﯿﺔ اﻟﻤﻜﻤﻠﺔ ﻟﻠﺮﻗﻢ

11111000 (reverse all bits)

1 + (add 1)

11111001 (two's-complement of 7, or -7)

Now the subtraction operations becomes

0000 1010 (= 10)

1111 1001 (= - 7)

0000 0011 (answer = 3)

2.Subtract (SUB) and Subtract With Borrow (SBB)

SUB and SBB are similar to their addition counterparts (ADD and ADC), but with

subtraction, the Carry Flag (CF) acts as a borrow indicator. SUB subtracts a source

operand from a destination operand and returns the result in the destination. That is,

ان اﯾﻌﺎزي اﻟﻄﺮح

SUB and SBB

ﻊـﻤـﺠﻟا يزﺎﻌﯾا ﻰﻟا ﺔﮭـﺑﺎـﺸﻣ

(ADD and ADC)

وﻟﻜﻦ ﺑﺎﻟﻄﺮح ﻓﺎن

Carry Flag (CF)

ﯾﻌﺎﻣﻞ ﻛﻤﺆﺷﺮ

borrow

واﯾﻌﺎز

SUB

ﯾﻘﻮم ﺑﻄﺮح اﻟﻤﺼﺪر ﻣﻦ اﻟﻤﺴﺘﻘﺮ وﯾﺨﺰن اﻟﻨﺎﺗﺞ ﻓﻲ

اﻟﻤﺴﺘﻘﺮ

destination == destination - source

SBB does the same thing, except it also subtracts out the Carry Flag (CF), like this:

destination == destination - source - carry

As with addition, the subtraction instructions perform two separate functions. One

instruction, SUB, subtracts single-byte or single-word numbers, or the low-order terms

of 32-bit numbers. The other, SBB, subtracts the higher-order terms of two 32-bit

numbers. For example, the instruction

٦٥

ﻓﺎن اﯾﻌﺎز، ﻊﻤﺠﻟا ﺔﻟﺎﺣ ﻲﻓ بﻮﻠﺳﻻا ﺲﻔﻨﺑ

اﻻﯾﻌﺎز اﻻول، ﺔﻠﺼﻔﻨﻤﻟا لاوﺪﻟا ﻦﻣ ﻦﯿﻋﻮﻧ مﺪﻘﺗ حﺮﻄﻟا تا

SUB

ﯾﻄﺮح رﻗﻤﯿﻦ ﺑﻄﻮل ﺑﺎﯾﺖ او

رﻗﻤﯿﻦ ﺑﻄﻮل

word

او اﻟﺠﺰء اﻻدﻧﻰ ﻣﻦ رﻗﻤﯿﻦ ﺑﻄﻮل

32-bit

واﻻﯾﻌﺎز اﻻﺧﺮ ھﻮ

SBB

ﯾﻄﺮح اﻟﺠﺰء اﻻﻋﻠﻰ ﻣﻦ رﻗﻤﯿﻦ ﺑﻄﻮل

32-bit

وﻛﻤﺜﺎل ﻓﺎن اﻻﯾﻌﺎز

SUB AX , CX

subtracts CX from AX and returns the result in AX.

For operands longer than 16 bits, use this kind of sequence:

SUB AX,CX ;Subtract low – order 16 bits,

SUB BX,DX ;then high – order 16 bits,

Here we subtract the 32-bit number in CX and DX from the one in AX and BX. When

SBB subtracts DX from BX, it includes any borrow out of the first subtraction.

You may also subtract a memory operand from a register (or vice versa) or an

immediate value from a register or memory location. You may not subtract one memory

value from another directly, nor may you use an immediate value as a destination.

Examples of legal forms are:

وﯾﺘﻢ ﻃﺮح اﻟﺮﻗﻢ

32-bit

اﻟﻤﻮﺟﻮد ﻓﻲ

CX and DX

ﻣﻦ اﻟﺮﻗﻢ اﻟﻤﻮﺟﻮد ﻓﻲ

AX and BX

وﯾﻄﺮح اﻻﯾﻌﺎز

SBB

اﻟﻤﺴﺠﻞ

ﻣﻦ اﻟﻤﺴﺠﻞ

وﺗﻘﻮم ﺑﺎﻟﺘﻌﺎﻣﻞ ﻣﻊ اي

borrow

ﻗﺪ ﯾﻨﺘﺞ ﻣﻦ ﻋﻤﻠﯿﺔ اﻟﻄﺮح اﻻوﻟﻰ

وﯾﻤﻜﻦ ان ﯾﻄﺮح ﻣﻮﻗﻊ ذاﻛﺮة ﻣﻦ ﻣﺴﺠﻞ ﻋﺎم واﻟﻌﻜﺲ ﺑﺎﻟﻌﻜﺲ او ﻃﺮح اﻟﻘﯿﻤﺔ اﻟﻤﺒﺎﺷﺮة

ﻣﻦ ﻣﺴﺠﻞ او ﻣﻮﻗﻊ اﻟﺬاﻛﺮة

وﯾﺠﺐ ان ﻧﻼﺣﻆ اﻧﮫ ﻻﯾﻤﻜﻦ ان ﻧﻄﺮح ﻣﻮﻗﻌﯿﻦ ذاﻛﺮة ﻣﺒﺎﺷﺮة او اﺳﺘﺨﺪام اﻟﻘﯿﻤﺔ اﻟﻤﺒﺎﺷﺮة ﻛﻤﺴﺘﻘﺮ

وھﺬه ﺑﻌﺾ اﻻﻣﺜﻠﺔ

ﻋﻠﻰ ھﺬا اﻟﻤﻮﺿﻮع

SUB AX , MEM_WORD; subtract memory from a register

SUB MEM_WORD , AX ; or vice versa

SUB AL , 10 ; subtract constant from a register

SUB MEM_BYTE , 0FH ; or from a memory location

3. Decrement Destination by One (DEC)

The DEC instruction subtracts 1 from a register or memory operand but unlike

SUB, does not affect the Carry Flag (CF). DEC is often used to decrement a loop

counter until it becomes zero or negative .It can also be used to decrease an

index register or pointer when you are accessing consecutive memory locations.

Some Examples are:

٧٥

ان اﯾﻌﺎز

DEC

ﯾﻄﺮح واﺣﺪ ﻣﻦ ﻣﺤﺘﻮى ﻣﺴﺠﻞ او ﻣﻮﻗﻊ اﻟﺬاﻛﺮة وﻟﻜﻨﮫ ﻟﯿﺲ ﻣ

ﺜﻞ اﯾﻌﺎز

SUB

ﻓﺎﻧﮫ ﻻﯾﺆﺛﺮ ﻋﻠﻰ

وھﻮ ﯾﺴﺘﺨﺪم ﻓﻲ ﻧﻘﺼﺎن ﻋﺪادات اﻟﺤﻠﻘﺎت اﻟﺘﻜﺮارﯾﺔ وﯾﻤﻜﻦ ان ﯾﻘﻠﻞ ﻗﯿﻤﺔ ﻣﺴﺠﻞ ﻓﮭﺮﺳﺔ او ﻣﺆﺷﺮ وذﻟﻚ ﻋﻨﺪ ﻣﻌﺎﻟﺠﺔ

ﻣﻮاﻗﻊ ﻣﺘﺴﻠﺴﻠﺔ ﻓﻲ اﻟﺬاﻛﺮة واﻻﻣﺜﻠﺔ ﻟﻠﻤﻮﺿﻮع ھﻲ

DEC CX ;Decrement a word register

DEC AL ;Decrement a byte register

DEC MEM_BYTE ; Decrement a byte in memory

DEC MEM_WORD ; Decrement a word in memory

4.Negate (NEG)

The NEG instruction subtracts the destination operand from zero, thus forming the

operand's two's – complement.

ان

اﯾﻌﺎز

NEG

ﯾﻄﺮح اﻟﻤﺴﺘﻘﺮ ﻣﻦ اﻟﺼﻔﺮ وھﺬا ﯾﻮﻟﺪ

two's – complement

ﻟﻠﺮﻗﻢ ﻓﻲ اﻟﻤﺴﺘﻘﺮ

MOV AL,0FF

NEG

AL= 1111 1111

0000 0000 COMPLEMENT

1 ADD 1

0000 0001

AL=01

٨٥

5. Compare Destination to Source (CMP

)

Like the SUB instruction, CMP subtracts a source from a destination and sets or clears

the flags based on the result (see Table 3-5). But unlike SUB, CMP does not save the

result of the subtraction; that is, it doesn't alter the destination. CMP's sale purpose is to

