assemblylanguage8086intermediate-120317144353-phpapp01
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INTERMEDIATE 8086 ASSEMBLY
LANGUAGE PROGRAMMING
Cutajar & Cutajar
2012
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Machine Code 2
There are occasions when the programmer must program at the machines own level.
Machine Code programs are tedious to write and highly error prone.
0000111100001111
0010010101010100
1010101010100101
In situations where a high-level language is inappropriate we avoid working in machine code most of the time by making the computer do more of the work. Thus we write in assembly language and then the computer converts this assembly language program into machine code.
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Assembly Language 3
In assembly language, a mneumonic (i.e. memory aid) is used as a short notation for the instruction to be used.
Assembly Machine Code
Language
SUB AX,BX 001010111000011
MOV CX,AX 100010111001000
MOV DX,0 10111010000000000000000
Assembly language is an intermediate step between high level languages and machine code. Most features present in HLL are not present in Assembly Language as type checking etc.
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Compilers / Assemblers 4
High-level Languages such as Pascal programs are sometimes converted firstly to assembly language by a computer program called compiler and then into machine code by another program called assembler
Pascal Program
Assembler language Program
Machine Code Program
This version is actually loaded and
executed
Compiler
Assembler
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General Purpose Registers
There are 4 general
purpose registers in
the 8086.
They are all 16-bit
registers
Each byte can be
addressed individually
by specifying the High
order or the Low order
byte of the register.
5
AX
AH AL
BX
BH BL
CX
CH CL
DX
DH DL
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Some Simple Commands 6
MOV AX,3 ; Put 3 into register AX ADD AX,2 ; Add 2 to the contents of AX MOV BX,AX ; Copy the contents of AX in BX INC CX ; Add 1 to the contents of CX DEC DX ; Subtract 1 from the contents of DX SUB AX,4 ; Subtract 4 from the contents of AX MUL BX ; Multiply the contents of AX with BX leaving
; the answer in DX-AX DIV BX ; Divide the contents of DX-AX by BX leaving
; the quotient in AX and remainder in DX.
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Number Formats 7
0 1 0 1 0 1 0 1
AX
0 0 1 0 0 1 1 1 0 1 0 1 0 1 0 1 MOV AH,01010101B MOV AL,00100111B
0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 MOV AX,3 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 MOV AH,AL 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 1 MOV AL,10D 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 1 MOV AL,10H
In case a number is moved (copied) into the register the base of a is specified by a letter B for Binary, D for Decimal and H for Hex.
AH AL
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AMBIGUITY 8
Consider the instruction MOV DL, AH Does it mean copy the contents of register AH to DL or Does it mean copy A in hexadecimal into register DL
To avoid this ambiguity all hexadecimal numbers must start with a number. This can always be done by preceding a number starting with A,B,C,D,E and F with a preceding zero to remove ambiguity. Thus MOV DL, AH means copy AH to DL whilst MOV DL, 0AH means sore hexadecimal A to DL
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The Flags Register 9
Some of the instructions (but not all) affect the flag register. The flag register signals the status of the CPU after the last operation performed. For example if SUB AX,2 results in zero the ZF get 1 (lights on) indicating that
the result of the last operation was zero.
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JUMPS 10
Jump instructions allow the 8086 to take decisions according to information provided by the flag register.
For example, if AX and BX contain the ASCII code for the same letter then do one thing, if not then do another.
`
CMP AX,BX ; Compares the contents of BX with that of AX JE SAME ; Jump if they are equal to the point ; in the code labeled SAME ; Obey these instructions if the contents of AX ; is not equal to that of BX SAME: MOV CX,AX ; Program continues from here if AX = BX.
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Labels 11
We saw that the jump instruction has a general format JE where is a facility offered by the assembler.
These labels are converted by the assembler to exact address where the program is to continue. Labels must start with a letter and can contain thereafter letters, numbers and
underscores (_). Spaces and punctuation marks are not permitted Avoid using keywords in labels Once_again, Next, Name34, this_37 are permitted as labels 3rdday, tues+wed and semi;colons are not permitted as labels.
