Microprocessor Chapter3 hawassa Univetsi

Microprocessor Chapter3 hawassa Univetsity

1. OUTLINE OF THE CHAPTER HAWASSA UNIVERSITY INISTITUTE OF TECHNOLOGY DEPARTMENT OF COMPUTER SCIENCE CHAPTER THREE Intel 8086 PROCESSOR PROGRAMING & INSTRUCTION SETS

2. Addresing mode The 8086 has about 117 different instructions with about 300 opcodes. The 8086 instruction sets can contain no operand , single operand , and two operand instructions. The processor can access memory in different ways that are collectively called addressing mode . The addressing modes describe the types of operands and the way they are accessed for executing an instruction. The number of addressing modes is determined when the microprocessor is designed and cannot be changed .

3. 8086 ADDRESSING MODES The 8086 provides a total of seven distinct addressing modes: Register addressing modes Immediate addressing modes Direct addressing modes Register indirect addressing modes Based relative addressing modes Indexed relative addressing modes Based indexed relative addressing modes MOV instructions are used to explain addressing modes.

4. 8086 ADDRESSING MODES (CONT..) REGISTER ADDRESSING MODE: MOV REG1, REG2; The register addressing mode involves the use of registers to hold the data to be manipulated. Memory is not accessed when this addressing mode is executed; Relatively fast transfer since memory is not accessed. Examples: MOV BX, DX ; copy the contents of DX into BX MOV ES, AX ; copy the contents of AX into ES ADD AL, BH ; add the contents of BH to Contents of AL. The size of reg1 and reg2 must be the same. MOV CL, AX is illegal for instance.

5. 8086 ADDRESSING MODES (CONT..) B) IMMEDIATE ADDRESSING MODE :- MOV REG, CONSTANT In the immediate addressing mode, the source operand is a constant. It can be used to load information into any of the registers except the segment registers and flag register. Examples: MOV AX, 2550H ; move 2550H into AX MOV CX, 625 ; load the decimal value 625 into CX MOV BL, 40H ; load 40H into BL MOV DS, 0123H; is illegal! Instead we can use: MOV AX, 0123H MOV DS, AX

6. 8086 ADDRESSING MODES (CONT..) C) DIRECT ADDRESSING MODE: MOV REG, [constant] or MOV [constant], REG Here constant is not operand but it is an offset or EA in memory of operand. In the direct addressing mode the data is in some memory location(s) and the address of the data in memory comes immediately after the instruction. This address is the offset address and one can calculate the physical address by shifting left the DS register and adding it to the offset as follows: Example: MOV DL, [2400H] ; move contents of DS: 2400H into DL EXERCISE 3-1: Find the physical address of the memory location and its contents after the execution of the following, MOV AL, 99H MOV [3518H], AL ;Assuming that DS = 1512H.

7. Cont… Direct addressing mode: address of the data in memory comes immediately after the instruction operand is a constant The address is the offset address. The offset address is put in a rectangular bracket Ex: MOV DL,[2400] ; move contents of DS:2400H into DL Physical address of DS:3518 => 15120+3518=18638H The memory location 18638H will contain the value 99H

8. 8086 ADDRESSING MODES (CONT..) D) REGISTER INDIRECT ADDRESSING MODE : MOV REG1, [REG2] or MOV [REG2], REG1 ; Here the address of the memory location where the operand resides is held by a register, REG2. REG1 can be any general purpose register and REG2 can be either of SI, DI, or BX and they must be combined with DS in order to generate the 20-bit physical address. Example: MOV AL, [BX] ; move contents of DS:BX into AL MOV CL, [SI] ; move contents of DS:SI into CL MOV [DI], AH ; move contents of AH into DS:DI MOV DX, [BX] ; move contents of DS:BX into DL and ; contents of DS:BX+1 into DH

9. 8086 ADDRESSING MODES (CONT..) Exercise: Assume that DS = 1120H, SI = 2498H, and AX = 17FEH. Show the contents of memory locations and its contents after the execution of MOV [SI], AX. Solution: The contents of AX are moved into memory locations with logical address DS: SI and DS: SI + 1; Therefore, the physical address starts at: PA = DS (shifted left) + SI = 13698H. According to the little endian convention, low address 13698H contains FEH, the low byte, and high address 13699H will contain 17H, the high byte.

