microprocessor architecture and programming

MICRO PROCESSOR ARCHITECTURE AND MICROPROGRAMMING

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The word micro is used in microscopes, microphones, microwaves, microprocessors, microcomputers, microprogramming, microcodes etc. It means small. A microprocessor is an example of VLSI bringing the whole processor to a single small chip. With the popularity of distributed processing, the emphasis has shifted from the single mainframe system to independently working workstations or functioning units with their own CPU, RAM, ROM and a magnetic or optical disk memory. Thus, the advent of the microprocessor has transformed the mainframe environment to a distributed platform.

Basic components of a microprocessor:

·         ROM stores the boot program.

·         The path from CPU to devices is through Buses. But what would be the size of these Buses?

Bus Sizes

1. The Address bus: 8085 microprocessor has 16 bit lines. Thus, it can access up to 2^16 = 64K Bytes. The address bus of 8086 microprocessor has a 20 bits address bps. Thus it can access up to 2^20 = 1M Byte size of RAM directly.

2. Data bus is the number of bits that can be transferred simultaneously. It is 16 bits in 8086.

Microprocessors

The microprocessor is a complete CPU on a single chip. The main advantages of the microprocessor are:

·         compact but powerful

·         Can be micro programmed for user's needs;

·         easily programmable and maintainable due to small size; and

·         Useful in distributed applications.

A microprocessor must demonstrate:

·         More throughput

·         More addressing capability

·         Powerful addressing modes

·         Powerful instruction set

·         Faster operation through pipelining

·         Virtual memory management.

However, RISC machine do not agree with above principles. Some of the most commercially available microprocessors are: Pentium, Xeon, G4 etc.

The assembly language for more advanced chips subsumes the simplest 8086/ 8088 assembly language. Therefore, we will confine our discussions to Intel 8086/8088 assembly language. You must refer to the further readings for more details on assembly language of Pentium, G4 and other processors. All microprocessors execute a continuous loop of fetch and execute cycles.

while (1)

{

fetch (instruction);

execute (using date);

}

STRUCTURE OF 8086 CPU

The 8086 microprocessor consists of two independent units:

1. The Bus Interface unit, and

2. The Execution unit.

The CPU of INTEL 8086 Microprocessor

The word independent implies that these two units can function parallel to each other. In other words they may be considered as two stages of the instruction pipeline.

The Bus Interface Unit

The BIU (Bus Interface Unit) primarily interacts with the system bus. It performs almost all the activities relating to fetch cycle such as:

·         Calculating the physical address of the next instruction

·         Fetching the instruction

·         Reading or writing data memory or 110 ports from memory or Input1 Output.

The instruction1 data is then passed to the execution unit. This BIU consists of:

(a) The Instruction Queue

The instruction queue is used to store the instruction "bytes" fetched. Please note two points here: that it is (1) A Byte (2) Queue. This is used to store information in byte form, with the underlying queue data structure. The advantage of this queue would only be if the next expected instructions are fetched in advance, thus, allowing a pipeline of fetch and execute cycles.

(b) The Segment Registers

These are very important registers of the CPU. Why? We will answer this later. In 8086 microprocessor, the memory is a byte organized, that is a memory address is byte address. However, the number of bits fetched is 16 at a time. The segment registers are used to calculate the address of memory location along with other registers. A segment register is 16 bits long.

The BIU contains four sixteen-bit registers, viz., the CS: Code Segment, the DS: Data Segment, the SS: Stack Segment, and the ES: Extra Segment. But what is the need of the segments: Segments logically divide a program into logical entities of Code, Data and Stack each having a specific size of 64 K. The segment register holds the upper 16 bits of the starting address of a logical group of memory, called the segment. But what are the advantages of using segments? The main advantages of using segments are:

·         Logical division of program, thus enhancing the overall possible memory use and minimize wastage.

·         The addresses that need to be used, in programs are re-locatable as they are the offsets. Thus, the segmentation supports re-locatability.

·         Although the size of address is 20 bits, yet only the maximum segment size, that is 16 bits, needs to be kept in instruction, thus, reducing instruction length.

The 8086 microprocessor uses overlapping segments configuration. The typical memory organization for the 8086 microprocessor may be as per the following diagram.

Logical Organization of Memory in INTEL 8086 Microprocessor

Although the size of each segment can be 64K, as they are overlapping segments we can create variable size of' segments, with maximum as 64K. Each segment has a specific function. 8086 supports the following segments:

As per model of assembly program, it can have more than one of any type of segments. However, at a time only four segments one of each type, can be active. The 8086 supports 20 address lines, thus supports 20 bit addresses. However, all the registers including segment registers are of only 16 bits. So, how may this mapping of 20 bits to 16 bits is performed? Let us take a simple mapping procedure:

The top four hex digits of initial physical address constitute segment address. We can add offset of 16 bits(4 Hex digits) from 0000h to FFFFh to it. Thus, a typical segment which starts at a physical address 10000h will range from 10000h to 1FFFFh. The segment register for this segment will contain 1000h and offset will range from 0000h to FFFFh. But, how will the segment address and offset be added to calculate physical address? Let us explain using the following examples

Example 1 (In the Figure above)

The value of the stack segment register (SS) = 6000h

The value of the stack pointer (SP) which is Offset = 0010h

Thus, Physical address of the top of the stack is:

This calculation can be expressed as:

Physical address = SS (hex) x 16 + SP (hex)

Example 2

The offset of the data byte = 0020h

The value of the data segment register (DS) = 3000h

Physical address of the data byte

This calculation can be expressed as physical address = DS (Hex) x 16 + Data byte offset (hex).

