Successfully reported this slideshow.
Your SlideShare is downloading. ×

Assembly language 8086 intermediate

Loading in …3

Check these out next

1 of 37 Ad

More Related Content

Slideshows for you (20)

Viewers also liked (20)


Similar to Assembly language 8086 intermediate (20)


Recently uploaded (20)

Assembly language 8086 intermediate

  2. 2. Machine Code 2  There are occasions when the programmer must program at the machine’s own level.  Machine Code programs are tedious to write and highly error prone. 0000111100001111 In situations where a high-level 0010010101010100 language is inappropriate we 1010101010100101 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.
  3. 3. 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.
  4. 4. Compilers / Assemblers 4  High-level Languages such as Pascal Program Pascal programs are sometimes Compiler converted firstly to assembly language by a computer program Assembler language Program called compiler and then into machine code by another Assembler program called assembler Machine Code Program This version is actually loaded and executed
  5. 5. General Purpose Registers 5 AH AL  There are 4 general purpose registers in AX the 8086. BH BL  They are all 16-bit registers  Each byte can be BX addressed individually CH CL by specifying the High order or the Low order byte of the register. CX DH DL DX
  6. 6. 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.
  7. 7. Number Formats 7 AX AH AL MOV AH,01010101B 0 1 0 1 0 1 0 1 MOV AL,00100111B 0 1 0 1 0 1 0 1 0 0 1 0 0 1 1 1 MOV AX,3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 MOV AH,AL 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 MOV AL,10D 0 0 0 0 0 0 1 1 0 0 0 0 1 0 1 0 MOV AL,10H 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 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.
  8. 8. 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
  9. 9. 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.
  10. 10. 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. …
  11. 11. Labels 11  We saw that the jump instruction has a general format JE <label> where <label> 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.
  12. 12. 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
  13. 13. 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
  14. 14. The Logical Family 14 AND NOT (Invert: One’s Complement) Contents of AX = 0000101011100011 Contents of AX = 0000101011100011 Contents of BX = 1001100000100001 Contents of AX = 1111010100011100 Contents of AX = 0000100000100001 after NOT AX is executed after AND AX,BX is executed OR TEST Contents of AX = 0000101011100011 Contents of AX = 0000101011100011 Contents of BX = 1001100000100001 Contents of BX = 1001100000100001 Contents of AX = 1001101011100011 Contents of AX = 0000101011100011 after OR AX,BX is executed after TEST AX,BX is executed XOR Similar to AND but the result is not stored in AX Contents of AX = 0000101011100011 but only the Z-flag is changed Contents of BX = 1001100000100001 NEG (Two’s Complement) Contents of AX = 1001001011000010 Contents of AX = 0000101011100011 after XOR AX,BX is executed Contents of AX = 1111010100011101 after NEG AX is executed
  15. 15. Use of Logical Family 15 Symbol ASCII (Dec) ASCII (Hex) By Making an AND between an ASCII value and 0FH 0 48 30 we can obtain the required number. 1 49 31 Say we AND 33H = 00110011B 2 50 32 with 0FH = 00001111B 3 51 33 We obtain = 00000011B (3) 4 52 34 5 53 35 By Making an OR between a number value and 30H we 6 54 36 can obtain its ASCII code. 7 55 37 Say we OR 05H = 00000101B 8 56 38 with 30H = 00110101B We obtain = 00110101B 9 57 39 (ASCII value for ‘5’)
  16. 16. 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. Suppose we want to test the if the AX = 0101010100011001 6th. bit of AX is 1 or 0: 04H = 0000000000000100 AX = 0101010100011001 OR AX,04H = 0101010100011101 20H = 0000000000100000 AND AX,20H = 0000000000000000 Suppose we want to set the 5th. So if the result is 0 then that bit of AX to 0 leaving the particular bit was 0, 1 otherwise others unchanged. AX = 0101010100011001 0FFEFH = 1111111111101111 AND AX,0FFEFH = 0101010100001101
  17. 17. 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) Data can be retrieved from memory, one or [0000] two bytes at a time: [0001] [0002] MOV AL, [20H] will transfer the [0003] Contents of location 20H to AL. [0004] [0005] MOV BX, [20H] will transfer the contents of [0006] [0007] locations 20H and 21H to BX. [0008] MOV [20H], AL will transfer the contents of [0009] [000A] AL to memory location 20H [000B] Location ADDRESS [000C] Location CONTENTS
  18. 18. 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
  19. 19. 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
  20. 20. The Instruction Pointer (IP) 20  The computer keeps track of the next line to be executed by keeping its address in a special START . This is the register called the Instruction Pointer (IP) or line which is Program Counter. . . executing  This register is relative to CS as segment register and points to the next instruction to MOV AX,BX be executed. MOV CX,05H  The contents of this register is updated with MOV DX,AX IP every instruction executed. .  Thus a program is executed sequentially line . by line .