10. 8086 ADDRESSING MODES (CONT..) E) BASED RELATIVE ADDRESSING MODE: MOV REG1, [REG2] + CONST or MOV [REG2] + CONST, REG1; CONST is an 8-bit displacement value. REG1 can be any general purpose register and REG2 can only be either of BP or BX In the based relative addressing mode, base registers BX and BP, as well as a displacement value, are used to calculate what is called the effective address. The default segments used for the calculation of the physical address (PA) are DS for BX and SS for BP. PA = DS*10H + BX + const ; EA = BX + const PA = SS*10H + BP + const ; EA = BP + const

11. 8086 ADDRESSING MODES (CONT..) EXAMPLES: Determine PA and EA MOV CX, [BX]+10 ; move DS:BX + 10 and DS:BX+10+1 into CX. PA = DS*10H + BX + 10, EA = BX + 10 MOV AL, [BP] + 5 ; PA = SS*10H + BP + 5 EA = BP + 5 Alternative codings for MOV reg1, [reg2] + constant is MOV reg1, [reg2 + constant] or MOV reg1, const[reg2] For instance, MOV CX, [BX]+10 is same as MOV CX, [BX+10] or MOV CX, 10[BX]

12. 8086 ADDRESSING MODES (CONT..) F) INDEXED RELATIVE ADDRESSING MODE: MOV REG1, [REG2] + CONST OR MOV [REG2] + CONST, REG1; The indexed relative addressing mode works the same as the based relative addressing mode, except that registers DI and SI hold the offset address. Constant is an 8-bit displacement value. REG1 can be any general purpose register and REG2 can only be either of DI or SI PA = DS*10H + DI + const ; EA=DI + const or PA = DS*10H + SI + const ; EA= SI + const Examples: MOV DX, [SI]+5 ; PA = DS (shifted left) + SI + 5 MOV CL, [DI]+20 ; PA = DS (shifted left) + DI + 20

13. 8086 ADDRESSING MODES (CONT..) EXERCISE 3-3 : Assume that DS = 4500, SS = 2000, BX = 2100, SI = 1486, DI = 8500, BP = 7814, and AX = 2512. Show the exact physical memory location where AX is stored in each of the following. All values are in hex. MOV [BX]+20, AX MOV [SI]+10, AX MOV [DI]+4, AX MOV [BP]+12, AX Solution: In each case PA = segment register (shifted left) + offset register + displacement. DS:BX+20 ;location 47120 = (12) and 47121 =(25) DS:SI+10 ;location 46496 = (12) and 46497 = (25) DS:DI+4 ;location 4D504 = (12) and 4D505 = (25) SS:BP+12 ;location 27826 = (12) and 27827 = (25)

14. 8086 ADDRESSING MODES (CONT..) G) BASED INDEXED ADDRESSING MODE : MOV REG1, [REG2][REG3] + CONST OR MOV [REG2][REG3] + CONST, REG1; In this mode, one base register and one index register are used. REG1 can be any general purpose register and REG2 can only be either of DI or SI and reg3 can only be either of BX or BP. PA= DS*10H+BX+DI + const; EA=DI+BX+ const or PA=SS*10H+BP+SI+const and EA= SI + BP + const EXAMPLES: MOV CL, [BX][DI]+8 ;PA = DS (shifted left) + BX + DI + 8 MOV CH, [BX][SI]+20 ;PA = DS (shifted left) + BX + SI + 20 MOV AH, [BP][SI]+29 ;PA = SS (shifted left) + BP + SI + 29 MOV AH, [BP][DI]+29 ;PA = SS (shifted left) + BP + DI + 29 Note that "MOV AX, [SI][DI]+displacement" is illegal.

15. SUMMARY OF ADDRESSING MODES The following table, summarizes the possible offset registers for various segments.

16. SUMMARY OF ADDRESSING MODES Table 3-2 summarizes sample segment overrides. As seen in chapter two, the Intel 8086 allows the program to override the default segment and use any segment register. To do that, specify the segment in the code.

17. SUMMARY OF ADDRESSING MODES .

18. 8086 ADDRESSING MODES (CONT..) EXAMPLE: Assume that the registers have the following values (all in hex) and that CS =1000, DS = 2000, SS = 3000, SI = 4000, DI = 5000, BX = 6080, BP = 7000, AX = 25FF, CX = 8791, and DX = 1299. Calculate the physical address of the memory where the operand is stored and the contents of the memory locations in each of the following addressing examples

19. Comparison of addressing modes Mode Algorithm Advantage Disadvantage Register EA= R No memory reference Limited address space Immediate Operand =A No memory reference Limited operand magnitude Direct EA=A simple Limited address space Register Indirect EA=[R] Large address space Extra memory reference Displacement EA= A+[R] flexibility complexity

20. Quiz-1 1. Show how the flag register is affected by: Mov BX, AAA9H Add BX, 5557H 2. Assume that SS=4567H, DS = 1120H, ES=4567H, SI = 2498H, and AX = 17FEH. Show the contents of memory locations and its contents after the execution of MOV [SI], AX.