Example 3

The value of the Instruction Pointer, holding address of the instruction = 1234h

The value of the code segment register (CS) = 448Ah

Physical address of the instruction

Physical Address = CS (Hex) x 16 + IP

(c) Instruction Pointer

The instruction pointer points to the offset of the current instruction in the code segment. It is used for calculating the address of instruction as shown above.

Execution Unit (EU)

Execution unit performs all the ALU operations. The execution unit of 8086 is of 16 bits. It also contains the control unit, which instructs bus interface unit about which memory location to access, and what to do with the data. Control unit also performs decoding and execution of the instructions. The EU consists of the following:

(a)! Control Circuitry, Instruction Decoder and ALU

The 8086 control unit is primarily micro-programmed control. In addition it has an instruction decoder, which translates an instruction into sequence of micro operations. The ALU performs the required operations under the control of CU which issues the necessary timing and control sequences.

(b) Registers

All CPUs have a defined number of operational registers. 8086 has several general purposes and special purpose registers.

REGISTER SET OF 8086

The 8086 registers have five groups of registers. These groupings are done on the basis of the main functions of the registers. These groups are:

General Purpose Register

8086 microprocessors have four general purpose registers namely, AX, BX, CX, DX. All these registers are 16 - bit registers. However, each register can be used as two general purpose byte registers also. These byte registers are named AH and AL for AX, BH and BL for BX, CH and CL for CX, and DH and DL for DX. The H in register name represents higher byte while L represents lower byte of the 16 bits registers. These registers are primarily used for general computation purposes. However, in certain instruction executions they acquire a special meaning.

AX register is also known as accumulator. Some of the instructions like divide, rotate, shift etc. require one of the operands to be available in the accumulator. Thus, in such instructions, the value of AX should be suitably set prior to the instruction.

BX register is mainly used as a base register. It contains the starting base location of a memory region within a data segment.

CX register is a defined counter. It is used in loop instruction to store loop counter.

DX register is used to contain I10 port address for 110 instructions.

Segment Registers

Segment Registers are used for calculating the physical address of the instruction or memory. Segment registers cannot be used as byte registers.

Pointer and Index Registers

The 8086 microprocessor has three pointer and index registers. Each of these registers is of 16 bit and cannot be accessed byte wise. These are Base Pointer (BP), Source Index (SI) and Destination Index (DI). Although they can be used as general purpose registers, their main objective is to contain indexes. BP is used in stack segment, SI in Data segment and DI in Extra Data segment.

Special Registers

A Last in First Out (LIFO) stack is a data structure used for parameter passing, return address storage etc. 8086 stack is 64K bytes Base of the stack is pointed to by the stack segment (SS) register while the offset or top of the stack is stored in Stack Pointer (SP) register. Please note that although the memory in 8086 has byte addresses, stack is a word stack, which is any push operation, will occupy two bytes.

Flags Register

A flag represents a condition code that is 0 or 1. Thus, it can be represented using a flip- flop. 8086 employs a 16-bit flag register containing nine flags. The following table shows the flags of 8086.

Conditional Flags represent result of last arithmetic or logical instruction executed. Conditional flags are set by some condition generated as a result of the last mathematical or logical instruction executed. The conditional flags are:

Flags	Meaning	Comments
CF	Carry Flag	1 if there is a carry bit
PF	Parity Flag	1 on even parity 0 on odd parity
AF	Auxiliary Flag	Set (1) if auxiliary carry for BCD occurs
ZF	Zero Flag	Set if result is equal to zero
SF	Sign Flag	Indicates the sign of the result (1 for minus , 0 for plus)
OF	Overflow Flag	set whenever there is an overflow of the result

Control flags, which are set or reset deliberately to control the operations of the execution unit. The control flags of 8086 are as follows:

Flags	Meaning	Comments
TF	Single step trap flag	Used for single stepping through the program
IF	Interrupt Enable Flag	Used to allow/inhibit the interruption of the program
DF	String direction Flag	Used with string instruction

INSTRUCTION SET OF 8086

After discussing the basic organization of the 8086 micro-processor, let us now provide an overview of various instructions available in the 8086 microprocessor. The instruction set is presented in the tabular form. An assembly language instruction in the 8086 includes the following:

Label: Op-code           Operand(s); Comment

For example, to add the content of AL and BL registers to get the result in AL, we use the following assembly instruction.

NEXT: ADD AL,BL   ;AL + BL

Please note that NEXT is the label field. It is giving an identity to the statement. It is an optional field, and is used when an instruction is to be executed again through a LOOP or GO TO. ADD is symbolic op-code, for addition operation. AL and BL are the two operands of the instructions. Please note that the number of operands is dependent upon the instructions. 8086 instructions can have zero, one or two operands. An operand in 8086 can be:

• 1.      A register
• 2.       A memory location
• 3.       A constant called literal
• 4.       A label.

Comments in 8086 assembly start with a semicolon, and end with a new line. A long comment can be extended to more than one line by putting a semicolon at the beginning of each line. Comments are purely optional, however recommended as they provide program documentation. In the next few sections we look at the instruction set of the 8086 microprocessor. These instructions are grouped according to their functionality.

Data Transfer Instructions

These instructions are used to transfer data from a source operand to a destination operand. The source operand in most of the cases remains unchanged. The operand can be a literal, a memory location, a register, or even an 110 port address, as the case may be. Let us discuss these instructions:

Note: If you want to run these examples, please use 8086 emulator, WinAsm ect. You can search from search engines.

Note: Be careful check “;”. Semicolon means comments. All comments are written after Semicolon.