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JUMP Conditions 12
JA/JNBE (CF and ZF) = 0 Above / Not Below or Equal JAE/JNB CF = 0 Above or Equal / Not Below JB/JNAE/JC CF = 1 Below / Not Above or Equal / Carry JBE/JNA (CF or ZF) = 1 Below or Equal / Not Above JE/JZ ZF = 1 Equal / Zero JMP none Unconditionally JNC CF = 0 No Carry JNE/JNZ ZF = 0 Not Equal / Not Zero JNO OF = 0 No Overflow JNP/JPO PF = 0 No Parity / Parity Odd JNS SF = 0 No Sign / Positive JO OF = 1 Overflow JP/JPE PF = 1 Parity / Parity Even JS SF = 1 Sign JG/JNLE ZF = 0 and SF = OF Greater / Not Less nor Equal JGE/JNL SF = OF Grater or Equal / Not Less JL / JNGE SF OF Less / Not Greater nor Equal JLE/JNG (ZF = 1) or (SF OF) Less or equal / not greater JCXZ Register CX = 0 CX is equal to zero
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Example using Jumps 13
MOV CX, AX ; Keep a copy of AX before modification SUB AX,BX ; AX := AX BX JZ MAKE1 ; This is instruction will cause execution ; to continue from MAKE1 if AX was ; equal to BX (subtraction resulted in Zero) MOV DX, 0 ; Otherwise store 0 in DX JMP RESET ; Jump to RESTORE where AX is restored ; thus avoiding the next instruction MAKE1: MOV DX, 1 ; If AX = BX then we set DX to 1 RESET: MOV AX, CX ; Restore the old value of AX
Note that in the Code a colon ends a label position
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The Logical Family 14
OR
Contents of AX = 0000101011100011
Contents of BX = 1001100000100001
Contents of AX = 1001101011100011
after OR AX,BX is executed
AND
Contents of AX = 0000101011100011
Contents of BX = 1001100000100001
Contents of AX = 0000100000100001
after AND AX,BX is executed
XOR
Contents of AX = 0000101011100011
Contents of BX = 1001100000100001
Contents of AX = 1001001011000010
after XOR AX,BX is executed
NOT (Invert: Ones Complement)
Contents of AX = 0000101011100011
Contents of AX = 1111010100011100
after NOT AX is executed
TEST
Contents of AX = 0000101011100011
Contents of BX = 1001100000100001
Contents of AX = 0000101011100011
after TEST AX,BX is executed
Similar to AND but the result is not stored in AX but only the Z-flag is changed
NEG (Twos Complement)
Contents of AX = 0000101011100011
Contents of AX = 1111010100011101
after NEG AX is executed
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Use of Logical Family 15
Symbol ASCII (Dec) ASCII (Hex) 0 48 30 1 49 31 2 50 32 3 51 33 4 52 34 5 53 35 6 54 36 7 55 37 8 56 38 9 57 39
By Making an AND between an ASCII value and 0FH we can obtain the required number. Say we AND 33H = 00110011B
with 0FH = 00001111B
We obtain = 00000011B (3)
By Making an OR between a number value and 30H we can obtain its ASCII code. Say we OR 05H = 00000101B
with 30H = 00110101B
We obtain = 00110101B
(ASCII value for 5)
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Masking 16
By the use of masking we can set or test individual bits of a register
Suppose we want to set the 3rd.
bit of AX to 1 leaving the
others unchanged.
AX = 0101010100011001
04H = 0000000000000100
OR AX,04H = 0101010100011101
Suppose we want to set the 5th.
bit of AX to 0 leaving the
others unchanged.
AX = 0101010100011001
0FFEFH = 1111111111101111
AND AX,0FFEFH = 0101010100001101
Suppose we want to test the if the
6th. bit of AX is 1 or 0:
AX = 0101010100011001
20H = 0000000000100000
AND AX,20H = 0000000000000000
So if the result is 0 then that
particular bit was 0, 1 otherwise
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Instructions which affect Memory 17
Computer memory is best thought of numbered pigeon holes (called locations), each capable of storing 8 binary digits (a byte)
[0000] [0001] [0002] [0003] [0004] [0005] [0006] [0007] [0008] [0009] [000A] [000B] [000C]
Data can be retrieved from memory, one or two bytes at a time: MOV AL, [20H] will transfer the Contents of location 20H to AL. MOV BX, [20H] will transfer the contents of locations 20H and 21H to BX. MOV [20H], AL will transfer the contents of AL to memory location 20H
Location ADDRESS Location CONTENTS
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Changing addresses 18
Varying an address whilst a program is running involves specifying the locations concerned in a register.