21. INTRODUCTION TO ASSEMBLY LANGUAGE PROGRAMMING Program execution in any microprocessor system consists of fetching binary information from memory and decoding that information to determine the instruction represented. For us it is much easier to remember the mnemonic SUB AX,AX than the corresponding machine code 29C0H. For this reason, we write source files containing all the instruction mnemonics needed to execute a program. The source file is converted into an object file, containing the actual binary information the machine will understand, by a special program called an assembler.

22. INTRODUCTION TO ASSEMBLY LANGUAGE PROGRAMMING An Assembly language program is : A series of statements, or lines which is, either Assembly language instructions, or Pseudo-instruction called Directives. Directives (pseudo-instructions): Give directions to the assembler about how it should translate the Assembly language instructions into machine code. Assembly language instructions consist of four fields: [label:] mnemonic [operands] [;comment] Brackets indicate that the field is optional, do not type in the brackets. The comment field begins with a ";" and may be at the end of a line. The assembler ignores comments. Comments are optional, but highly recommended to make it easier to read and understand the program.

23. DIRECTIVES AND A SAMPLE PROGRAM The label field allows the program to refer to a line of code by name. The label field can be any character and cannot exceed 31 characters. A label must end with a colon when it refers to an opcode and, end without colon when it refer to directives. The mnemonic (instruction) and operand fields together accomplish the tasks for which the program was written. The mnemonic opcodes are ADD and MOV, and "AL,BL" and "AX, 6764" are the operands. Instead of a mnemonic and operand, these fields could contain assembler pseudo-instructions, or directives. Directives do not generate machine code and are used only by the assembler as opposed to instructions.

24. DIRECTIVES AND A SAMPLE PROGRAM Examples of directives are DB , PROC, END , and ENDP .

25. MODEL DEFINITION After the first two comments is the MODEL directive. This directive selects the size of the memory model.

26. MEMORY MODEL DEFINITION Among the options for the memory model are SMALL,MEDIUM, COMPACT, and LARGE.

27. SEGMENT DEFINITION Every line of an Assembly language program must correspond to one of an x86 CPU segment register. CS (code segment) DS (data segment) SS (stack segment) ES (extra segment). The simplified segment definition format uses three simple directives: .CODE : w hich correspond to the CS register. .DATA : w hich correspond to the DS register. .STACK : w hich correspond to the SS register.

28. SEGMENT DEFINITION .STACK Marks the beginning of the stack segment The stack segment defines storage for the stack. .DATA Marks the beginning of the data segment The data segment defines the data the program will use .CODE Marks the beginning of the code segment The code segment contains Assembly language instructions.

29. STACK SEGMENT This directive reserves 64 bytes of memory for the stack:

30. DATA SEGMENT The data segment defines three data items: DATA1, DATA2, and SUM.

31. DATA SEGMENT The DB directive is used by the assembler to allocate memory in byte-sized chunks. Each is defined as DB (define byte). Memory can be allocated in different sizes. Data items defined in the data segment will be accessed in the code segment by their labels. DATA1 and DATA2 are given initial values in the data section. SUM is not given an initial value, but storage is set aside for it.

32. CODE SEGMENT DEFINITION The first line of the segment after the .CODE directive is the PROC directive.

33. CODE SEGMENT DEFINITION A procedure is a group of instructions designed to accomplish a specific function. A code segment is organized into several small procedures to make the program more structured. Every procedure must have a name defined by the PROC directive, followed by the assembly language instructions, and closed by the ENDP directive. The PROC and ENDP statements must have the same label. The PROC directive may have the option FAR or NEAR . The OS requires the entry point to the user program to be a FAR procedure.

34. CODE SEGMENT DEFINITION Before the OS passes control to the program so it may execute, it assigns segment registers values. When the program begins executing, only CS and SS have the proper values. DS (and ES) values are initialized by the program.

35. CODE SEGMENT DEFINITION The program loads AL & BL with DATA1 & DATA2 , ADDs them together, and stores the result in SUM .

36. CODE SEGMENT DEFINITION The last instructions, "MOV AH, 4CH" & "INT 21H“ return control to the operating system.

37. CODE SEGMENT DEFINITION The last two lines end the procedure & program, and The label for ENDP(MAIN) matches the label for PROC.

38. CODE SEGMENT DEFINITION It is handy to keep a sample shell & fill it in with the instructions and data for your program.