MNEMONIC	DESCRIPTION			EXAMPLE
MOV des, src	DES ← SRC; Both the operands should be byte or word. SRC operand can be register, memory location or an immediate operand DES can be register or memory operand. Restriction: Both source and destination cannot be memory operands at the same time.			MOV CX,037AH MOV AX,BX
PUSH operand	Pushes the operand into a stack. SP ← SP-2; Value [TOS] ← operand. Initialize stack segment register, and the stack pointer properly before using this instruction. No flags are affected by this instruction. The operand can be a general purpose register, a segment register, or a memory location. Please note it is a word stack and memory address is a byte address, thus, you decrement by 2 Also you decrement as SP is initialized to maximum offset and condition of stackful is a zero offset (so, it is a reversed stack)			PUSH BX ; decrement stack pointer ; by; two, and copy BX to ; stack. ; decrement stack pointer ; by two, and copy ; BX to stack
POP des	POP a word from stack. The des can be a general-purpose register, a segment register (except for CS register), or a memory location. Steps are: des ← value [TOS] SP ← SP+2			POP AX ; Copy content for top ; of stack to AX
XCHG des, SRC	Used to exchange bytes or words of SRC and des. It requires at least one of the operands to be a register operand. The other can be a register or memory operand. Thus, the instruction cannot exchange two memory locations directly. Both the operands should be either byte m e or word type. The segment registers cannot be used as operands for this instruction.			XCHG DX,AX
XLAT	Translate a byte in AL using a table stored in the memory. The instruction replaces the AL register with a byte from the lookup table. This instruction is a complex instruction
IN accumulator, port address	It transfers a byte or word from specified port to accumulator register In case an 8-bit port is supplied as an operand then the data byte read from that part will be transferred to AL register. If a 16-bit port is read then the AX will get 16 bit word that was read. The port address can be an immediate. Operand, or contained in DX register. This instruction does not change any flags.			IN AL,028h
OUT port address, Accumulator	It transfers a byte or word from accumulator register to specified port. This instruction is used to output on devices like the monitor or the printer.
LEA register source	Load "effective address" (refer to this term in block 2, Unit 1 in addressing modes) of operand into specified 16 bit register. Since, an address is an offset in a segment and maximum can be of 16 bits, therefore, the register can only be a 16-bit register. LEA instruction does not change any flags. The instruction is very useful for array processing.			LEA BX, PRICES ; Assume PRICES is ; an array in the data ; segment
LDS des-reg	It loads data segment register and other specified register by using consecutive memory locations.			LDS SI, DATA ; DS ← content of memory ; location DATA & ; DATA + 1 ; SI  content of ; memory locations ;DATA+2&DATA+ : 3
LES des-reg	It loads ES register and other specified register by using consecutive memory locations. This instruction is used exactly like the LDS except in this case ES & other specified registers are initialized.
LAHF	Copies the lower byte of flag registers to AH. The instruction does not change any flags and has no operands.
SAHF	Copies the value of AH register to low byte of flag register. This instruction is just the opposite of LAHF instruction. This instruction has no operands.
PUSHF	Pushes flag register to top of stack. SP ← SP - 2; stack [SP] ← Flag Register.
POPF	Pops the stack top to Flag register. Flag register ← stack [SP] SP ← SP+2
Arithmetic Instructions:
ADD	Adds byte to byte, or word to word. The source may be an immediate operand, a register or a memory location. The rules for operands are the same as that of MOV instruction. To add a byte to a word, first copy the byte to a word location, then fill up the upper byte of the word with zeros. This instruction affects the following flags: AF, CF, OF, PF, SF. ZF.		ADD AL,74H ; Add the number 74H to ; AL register, and store the ; result back in AL ADD DX,BX ; Add the contents of DX to ; BX and store the result in ; DX, BX remains ; Unaffected.
ADC des, src	Add byte + byte + carry flag, or word + word + carry flag. It adds the two operands with the carry flag. Rest all the details are the same as that of ADD instruction.
INC des	It increments specified byte or word operand by one. The operand can be a register or a memory location. It can effect AF, SF, ZF, PF, and OF flags. It does not affect the carry flag, that is, if you increment a byte operand having OFFH, then it will result in 0 values in register and no carry flag.		INC BX ; Add 1 to the contents of BX INC BL Add 1 to the contents of BL