From all the general purpose registers BX is the only capable of storing such addresses. Thus MOV AX, [CX] is illegal Whilst MOV CL, [BX] copies the contents of memory location whose address is
specified by BX into the register CL. And MOV [BX], AL copies the contents of AL in the memory location whose
address is specified in BX
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Examples Affecting Memory 19
Consider the checkerboard memory test where a section of memory is filled with alternate 01010101 and 10101010.
The following program does the checkerboard test on locations 200H-300H inclusive.
MOV BX,200H
MOV AX,1010101001010101B
NEXT: MOV [BX],AX
INC BX
CMP BX,300H
JLE NEXT
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The Instruction Pointer (IP) 20
The computer keeps track of the next line to be executed by keeping its address in a special register called the Instruction Pointer (IP) or Program Counter.
This register is relative to CS as segment register and points to the next instruction to be executed.
The contents of this register is updated with every instruction executed.
Thus a program is executed sequentially line by line
MOV AX,BX
MOV CX,05H
MOV DX,AX
START
.
.
.
.
.
.
This is the
line which is
executing
IP
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The Stack 21
The Stack is a portion of memory which, like a stack of plates in a canteen, is organized on a Last-In-First-Out basis.
Thus the item which was put last on the stack is the first to be withdrawn
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The Stack Pointer 22
The Stack pointer keeps track of the position of the last item placed on the stack (i.e. the Top Of Stack)
The Stack is organized in words, (i.e. two
bytes at a time). Thus the stack pointer is incremented or decremented by 2.
The Stack Pointer points to the last occupied locations on the stack
[0000] [0002] [0004] [0006] [0008] [000A] [000C] [000E] [0010] [0012] [0014] [0016] [0018]
SP
Note that on placing items on the stack the address decreases
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PUSH & POP 23
The two set of instructions which explicitly modify the stack are the PUSH (which places items on the stack) and the POP (which retrieves items from the stack). In both cases, the stack pointer is adjusted accordingly to point always to the top of stack.
Thus PUSH AX means SP=SP-2 and AX -> [SP]
POP AX means [SP] -> AX and SP=SP+2.
[0000] [0002] [0004] [0006] [0008] [000A] [000C] [000E] [0010] [0012] [0014] [0016] [0018]
OLD SP
PUSH AX
OLD SP NEW SP
AX
[0000] [0002] [0004] [0006] [0008] [000A] [000C] [000E] [0010] [0012] [0014] [0016] [0018]
OLD SP
POP AX
OLD SP NEW SP
AX
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Subroutines 24
In high-level languages, procedures make it possible to break a large program down into smaller pieces so that each piece can be shown to work independently. In this way the final program is built up of a number of trusty bricks and is easier to debug because the error is either localized to one subprogram or its interlinking. This has also the advantage of re-usability of bricks.
CALL SUB1
START SUB1 PROC
RET
.
.
.
.
.
.
.
.
.
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The CALL Mechanism 25
Although at first sight the CALL and RET mechanism can be implemented by using two JMPs. In fact this cannot be done since the CALL mechanism remembers the place where it was called from and returns to the line following it. Thus this is not a fixed address.
CALL SUB1
START SUB1 PROC
RET
.
.
.
.
.
.
.
.
.
CALL SUB1 .
.
.
1
2
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The Return Mechanism 26
When a CALL is encountered the current value of the instruction pointer is pushed on the stack and the it is filled with the address stated by the call.
Since the fetch cycle goes to search for the instruction pointed at by the instruction pointer, the program continues its execution from the first statement in the
subroutine. On encountering the RET instruction the contents of the IP is popped from the stack
thus continuing the execution where it was suspended. Thus care must be taken to leave the return address intact before leaving a
subroutine. (i.e. a symmetrical number of pushes and pops within the subroutine)
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Software Interrupts 27
Software interrupts are like hardware interrupts which are generated by the program itself. From the interrupt number, the CPU derives the address of the Interrupt service routine which must be executed.
Software interrupts in assembly language can be treated as calls to subroutines of other programs which are currently running on the computer.