39. ASSEMBLE, LINK, AND RUN A PROGRAM MASM & LINK are the assembler & linker programs. Many editors or word processors can be used to create and/or edit the program, and produce an ASCII file . The steps to create an executable Assembly language program are as follows:

40. ASSEMBLE, LINK, AND RUN A PROGRAM .

41. Introduction to Assembly Language Programming The source file must end in ". asm “. The ". asm " file is assembled by an assembler, like MASM. The assembler will produce an object file and a list file, along with other files useful to the programmer. The ". lst " file, which is optional, is very useful to the programmer because it lists all the opcodes and offset addresses as well as errors that MASM detected. MASM assumes that the list file is not wanted (NUL.LST indicates no list) The extension for the object file must be ".obj". Before feeding the ".obj" file into LINK, all syntax errors must be corrected.

42. ASSEMBLE, LINK, AND RUN A PROGRAM This object file is input to the LINK program, to produce the executable program that ends in ".exe". The ".exe" file can be run (executed) by the microprocessor. MASM produces another optional file, the crossreference , which has the extension ".crf". An alphabetical list of all symbols & labels in the program. Also program line numbers in which they are referenced. The assembler (MASM) creates the opcodes, operands & offset addresses under the ".obj" file.

43. ASSEMBLE, LINK, AND RUN A PROGRAM The LINK program produces the ready-to-run program with the ".exe" (Executable) extension. The LINK program sets up the file so it can be loaded by the OS and executed. The program can be run at the OS level, using the following command: C>myfile When the program name is typed in at the OS level, the OS loads the program in memory. Referred to as mapping, which means that the program is mapped into the physical memory of the PC.

44. ASSEMBLE, LINK, AND RUN A PROGRAM The following figure shows how an executable program is created & run by following the steps outlined above.

45. ASSEMBLE, LINK, AND RUN A PROGRAM Cont ……

46. DATA TYPES AND DATA DEFINITION The 8088/86 processor supports many data types. Data types can be 8- or 16-bit , positive or negative. A number less than 8 bits wide must be coded as an 8-bit register with the higher digits as zero. A number is less than 16 bits wide must use all 16 bits. ORG is used to indicate the beginning of the offset address. The number after ORG can be either in hex or in decimal. If the number is not followed by H, it is decimal and the assembler will convert it to hex.

47. DATA TYPES AND DATA DEFINITION DB Define Byte: One of the most widely used data directives, it allows allocation of memory in byte-sized chunks. This is the smallest allocation unit permitted. DB can define numbers in decimal, binary, hex, & ASCII. D after the decimal number is optional. B (binary) and H (hexadecimal) is required. To indicate ASCII, place the string in single quotation marks. DB is the only directive that can be used to define ASCII strings larger than two characters. It should be used for all ASCII data definitions.

48. DATA TYPES AND DATA DEFINITION Some examples:

49. DATA TYPES AND DATA DEFINITION List file for DB examples

50. DATA TYPES AND DATA DEFINITION DUP duplicate: DUP will duplicate a given number of characters. Two methods of filling six memory locations with FFH.

51. DATA TYPES AND DATA DEFINITION List file of DUP example:

52. DATA TYPES AND DATA DEFINITION DW Define Word: DW is used to allocate memory 2 bytes (one word) at a time:

53. DATA TYPES AND DATA DEFINITION DW Define Word: List file for DW examples.

54. DATA TYPES AND DATA DEFINITION EQU equate: This is used to define a constant without occupying a memory location. EQU doesn’t set aside storage for data item but associates a constant value with a data label. So that when the label appears in the program, its constant value will be substituted for the label. When EQU is used for the counter constant: COUNT EQU 25, and when executing the instructions "MOV CX, COUNT", the register CX will be loaded with the value 25, it will be in the immediate addressing mode. In contrast to using DB: COUNT DB 25, and when executing the same instruction "MOV CX, COUNT" it will be in the direct addressing mode.

55. DATA TYPES AND DATA DEFINITION EQU can also be used in the data segment: Assume a constant (a fixed value) used in many different places in the data and code segments. By use of EQU, one can change it once and the assembler will change all of them.

56. DATA TYPES AND DATA DEFINITION DD define doubleword: The DD directive is used to allocate memory locations that are 4 bytes (two words) in size. Data is converted to hex & placed in memory locations Low byte to low address and high byte to high address. List file for DD examples.

57. DATA TYPES AND DATA DEFINITION DQ define quadword: DQ is used to allocate memory 8 bytes (four words) in size, to represent any variable up to 64 bits wide: List file for DQ examples.

58. DATA TYPES AND DATA DEFINITION DT define ten bytes: DT is used for memory allocation of packed BCD numbers. This directive allocates 10 bytes. The " H " after the data is not needed.