AAA	ASCII adjusts after addition. The data entered from the terminal is usually in ASCII format. In ASCII 0-9 are represented by codes 30-39. This instruction allows you to add the ASCII codes instead of first converting them to decimal digit using masking of upper nibble. AAA instruction is then used to ensure that the result is the correct unpacked BCD.		ADD AL,BL ; AL=00 1 10 10 1, ASCII 05 ; BL=00 1 1 100 1, ASCII 09 ; after addition ; AL = 01 101 110, that is, ; 6EH- incorrect ; temporary result AAA ; AL = 00000 100. ; Unpacked BCD for 04 ; carry = 1, indicates ; the result is 14
DAA	Decimal (BCD) adjust after addition. This is used to make sure that the result of adding two packed BCD numbers is adjusted to be a correct BCD number. DAA only works on AL register.		; AL = 0 10 1 100 1 (59BCD) ; BL = 001 1 0101 (35BCD) ADD AL, BL ; AL = 10001 101 or ; 8Eh (incorrect BCD) DAA ; AL = 1001 0100 ; =94 BCD : Correct
SUB des, src	Subtract byte from byte, or word from word. (des ← des - src). For subtraction the carry flag functions as a borrow flag, that is, if the number in the source is greater than the number in the destination, the borrow flag is to set 1. Other details are equivalent to that of the ADD instruction.		SUB AX, 3427h ; Subtract 342% from AX ; register, and store the ; result back in AX
SBB des, src	Subtract operands involving previous carry if any. The instruction is similar to SUB, except that it allows us to subtract two multi byte numbers because any borrow produced by subtracting less-significant byte can be included in the result using this instruction.		SBB AL,CH ; subtract the contents ; of CH and CF from AL ; and store the result ; back in AL.
DEC src	Decrement specified byte or specified word by one. Rules regarding the operands and the flags that are affected are same as INC instruction. Please note that if the contents of the operand is equal to zero then after decrementing the contents it becomes 0FFH or 0FFFFH, as the case may be. The carry flag in this case is not affected.		DEC BP ; Decrement the contents of BP ; register by one.
NEG src	Negate - creates 2's complement of a given number, this changes the sign of-a number. However, please note that if you apply this instruction on operand having value -128 (byte operand) or -32768 (word operand) it will result in overflow condition. The overflow (0F) flag will be set to indicate that operation could not be done		NEG AL ; Replace the number in AL with its 2's ; complement
CMP des,src	It compares two specified byte operands or two specified word. The source and destination operands can be an immediate number, a register or a memory location. But, both the operands cannot be memory locations at the same time. The comparison is done simply by internally subtracting the source operand from the destination operand. The value of source and the destination, operand is not changed but the flags are set to indicate the results of the comparison.		CMP CX,BX ; Compare the CX register ; with the BX register ; In the example above, the ;CF, ZF, and the SF flags
AAS	ASCII adjusts after subtraction. This instruction is similar to AAA (ASCII adjust after addition) instruction. The AAS instruction works on the AL register only. It updates the AF and CF flags, but the OF, PF, SF and the ZF flags remain undefined		SUB AL,BL AAS ;result
DAS	Decimal adjust after subtraction. This instruction is used after subtracting two packed BCD numbers to make sure the result is the packed BCD. DAS only works on the AL register. The DAS instruction updates the AF, CF, SF, PF and ZF flags. The overflow (OF) is undefined after DAS.		SUB AL,BH DAS
MUL Src	This is an unsigned multiplication instruction that multiplies two bytes to produce a word operand or two words to produce a double word such as: AX ← AL* src (byte multiplication src is also byte) DX or AX ← AX * Src (word multiplication is two word). .This instruction assumes one of the operand in AL (byte) or AX (word): the src operand can be register or memory operand. If the most significant word of the result is zero then, the CF and the OF flags are both made zero. The AF, SF, PF, ZF flags are not defined after the MUL instruction. If you want to multiply a byte with a word, then first convert - byte to a word operand.		MOV AX,05 ;AX=05 MOV CX,02 ;CX=02 MUL CX ; result in DX=0 ;AX=0AH
AAM	ASCII adjusts after multiplication. Note: Two ASCII numbers cannot be multiplied directly. To multiply first convert the ASCII number to numeric digits by masking of the upper nibble of each byte. This leaves unpacked BCD in the register. AAM instruction is used to adjust the product to two unpacked BCD digits in AX after the multiplication has been performed. AAM defined by the Instruction whiles the CF, OF and the AF flags are left undefined.		MUL BH ; AX=AL * BH=002Dh AAM ; AX=00000100 00000101 ;BCD 45 : Correct result
DIV src	This instruction divides unsigned word by byte, or unsigned double word by word. For dividing a word by a byte, the word is stored in AX register, divisor the src operand and the result is obtained in AH : remainder AL: quotient. It can be represented as: AH: Remainder AL: Quotient Similarly for double word division by a word we have DX: Remainder AX: Quotient A division by zero results in run time error. The divisor src can be either in a register or a memory operand.		; AX = 37D7h = 14295 ; decimal ; BH = 97h = 15 1 decimal DIV BH ; AX / BH quotient ;AL=5Eh=94 ; decimal Reminder AH =65h = 101 ; decimal