set up the flags for decision-making by conditional jump instructions

اﯾﻌﺎز اﻟﻤﻘﺎرﻧﺔ

CMP

ﯾﺸﺒﮫ اﯾﻌﺎز اﻟﻄﺮح

SUB

ﺑﺎن ﯾﻄﺮح اﻟﻤﺼﺮ ﻣﻦ اﻟﻤﺴﺘﻘﺮ وﯾﻘﻮم ﺑﻮﺿﻊ واﺣﺪ او ﺻﻔﺮ ﻓﻲ

flags

وﺑﺎﻻﻋﺘﻤﺎد ﻋﻠﻰ اﻟﻨﺘﯿﺠﺔ ﻛﻤﺎ ﻓﻲ اﻟﺠﺪول

3-5

وﻟﻜﻦ ﻟﯿﺲ ﻣﺜﻞ،

SUB

ﻓﺎن اﯾﻌﺎز

CMP

ﻻﯾﺨﺰن ﻧﺎﺗﺞ اﻟﻄﺮح وﻻﯾﻐﯿﺮ

اﻟﻤﺴﺘﻘﺮ وھﻮ ﯾﺴﺘﺨﺪم ﻓﻲ ﻋﻤﻠﯿﺔ ﺗﻐﯿﯿﺮ ﻗﯿﻢ

flags

وذﻟﻚ ﻟﺘﺤﺪﯾﺪ ﻗﺮار ﺑﺎﺣﺪ اﯾﻌﺎزات اﻟﻘﻔﺰاﻟﻤﺸﺮوﻃﺔ

٩٥

Multiplication Instructions

Multiply (MUL) multiplies unsigned numbers and Integer Multiply (IMUL) multiplies

signed numbers. Both can multiply bytes or words

اﯾﻌﺎز اﻟﻀﺮب

MUL

ﯾﻀﺮب ارﻗﺎم ﺑﺪون اﺷﺎرة واﻻرﻗﺎم اﻟﺼﺤﯿﺤﺔ

اﻻﯾﻌﺎز

IMUL

ﯾﻀﺮب اﻻرﻗﺎم ﺑﺎﻻﺷﺎرة

وﻛﻼھﻤﺎ ﯾﻀﺮب ارﻗﺎم ﺑﻄﻮل ﺑﺎﯾﺖ او ﻛﻠﻤﺔ

Multiply, Unsigned (MUL) and Integer Multiply, Signed (IMUL)

The 8086 multiply instructions have the general forms

MUL SOURCE

IMUL SOURCE

where source is a byte- or word-length general register or memory location. For the second

operand, MUL and IMUL use the contents of AL (for byte operations) or AX (for word

operations). They return double-length products as follows:

ﻌﺎﺯﻱـﻴﻻﺍ ﻥﺎﻓ ﻲﻨﺎـﺜﻟﺍ لﻤﺎﻌـﻤﻟﺍﻭ ﺓﺭﻜﺍﺫ ﻊـﻗﻭـﻤ ﻭﺍ ﻡﺎـﻋ لـﺠﺴﻤ ﻲﻓ ﺔﻤـﻠﻜ ﻭﺍ ﺕﻴﺎﺒ ﻥﻭﻜﻴ ﺩﻗ ﺭﺩﺼﻤﻟﺍ ﻥﺍ ﺙﻴﺤ

MUL and IMUL

ﺘﺴﺘﺨﺩﻡ ﻤﺤﺘﻭﻯ ﺍﻟﻤﺴﺠل

ﻟﻌﻤﻠﻴﺎﺕ ﺍﻟﺒﺎﻴﺕ ﻭﻤﺴﺠل

ﻟﻌﻤﻠﻴﺎﺕ ﺍﻟﻜﻠﻤﺔ ﻭﺍﻟﻘﻴﻤﺔ ﺍﻟﻨﺎﺘﺠﺔ ﺴﺘﻜﻭﻥ

ﻀﻌﻑ ﻁﻭل ﺍﻻﺭﻗﺎﻡ ﺍﻟﺘﻲ ﺘﻡ ﻀﺭﺒﻬﺎ

. Multiplying bytes produces a 16-bit product in AH (high byte) and AL (low byte)

. Multiplying words produces a 32-bit product in DX (high word) and AX (low word)

Upon completion, the Carry and Overflow Flags (CF and OF) tell how much of the

product is relevant. For MUL if the high-order half of the product is zero, CF and OF are

0; otherwise, they are 1. For IMUL, if the high-order half of the product is just a sign-

extension of the low-order half, CF and OF are 0; otherwise, they are 1.