  21. 21. 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
  22. 22. The Stack Pointer 22 [0000] [0002]  The Stack pointer keeps track of the [0004] position of the last item placed on the [0006] [0008] stack (i.e. the Top Of Stack) [000A] [000C] SP [000E] [0010] [0012] [0014]  The Stack is organized in words, (i.e. two [0016] bytes at a time). Thus the stack pointer is [0018] incremented or decremented by 2. Note that on placing items on the  The Stack Pointer points to the last stack the address decreases occupied locations on the stack
  23. 23. PUSH & POP 23 PUSH AX AX  The two set of instructions which [0000] [0002] [0004] explicitly modify the stack are the [0006] [0008] NEW SP PUSH (which places items on the OLD SP [000A] [000C] stack) and the POP (which [000E] [0010] retrieves items from the stack). In [0012] [0014] both cases, the stack pointer is [0016] [0018] adjusted accordingly to point POP AX always to the top of stack. [0000] AX [0002]  Thus PUSH AX means SP=SP-2 [0004] [0006] and AX -> [SP] [0008] [000A] OLD SP NEW SP [000C]  POP AX means [SP] -> AX and [000E] [0010] SP=SP+2. [0012] [0014] [0016] [0018]
  24. 24. Subroutines 24  In high-level languages, procedures START SUB1 PROC . make it possible to break a large . . program down into smaller pieces so . . . RET that each piece can be shown to work independently. In this way the final CALL SUB1 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.
  25. 25. The CALL Mechanism 25  Although at first sight the CALL START SUB1 PROC and RET mechanism can be . . . implemented by using two JMP’s. . . In fact this cannot be done since . RET the CALL mechanism remembers CALL SUB1 the place where it was called from 1 . and returns to the line following it. . Thus this is not a fixed address. . CALL SUB1 . 2 . .
  26. 26. 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 it’s 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)
  27. 27. 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
  28. 28. Some INT 21H functions 28 Function Description Explanation Number 1 Keyboard Waits until a character is typed at the keyboard and then puts the ASCII Input code for that character in register AL and echoed to screen (echoed) 2 Display Prints the character whose ASCII code is in DL Output 8 Keyboard Waits until a character is typed at the keyboard and then puts the ASCII Input code for that character in register AL and NOT echoed to screen (No echo) 9 Display Prints a series of characters stored in memory starting with the one in the String address given in DX (relative to DS).Stop when the ASCII code for $ is encountered
  29. 29. 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 This is only a INT 21H program fragment to illustrate the use of GET: MOV AH,01H interrupt 21H – For INT 21H full details consult the MASM notes CMP AL,01H JE NEXT MOV DX, OFFSET Song1 MOV AH,09H INT 21H
  30. 30. Addition and Subtraction with carry or 30 borrow  In assembly language there are two versions of addition and two versions of subtraction. CF CF  ADD - Simple addition of two numbers 0  ADC - Adds two numbers together with the carry flag 0  SUB – Simple subtraction of two numbers  SBB – Subtracts the second number and Last 0 1 1 the carry flag (borrow) addition in 00 01 98 41 +  This provides a means of adding numbers case of an 00 02 71 64 outgoing 00 04 70 05 greater than 32-bits. carry  CLC clears the carry for the first digit addition
  31. 31. 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
  32. 32. Addressing Modes 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 Computer Logic II
  33. 33. Register Addressing 33  This addressing mode General Purpose Segment Registers involves the contents of AX AH AL CS the register directly as BX BH BL DS for example: CX CH CL SS  MOV AX, BX DX DH DL ES  MOV CL, DL  Note that the IP and SI FLAGS Flags register cannot be DI IP accessed directly by the SP programmer BP AX BX Ex. MOV AX,BX
  34. 34. Immediate and Direct Addressing 34  In Immediate addressing – for example Ex. MOV CL,61H MOV CL,61H – the immediate operand CL 61H 61H is stored as part of the instruction. Thus the number 61H is loaded directly in CL.  Direct addressing is similar except that in Ex. MOV AL,[210H] this case the effective address of one of the AL operands is taken directly from the instruction. Thus in MOV AL, [210H] the contents of location 210H relative to DS is put in AL (DS:210H) 75H
  35. 35. Default Segment Register 35 Note that the default segment register can be changed using the segment override, i.e. stating the whole General Purpose address in the form DS: Offset AX AH AL BX BH BL Relative to DS by default CX CH CL DX DH DL NORMALLY FOR STRINGS SI Relative to DS by default DS DI Relative to DS by default ES SP Relative to SS by default BP Relative to SS by default IP Relative to CS by default
  36. 36. 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 (2’s Complement)  NOT: Compliment (1’s Complement)
  37. 37. 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