IDIV	Divide signed word by byte or signed double word by word. For this division the operand requirement, the general format of the instruction etc. are all same as the DIV instruction. IDIV instruction leaves all flags undefined.		; AL = 1 1001 01 0 = -26h =- 38 decimal ; CH = 0000001 1 = + 3h =3 decimal ; According to the operand ; rules to divide by a byte ; the number should be ; present in a word register, ; - i.e. AX. So, first convert ; the operand in AL to word ; Operand. This can be done ;by sign extending the ; AL register, ; this makes AX ; 11111111 11001010. IDN CH ; AXICH ;AL= 11110100=-OCH ;= - 12 Decimal ;AH= 11111110=-02H= ;-02 Decimal ; Although the quotient is ; actually closer to -13 ; (-'12.66667) than -12, but ; 8086 truncates the result to give-12.
AAD	ASCII adjusts after division. The BCD numbers are first unpacked, by masking off the upper nibble of each byte. Then ADD instruction is used to convert the unpacked BCD digits in AL and AH registers to adjust them to equivalent binary prior to division. Such division will result in unpacked BCD quotient and remainder. The PF, SF, ZF flags are updated, while the AF, CF, and the OF flags are left undefined.		AAD DIV CH
CBW	Fill upper-byte or word with copies of sign bit of lower bit. This is called sign extension of byte to word. This instruction does no1 change any flags. This operatio11 is done with AL register in the result being stored in AX.		CBW ;convert signed
CWD	Fill upper word or double word sign bit of lower word. This instruction is an extension of the CWD Previous instruction. This instruction results in sign extension of AX register to DX: AX double word.		; DX : 0000 0000 0000 0000 ;AX:1111000001010001 CWD ;DX:AX= 1111 1111 1111 1111: ; 1111 0000 0101 0001
Bits Manipulation Instructions These instructions are used at the bit level. These instructions can be used for testing a zero bit, set or reset a bit and to shift bits across registers. Let us look into some such basic instruction.
NOT des	Complements each bit to produce 1's complement of the specified byte or word operand. The operand can be a register or a memory operand.	; BX = 00 1 1 10 10 000 1 0000 NOT BX ;BX= 11000101 1110 1111
AND des, src	Bitwise AND of two byte or word operands. The result is des ← des AND src. The source can be an immediate operand a register, or a memory operand. Destination can be a register or a memory operand. Both operands cannot be memory operands at the same time. The CF and the OF flags are both zero after the AND operation. PF, SF and ZF area updated, AF is left undefined.	; BH = 001 1 1010 before AND BH, 0Fh ; BH = 0000 10 10 ; after the AND operation
OR des, src	OR each corresponding bits of the before byte or word operands. The other operands rules are same as AND des ← des OR src.	; BH = 001 l 1010 before OR BH, 0Fh ; BH = 001 1 1 1 1 1 after
XOR des,src	XOR each corresponding bit in a byte or word operands rules are two same as AND and OR. Des ← Des + src	;BX =0011110101101001 ; CX = 00000000 1 1 11 11 1 1 XOR BX,CX ;BX=0011110110010110 ; Please note, that the bits in ; The lower byte is inverted.
TEST des, src	AND the operands to update flags, but do not change operands value. It can be used to set and test conditions. CF and OF are both set to zero, PF, SF and ZF are all updated, AF is left undefined after the operation.	TEST AL, 80h
SHL / SAL des, count	Shift bits of word or byte left, by count. It puts zero(s) in LSB(s). MSB is shifted into the carry flag If more than one bits are shifted left, then the CF gels the most recently moved MSB. If the number of bits desired to be shifted is only 1, then the immediate number. 1 can be written as one of the operands. However, if the number of bits desired to be shifted is more than one, then the second operand is put in CL register.	SAL BX,01
SHR des, count	It shifts Bits of a byte or word to register put zero in MSB. LSB is moved into CF.	SHR BX,01 MOV CL,02 SHR BX,CL
SAR des, count	Shifts bits of word or byte right, but it retains the value of new MSB to that of old MSB. This also called arithmetic shift operation, as it does not change the MSB, Which is sign bit of a number.	SAR AL, 01 SAR BH,01
ROL des, count	Rotate bits of word or byte left, MSB is transferred to LSB and also to CF. The operation is called rotate as it Circulates bits. The operands can be register or memory operand
ROR ds, count	Rotate bits of word or byte right LSB is transferred to MSB and also to CF.	ROR BX,1
RCL des, count	Rotate bits of words or byte left, MSB to CF and CF to LSB. The operation is circular and involves carry flag in rotation.
RCR des, count	Rotate bits of word or byte right, LSB to CF to MSB. This instruction rotates left