وﺑﺎﻟﻨﺴﺒﺔ ﻟﻞ

flags

ﻓ

ﺎن

CF and OF

ﺗﻜﻮن ﻟﮭﺎ ﻋﻼﻗﺔ ﺑﺎﻟﻨﺎﺗﺞ

ﻓﻔﻲ ﺣﺎﻟﺔ اﻟﻀﺮب

MUL

اذاﻛﺎن ﺟﺰء

high-order

ﻣﻦ اﻟﻨﺎﺗﺞ ھﻮ ﺻﻔﺮ ﻓﺎن ﻛﻞ ﻣﻦ

CF and OF

ﺳﺘﻜﻮن ﻗﯿﻤﺘﮭﻤﺎ ﺻﻔﺮ واﻻ ﻓﮭﻲ واﺣﺪ وﻓﻲ ﺣﺎﻟﺔ

IMUL

اذاﻛﺎن ﺟﺰء

high-order

ﻣﻦ اﻟﻨﺎﺗﺞ ھﻮ ﻣﺠﺮد ﺗﻤﺪﯾﺪ ﻟﻼﺷﺎرة ﺻﻔﺮ ﻓﺎن ﻛﻞ ﻣﻦ

CF and OF

ﺳﺘﻜﻮن ﻗﯿﻤﺘﮭﻤﺎ ﺻ

ﻔﺮ واﻻ ﻓﮭﻲ واﺣﺪ

٠٦

MUL BX ;Unsigned multiply of BX times AX

MUL MEM_BYTE ; Unsigned multiply of memory times AL

IMUL DL ; signed multiply of BL times AL

IMUL MEM_WORD ; signed multiply of memory times AX

Division Instructions

Divide (DIV) performs an unsigned division, while Integer Divide (IDIV) performs a

signed division

ان اﯾﻌﺎز

DIV

ﯾﻘﺪم ﻋﻤﻠﯿﺔ اﻟﻘﺴﻤﺔ ﻻﻋﺪاد ﺑﺪون اﺷﺎرة ﺑﯿﻨﻤﺎ

IDIV

ﯾﻘﺴﻢ ارﻗﺎم ﺑﺎﺷﺎرة

Divide, Unsigned (DIV) and Integer Divide, Signed (IDIV)

DIV SOURCE

IDIV SOURCE

where source is a byte- or word-sized general register or memory location that contains

the divisor (the value by which you want to divide). The dividend is a double-length

operand contained in either AH and AL (for byte operations) or DX and AX (for word

operations).

ﻊ ﺫﺍﻜﺭﺓ ﺍﻟﺘﻲ ﺘﺤﺘﻭﻱ ﻋﻠﻰ ﻗﻴﻡ ﺍﻟﻤﻘﺴﻭﻡ ﻋﻠﻴﻪـﻗﻭـﻤ ﻭﺍ ﻡﺎـﻋ لـﺠﺴﻤ ﻲﻓ ﺔﻤـﻠﻜ ﻭﺍ ﺕﻴﺎﺒ ﻥﻭﻜﻴ ﺩﻗ ﺭﺩﺼﻤﻟﺍ ﻥﺍ ﺙﻴﺤ

ﻭﺍﻟﻤﻘﺴﻭﻡ ﻗﺩ ﻴﻜﻭﻥ ﻓﻲ ﻤﺴﺠل

AH and AL

ﻭﻗﺩ ﻴﻜﻭﻥ ﻓﻲ ﻤﺴﺠﻠﻲ

DX and AX

ﻓﻲ ﺤﺎﻟﺔ ﺍﻟﻜﻠﻤﺎﺕ

DIV and IDIV produce the following results:

. If the source operand is a byte, the quotient is returned in AL and the remainder in AH.

. If the source operand is a word, the quotient is returned in AX and the remainder in DX.

١٦

Both instructions leave the flags undefined. However, if the quotient cannot fit in the

destination register (AL or AX)-that is, if it overflows the destination the 8086 has a

dramatic way of telling you that the result is invalid: it generates a type 0 (divide by 0)

interrupt.

وﺑﺎﻟﻨﺴﺒﺔ ﻟﻞ

flags

ﻓﺎن اﻻﯾ

ﻌﺎزﯾﻦ ﻻﯾﺆﺛﺮان ﻋﻠﻰ اي

flags

وﻟﻜﻦ ﻓﻲ ﺣﺎﻟﺔ ﻛﻮن ﻧﺎﺗﺞ اﻟﻘﺴﻤﺔ ﻻﯾﻤﻜﻦ ﺧﺰﻧﮫ ﻓﻲ ﻣﺴﺠﻠﻲ

AL or AX

ﻓﺎن ﻋﻤﻠﯿﺔ

overflows

ﺳﻮف ﺗﻔﻌﻞ وﯾﻠﺪ ﻣﻔﺎﻃﻌﺔ اﻟﺘﻘﺴﯿﻢ ﻋﻠﻰ ﺻﻔﺮ

The following conditions cause a divide overflow:

وھﺬه اﻟﺤﺎﻻت ﺗﻮﻟﺪ

divide overflow

1. The divisor is zero.

2. For an unsigned byte divide, the dividend is at least 256 times larger than

the divisor.

3. For an unsigned word divide, the dividend is at least 65,536 times larger

than the divisor.

4. For a signed byte divide, the quotient exceeds + 127 or -128.

5. For a signed word divide, the quotient exceeds + 32,767 or -32,768.

Here are some typical division operations:

DIV BX ;Divide DX:AX By BX, unsigned

DIV MEM_BYTE ; Divide AH:AL By MEMORY, unsigned

IDIV DL ; Divide AH:AL By DL, signed

IDIV MEM_WORD ; Divide DX:AX By MEMORY, signed