One of the most famous software interrupt is Interrupt No. 21H, which branches in the operating system, and permits the use of PC-DOS functions defined there. The function required to be performed by DOS is specified in AH prior to the the
interrupt. The functions return and accept values in various registers. AN interrupt is called using the instruction INT followed by the interrupt number
. For example: INT 21H
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Some INT 21H functions 28
Function Number
Description Explanation
1 Keyboard Input (echoed)
Waits until a character is typed at the keyboard and then puts the ASCII code for that character in register AL and echoed to screen
2 Display Output
Prints the character whose ASCII code is in DL
8 Keyboard Input (No echo)
Waits until a character is typed at the keyboard and then puts the ASCII code for that character in register AL and NOT echoed to screen
9 Display String
Prints a series of characters stored in memory starting with the one in the address given in DX (relative to DS).Stop when the ASCII code for $ is encountered
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INT 21H Example 29
Prompt DB Please enter 1 or 2: ,13D,10D,$ Song1 DB So you think you can tell heaven from hell Song2 DB Blue Sky is in pain,13D,10D,$
ASK: MOV DX, OFFSET Prompt
MOV AH,09H
INT 21H
GET: MOV AH,01H
INT 21H
CMP AL,01H
JE NEXT
MOV DX, OFFSET Song1
MOV AH,09H
INT 21H
This is only a
program fragment to
illustrate the use of
interrupt 21H For full details consult the
MASM notes
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Addition and Subtraction with carry or
borrow 30
In assembly language there are two versions of addition and two versions of subtraction. ADD - Simple addition of two numbers ADC - Adds two numbers together with
the carry flag SUB Simple subtraction of two
numbers SBB Subtracts the second number and
the carry flag (borrow) This provides a means of adding numbers
greater than 32-bits. CLC clears the carry for the first digit
addition
CF CF
0
0
0 1 1
00 01 98 41 +
00 02 71 64
00 04 70 05
Last
addition in
case of an
outgoing
carry
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The Compare Instruction 31
The compare instruction does not change the contents of the registers involved but only sets the flag register accordingly.
The actual operation performed by the compare is a subtraction, leaving the source and destination registers intact
Consider CMP AX,BX : Flags are set according to the result of subtracting BX from AX: If AX = BX then the ZF is set to 1 If AX > BX then the ZF is set to 0 and CF is set to 0 too If AX < BX then we need an external borrow, which is reflected in CF = 1 These flags are tested in the ABOVE or BELOW jumps which test unsigned numbers The GREATER and LESS jumps are for signed numbers and work on the SF, OF
and the ZF instead
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Addressing Modes
Computer Logic II
32
The addressing modes deal with the source and destination of the data required by the instruction. This can be either a register or a location in memory, or even a port.
Various addressing modes exist: Register Addressing Immediate and Direct Addressing Indirect Addressing Indexed Addressing Based Addressing Based-Indexed Addressing
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Register Addressing 33
This addressing mode involves the contents of the register directly as for example: MOV AX, BX MOV CL, DL
Note that the IP and Flags register cannot be accessed directly by the programmer
AX AH AL
BX BH BL
CX CH CL
DX DH DL
SI
DI
SP
BP
IP
FLAGS
General Purpose
SS
CS
DS
ES
Segment Registers
Ex. MOV AX,BX AX BX
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Immediate and Direct Addressing 34
In Immediate addressing for example MOV CL,61H the immediate operand 61H is stored as part of the instruction. Thus the number 61H is loaded directly in CL.
Direct addressing is similar except that in this case the effective address of one of the operands is taken directly from the instruction. Thus in MOV AL, [210H] the contents of location 210H relative to DS is put in AL
CL
Ex. MOV CL,61H
61H
Ex. MOV AL,[210H]
AL
(DS:210H) 75H
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Default Segment Register 35
AX AH AL
BX BH BL
CX CH CL
DX DH DL
SI
DI
SP
BP
IP
General Purpose
Relative to DS by default
Relative to DS by default
Relative to SS by default
Relative to SS by default
Relative to CS by default
Relative to DS by default
NORMALLY FOR STRINGS
DS
ES
Note that the default segment register can be changed using the segment override, i.e. stating the whole address in the form DS: Offset
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Some other useful Instructions 36
CLC: Clear Carry Flag (CF = 0) STC: Set Carry Flag (CF = 1) CMC : Complement Carry Flag (CF = CF) CBW: Convert Byte to Word CWD: Convert Word to Double-Word NEG: Negate (2s Complement) NOT: Compliment (1s Complement)
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Reference Books 37
Programming the 8086/86 for the IBM PC and Compatibles . Michael Thorne Microprocessors and Interfacing Programming and Hardware Douglas V.Hall Microsoft Macro Assembler for the MS-DOS Operating Systems Reference
Manual
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