Program Execution Transfer Instructions

These instructions are the ones that cause change in the sequence of execution of instruction. This change can be through a condition or sometimes may be unconditional. The conditions are represented by flags. For example, an instruction may be jump to an address if zero flag is set, that is the last ALU operation has resulted in zero value. These instructions are often used after a compare instruction, or some arithmetic instructions that are used to set the flags, for example, ADD or SUB. LOOP is also a conditional branch instruction and is taken till loop variable is below a certain count.

MNEMONIC		DESCRIPTION		EXAMPLE
CALL Proc 1		This function results in a procedure1 function call. The return address is saved on the stack. There are two basic types of CALLS. NEAR or Intra-Segment, calls: if the call is made to a procedure in the same segment as the calling program. FAR or Inter segment call: if the call is made to a procedure in the segment, other than the calling program. The saved return address for NEAR procedure call is just the IP. For FAR Procedure call IP and CS are saved as return address. A procedure can also be called indirectly, by first initializing some 16-bit register, or some other memory location with the new addresses as follows.		CALL procl CALL proc2 The new instruction address is determined by name declaration proc 1 is a near procedure, thus, only IP is involved. proc2 involves new CS: IP pair. On call to proc 1 stack<- font="font"> IP IP <= address offset of proc 1 on call to proc2 Stack [top] <= CS Stack [top] <= IP CS <- font="font"> code segment of proc2 IP <- font="font"> address offset of proc2 Here we assume that procl is defined within the same segment as the calling procedure, while proc2 is defined in another segment. As far as the calling program is concerned, both the procedures have been called in the same manner. But while declaring these procedures, we declare procl as NEAR procedure and proc2 as FAR procedure, as follows: procl PROC NEAR proc2 PROC FAR LEA BX, procl CALL BX
RET number		It returns the control from procedure to calling program. Every CALL should be a RET instruction. A RET instruction, causes return from NEAR or FAR procedure call. For return from near procedure the values of the instruction pointer is restored from stack: While for far procedure the CS: IP pair get is restored. RET instruction can also be followed by a number.		RET 6 ; In this case, 8086 ; increments the stack ; pointer by this number ; after popping off the 1P ; (for new) or IP and CS ; registers (for far) from ; The stack. This cancels ; the local parameters, or ; temporary parameters ; created by the ; Programmer. RET ; instruction does not ; affect any flags.
JMP label		Unconditionally go to specified address and get next instruction from the label specified. The label assigns the instruction to which jump has to take place within the program, or it could be a register that has been initialized with the offset value. JMP can be a NEAR JMP or a FAR jump, Just like call.		JMP CONTINUE ; CONTINUE is the label ; given to the instruction ; where the control needs ; To be transferred. JMP BX ; initialize BX with the ; offset of the instruction, ; where the control needs ; To be transfer
Conditional Jump		All the conditional jumps follow some conditional statement, or any instruction that affects the flag.		MOV CX. 05 MOV BX, 04 CMP  CX, BX ; this instruction will set ; various flags like the ZF, ; and the CF. JE LABELl ; conditional jump can ; now be applied, which ; checks for the ZF, and if ; it is set implying CX = ; BX, it makes ; a jump to LABELl, ; otherwise the control ; simply falls ; through to next ; instruction ; in the above example as ; CX is not equal to BX ; the jump will not take ; place and the next ; instruction to conditional ;jump instruction will be ; executed. However, if ; JNE (Jump if not equal ; to) or JA (Jump if above), ; or JAE (Jump above or ; equal) jump instructions ; if applied instead of JE, ; will cause the conditional ;Jump to occur
All the conditional jump instructions which are given below are self-explanatory. Please note that a "/" is used to separate two mnemonics which represent the same instruction.
JA / JNBE	Jump if above / Jump if not neither below nor equal.
JAEIJNB	Jump if above or equal Jump if not below
JB/JNAE	Jump if below1 Jump if not above nor equal
JBE/JNA	Jump if below or equal / Jump if not above
JC	Jump if carry flag set
JE/JZ	Jump if equal / Jump if zero flag is set
JNC	Jump if not carry
JNE/JNZ	Jump if not equal / Jump if zero flag is not set
JO	Jump if overflow flag is set
JNO	Jump if overflow flag is not set
JP/JPE	Jump if parity flag is set / Jump if parity even
JNP/JPO	Jump if not parity /Jump if parity odd
JG/JNLE	Jump if greater then / Jump if nut less than nor equal
JA/JNL	Jump if above / Jump if not less than
JL/JNGE	Jump if less than / Jump if not greater than nor equal
JLE/JNG	jump if less than or equal to / jump if not greater than
JS	Jump if sign flag is set
JNS	jump if sign flag is not set
LOOP label	This is a looping instruction of assembly. The number of times the looping is required is placed in CX register. Each iteration decrements CX register by one implicitly, and the Zero Flag is checked to check whether to loop again. If the zero flag is not set (CX is zero) greater than the control goes back to the specified label in the instruction, or else the control falls through to the next instruction. The LOOP instruction expects the label destination at offset of - 128 to +127 from the loop Instruction offset.		; Let us assume we want to ; add 07 to AL register, ; Three times. MOV CX,03 ; count of iterations L1 : ADD AL, LOOP L1 ; loop back to L1 ; until CX ; set (CX is zero) greater than the ; becomes equal to zero ;control goes back to the ; Loop affects no flags.
LOOPE/ LOOPZ label	Loop through a sequence of instructions while zero flag = 1 and CX is not equal to zero. There are two ways to exit out of the loop, firstly, when the count in the CX register becomes equal to zero, or when the quantities that are being compared become unequal.		Let us assume we have an array of 20 bytes. We want to see if all the elements of that array are equal to 0FFh or not. To scan 20 elements of the array, we loop 20 times. And we come out of the loop, when either the count of iterations has become equal to 20, or in other words CX register has decremented to zero, which means all the elements of the array are equal to 0FFh, or an element in the array is found which is not equal to 0FFh. In this case, the CX register may still be greater than zero, when the control comes out. This can be coded as follows: (Please note here that you might not understand everything at this place, that is because you are still not familiar with the various addressing modes. Just concentrate on the LOOPE instruction): MOV BX, OFFSET ARRAY ; Point BX at the start ; of the ARRAY DEC BX ; put number of ; array elements in CX MOV CX, 10 L1: INC BX ; point to ; next element in array CMP [BX] ,0FFh ; compare array element ; with 0FFh LOOPE L1 ; When the control comes ; out of the loop, it has ; either scanned all the ; elements and found them ; to be all equal to 0FFh, or ; it is pointing to the first ; non-0FFh, element in the ; Array.
LOOPNE/LOOPNZ label	This instruction causes Loop through a sequence of instructions while zero flag = 0 and CX is not equal to zero. This instruction is just the opposite of the previous instruction in its functionality.
JCXZ label	Jump to specified address if CX=0. This instruction will cause a jump, if the value of CX register is zero. Otherwise it will proceed with the next instruction in sequence.		This instruction is useful when you want to check whether CX is zero even prior to entering into a loop. Please note that LOOP instruction executes the loop at least once before decrementing and checking the value of CX register. Thus, CX=O will execute the loop once and decrement the CX register, making it 0FFFFh, which is non-zero: This will cause FFFFh times. Execution of loop. To avoid such type of conditions you can proceed as follows: JCXZ SKIP -LOOP ; if CX is already 0, skip ; loop L1 : SUB [BX],07h INC BX LOOP L1 ; loop until CX=0 SKIP LOOP
In addition to these instructions, there are other interrupt handling instructions also, which too transfer the control of the program to some specified location.

String Instructions

These are a very strong set of 8086 instructions as these instructions process strings, in a compact manner, thus, reducing the size of the program by a considerable amount. "String" in assembly is just a sequentially stored bytes or words. A string often consists of ASCII character codes. A subscript B following the instruction indicates that the string of bytes is to be acted upon, while "W" indicates that it is the string of words that is being acted upon.

MNEMONIC	DESCRIPTION	EXAMPLES
REP	This is an instruction prefix. It causes repetition of the following instruction till CX becomes zero. REP. It is not an instruction, but it is an instruction prefix that causes the CX register to be decremented. This prefix causes the string instruction to be repeated, until CX becomes equal to zero.	REP MOVSB STR1, STR2 The above example copies byte by byte contents. The CX register is initialized to contain the length of source string. REP repeats the operation MOVSB that copies the source string byte to destination byte. This operation is repeated until the CX register becomes equal to zero.
REPE/REPZ	It repeats the instruction following until CX =O or ZF is not equal to one. REPEIREPZ may be used / with the compare string instruction or the scan string instruction. REPE causes the string instruction to be repeated, till compared bytes or words are equal, and CX is not yet decremented to zero.
REPNE/REPNZ	It repeats instruction following it until CX =O or ZF is equal to 1. This comparison here is just inverse of REPE except for CX, which is checked to be equal to zero.
MOVS/MOVSB/MOVSW	It causes moving of byte or word from one string to another. This instruction assumes that: ·         Source string is in Data segment. ·         Destination string is in extra data segment ·         SI stores offset of source string in extra segment ·         DI stores offset of destination string is in data segment ·         CX contains the count of operation A single byte transfer rewires: ·         One byte transfer from source string to destination. ·         Increment of SI and DI to next byte ·         Decrement count register that is CX register	Assumes both data and extra segment start at address 1000 in the memory. Source string starts at offset 20h and the destination string starts at offset 30h. Length of the source string is 10 bytes. To copy the source string to the destination string, proceed as follows: MOV AX, 1000h MOV DS,AX ; initialize data segment and MOV ES,AX ; extra segment MOV SI,20h MOV DI,30h ;load offset of start of ;source string to SI ;load offset of start of ;destination string to DI MOV CX, 10 ; load length of string to CX ; as counter REP MOVSB ; Decrement CX and ; MOVSB until ; CX =o ; after move SI will be one ; greater than offset of last ; byte in source string, DI ; will be one greater than ; offset of last destination ; String. CX will be equal : To zero.
CMPS/CMPSB/CMPSW	It compares two string bytes or words. The source string and the destination strings should be present in data segment and the extra segment respectively. SI and DI are used as in the previous instruction. CX is used if more than one bytes or words are to be compared; however for such a case appropriate repeating prefix like REP, PEPE etc. need to be used.	MOV cx,10 MOV S1,OFFSET SRC-STR ; offset of source ; string in SI MOV DI, OFFSET DES-STR ; offset of destination ; string in DI REPE CMPSB ; Repeat the comparison of ; string bytes until ; end of string or until ; compared bytes are not ; Equal.
SCAS/SCASB/SCASW	It scans a string. Compare a string byte with byte in AL or a string word with a word in AX. The instruction does not change the operands in AL (AX) or the operand in the string. The string to be scanned must be present in the extra segment, and the offset of the string must be contained in the DI register. You can use CX if operation is to be repeated using REP prefixes.	MOV AL, 0Dh ; Byte to be scanned ; for in AL MOV D1,OFFSET DES-STR MOV CX,l0 REPNE SCAS DES-STR ; Compare byte in DES-STR ; with byte in AL register ; Scanning is repeated while ;the bytes are not equal and ;it is not end of string. If a ; carriage return 0Dh is ; found, ZF = DI will point ; at the next byte after the ; Carriage return. If a ; carriage return is not ; found then, ZF = 0 and ; CX = 0. SCASB or ; SCASW can be used to ; explicitly state whether ; the byte comparison or the ;word comparison is ; required.
LODS/LODSB/ LODSW	It loads string byte into AL or a string word into AX. The string byte is assumed to be pointed to by SI register. After the load, the SI pointer is automatically adjusted to point to the next byte or word as the case may be. This instruction does not affect any flag.	MOV S1,OFFSET SRC-STR LODS SRC-STR ; LODSB or LODSW can ; be used to indicate to the ; assembler, explicitly ; whether it is the byte that ; is required to be loaded or : the word
STOS/STOSB/ STOSW	It stores byte from AL or word from AX into the string present in the extra segment with offset given by DI. After the copy, DI is automatically adjusted to point to the next byte or word as per the instruction. No flags are affected.	MOVDI,OFFSET DES_STR STOSB DES_STR

Processor Control Instructions

The objectives of these instructions are to control the processor. This raises two questions:

·         How can you control processor, as this is the job of control unit?

·         How much control of processor is actually allowed?

Well, 8086 only allows you to control certain control flags that causes the processing in a certain direction, processor synchronization if more than one processor are attached through LOCK instruction for buses etc.

MNEMONIC	DESCRIPTION	EXAMPLE
STC	It sets carry flag to 1.
CLC	It clears the carry flag to 0.
CMC	It complements the state of the carry flag from 0 to 1 or 1 to 0 as the case may be.	CMC; Invert the carry flag
STD	It sets the direction flag to 1. The string instruction moves either forward (increment SI, DI) or backward (decrement SI, DI) based on this flag value. STD instruction does not affect any other flag. The set direction flag causes stings to move from right to left.
CLD	This is opposite to STD, the string operation occurs in the reverse direction:	CLD ; Clear the direction flag ; so that the string pointers ; Auto-increment. MOV AX, 1 000h MOV DS, AX ; Initialize data segment ; and extra segment MOV ES, AX MOV SI, 20h ; Load offset of start of ; source string to SI MOV DI,30h ; Load offset of start of ; destination string to DI MOV CX,10 ; Load length of string to ; CX as counter REP MOVSB ; Decrement CX and ; increment ; SI and DI to point to next ; byte, then MOVSB until ;CX=0

Note: These instructions include instructions for setting and closing interrupt flag, halting the computer, LOCK (locking the bus), NOP etc.

ADDRESSING MODES

The basic set of operands in 8086 may reside in register, memory and immediate operand. How can these operands be accessed through various addressing modes? The answer to the question above is given in the following sub-section. Large number of addressing modes helps in addressing complex data structures with ease. Some specific Terms and registers roles for addressing:

·         Base register (BX, BP): These registers are used for pointing to base of an array, stack etc.

·         Index register (SI, DI): These registers are used as index registers in data and/or extra segment.

·         Displacement: It represents offset from the segment address.

·          

Addressing modes of 8086

Mode	Description	Example
Direct	Effective address is the displacement of memory variable.
Register Indirect	Effective address is the contents of a register.	[BX]
		[SI]
		[DI]
		[BP]
Based	Effective address is the sum of a base register and a displacement.	LIST [BX] (OFFSET LIST + BX)
		[BP + 1
Indexed	Effective address is the sum of an index register and a displacement.	LIST [SI]
		[LIST + DI]
		[DI +2]
Based Indexed	Effective address is the sum of a base and an index register.	[BX + SI]
		[BX] [DI]
		[BP + DI]
Based Indexed with displacement	Effective address is the sum of a base register, an index register, and a displacement.	[BX + SI + 2]

Register Addressing Mode

Operand can be a 16-bit register:

Addressing Mode	Description	Example
AX, BX, CX, DX, SI, DI,BP,IP,CS ,DS,ES,SS Or it may be AH, AL, BH, BL, CH, CL, DH, DL	In general, the register addressing mode is the most efficient because registers are within the CPU and do not require memory access.	MOV AL,CH MOV AX,CX

Immediate Addressing Mode

An immediate operand can be a constant expression, such as a number, a character, or an arithmetic expression. The only constraint is that the assembler must be able to determine the value of an immediate operand at assembly time. The value is directly inserted into the machine instruction.

MOV AL, 05

Mode	Description	Example
Immediate	Please note in the last examples the expression (2 + 3)/5, is evaluated at assembly time.	MOV AL, 10 MOV AL,'A' MOV AX,'AB’ MOV AX, 64000 MOV AL, (2 + 3)/5

Direct Addressing Mode

A direct operand refers to the contents of memory at an address implied by the name of the value.

Mode	Description	Example
DIRECT	The direct operands are also called as Re-locatable operands as they represent the offset of a label from the beginning of a segment. On reloading a program even in a different segment will not cause change in the offset that is why we call them re-locatable. Please note that a variable is considered in Data segment (DS) and code label in code segment (SS) by default. Thus, in the example, COUNT, by default will be assumed to be in data segment, while LABEL 1, will be assumed to be in code segment. If we specify, as a direct operand then the address is non-re-locatable. Please note the value of segment register will be known only at the run time.	MOV COUNT, CL ; move CL to COUNT (a byte variable) MOV AL,COUNT ; move COUNT to AL JMP LABEL1 ;jump to LABEL1 MOV AX,DS:5 ; segment register and ; offset MOV BX,CSEG:2Ch ; segment name and offset MOV AX,ES:COUNT ; segment register and ; Variable. ; The offsets of these ; variables are calculated ; with respect to the ; segment name (register) ; specified in the ; instruction.

Indirect Addressing Mode

In indirect addressing modes, operands use registers to point to locations in memory, So, it is actually a register indirect addressing mode. This is a useful mode for handling strings/ arrays etc. For this mode two types of registers are used. These are:

·         Base register BX, BP

·         Index register SI, DI

BX contain offset/ pointer in Data Segment

BP contains offset/ pointer in Stack segment.

-SI contains offset/pointer in Data segment.

DI contains offset /pointer in extra data segment.

There are five different types of indirect addressing modes:

1.       Register indirect

2.       Based indirect

3.       Indexed indirect

4.       Based indexed

5.       Based indexed with displacement.

Mode	Description	Example
Register indirect	Indirect operands are particularly powerful when processing list of arrays, because a base or an index register may be modified at runtime.	MOV BX, OFFSET ARRAY ; point to start of array MOV AL,[BX] ; get first element INC BX ; point to next MOV DL,[BX] ; get second element The brackets around BX signify that we are referring to the contents of memory location, using the address stored in BX. In the following example, three bytes in an array are added together: MOV S1,OFFSET ARRAY ; address of first byte MOV AL,[SI] * ; move the first byte to AL IMC SI ;point to next byte ADD AL, [SI] ;add second byte INC SI ;point to the third byte ADD AL, [SI] ;add the third byte
Based Indirect and Indexed Indirect	Based and indirect addressing modes are used in the same manner. The contents of a register are added to a displacement to generate an effective address. The register must be one of the following: SI, DI, BX or BP. If the registers used for displacement are base registers, BX or BP, it is said to be base addressing or else it is called indexed addressing. A displacement is either a number or a 1abel whose offset is known at assembly time. The notation may take several equivalent forms. If BX, SI or DI is used, the effective address is usually an offset from the DS register; BP on the other hand, usually contains an offset from the SS register.	; Register added to an offset MOV DX, ARRAY[BX] MOV DX,[DI + ARRAY] MOV DX,[ARRAY + SI] ; Register added to a constant MOV AX,[BP + 21 MW DL,[DI - 21 ; DI + (-2) MOV DX,2[SI]
Based Indexed	In this type of addressing the operand's effective address is formed by combining a base register with an index register.	MOV AL,[BP] [SI] MOV DX,[BX 1 SI] ADD CX,[DII [BX] ; Two base registers or two ; index registers cannot be ; combined, so the ; following would be ; incorrect: MOV DL,[BP + BX] ; error : two base registers MOV AX,[SI + DI] ; error : two index registers
Based Indexed with Displacement	The operand's effective address is formed by combining a base register, an index register, and a displacement.	MOV DX, ARRAY [BX][SI] MOV AX, [RX + SI + ARRAY] ADD DL,[BX + SI + 3] SUB CX, ARRAY [BP +SI] Two base registers or two index registers cannot be combined, so the following would be incorrect: MOV AX,[BP + BX + 21 MOV DX,ARRAY [SI + DI]

To summarize the features of 8086, we can say 8086 has:

·         a 16-bit data bus

·         A 20-bit address bus

·         CPU is divided into Bus Interface Unit and Execution Unit

·         6-byte instruction pre fetch queue

·         Segmented memory

·         4 general purpose registers (each of 16 bits)

·         Instruction pointer and a stack pointer

·         set of index registers

·         Powerful instruction set

·         Powerful addressing modes

·         designed for multiprocessor environment

·         Available in versions of 5 MHz and 8 MHz clock speed.

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