59. DATA TYPES AND DATA DEFINITION DT define ten bytes: List file for DT examples. DT can also be used to allocate 10-byte integers by using the "D" option:

60. FULL SEGMENT DEFINITION SEGMENT DEFINITION: The SEGMENT and ENDS directives indicate the beginning & ending of a segment, in this format: The label, or name, must follow naming conventions and be unique. The [ options ] field gives important information to the assembler for organizing the segment, but is not required. The ENDS label must be the same label as in the SEGMENT directive. In full segment definition, the " .MODEL " directive is not used.

61. FULL SEGMENT DEFINITION The directives " .STACK ", " .DATA ", and ". CODE " are replaced by SEGMENT and ENDS directives that surround each segment. The following Figure shows the full segment definition and simplified format, side by side, and followed by programs.

62. EXAMPLE OF FULL SEGMENT DEFINITION

63. EXAMPLE OF FULL SEGMENT DEFINITION

64. FULL SEGMENT DEFINITION Stack Segment Definition: The stack segment shown contains the line " DB 64 DUP (?) " to reserve 64 bytes of memory for the stack. The following three lines in full segment definition are comparable to " .STACK 64 " in simple definition: Data Segment Definition In full segment definition, the SEGMENT directive names the data segment and must appear before the data. The ENDS segment marks the end of the data segment:

65. FULL SEGMENT DEFINITION Code Segment Definition: The code segment also begins and ends with SEGMENT and ENDS directives: Immediately after PROC, the ASSUME directive, associates segments with specific registers. By assuming the segment register is equal to the segment labels used in the program. If an extra segment had been used, ES would also be included in the ASSUME statement. ASSUME tells the assembler which of the segments, defined by SEGMENT, should be used. Also helps the assembler to calculate the offset addresses from the beginning of that segment.

66. FULL SEGMENT DEFINITION In " MOV AL, [BX] " the BX register is the offset of the data segment. On transfer of control from OS to the program, of the three segment registers, only CS and SS have the proper values. The DS value (and ES) must be initialized by the program

67. MORE SAMPLE PROGRAMS Example1: Write, run, and analyze a program that adds 5 bytes of data and saves the result. The data should be the following hex numbers: 25, 12, 15, 1F, and 2B. Solution:

68. MORE SAMPLE PROGRAMS Example2:- Write and run a program that adds four words of data and saves the result. The values will be 234DH, 1DE6H, 3BC7H, and 566AH. Use DEBUG to verify the sum is D364. Solution:

69. MORE SAMPLE PROGRAMS Example3:- Write and run a program that transfers 6 bytes of data from memory locations with offset of 0010H to memory locations with offset of 0028H. Solution:

70. INSTRUCTION SET OF 8086 The instruction set of the 8086 microprocessor is divided into seven different groups: Data transfer instruction Strings instruction Loops and jumps instruction Arithmetic instruction Bit manipulation instruction Subroutine and interrupt instruction Processor control instruction The instructional groups are organized in such a way that the more commonly used instructions are presented first, followed by less frequently used instructions.

71. DATA TRANSFER INSTRUCTIONS: This group of instructions makes it possible to move (copy) data around inside the processor and between the processor and its memory. A. MOV DESTINATION, SOURCE: Transfer can be from register to register, register to memory or from memory to register but not from memory to memory . The source and destination must be of same type i.e. either both must be byte or word. In this instruction, the assembler will look at the size of the specified register in the operand field to determine if the immediate data is a 1-, or 2-byte number. MOV instruction does not affect any flags.

72. DATA TRANSFER INSTRUCTIONS: EXAMPLE: MOV AL, 30H MOV AX, 30H In the first instruction, the 30H is coded as a byte value because it is being MOVed into AL. In the second instruction, the 30H is coded as a word value because it is being MOVed into AX. This is clearly shown by the resulting code for both instructions. MOV AL, 30H is coded as B0 30 and MOV AX, 30H is coded as B8 30 00. Note that the second two bytes represent the byte-swapped value 0030H.

73. Cont.. .

74. Cont.…

75. DATA TRANSFER INSTRUCTIONS: Some times, For example, in MOV [SI], 0 the processor does not know if the operand should be coded as a byte value , or as word value . For cases like this, we use BYTE PTR and WORD PTR directives to indicate the size of data. If you wish to MOV a byte value into memory, use: MOV BYTE PTR [SI], 0 MOV WORD PTR [SI], 0 The byte ptr , and word ptr assembler directives stand for "byte pointer," and "word pointer."

76. Cont… The INT Instruction The INT instruction is the instruction which does the most work in any assembler program. What it does is it calls a DOS interrupt (like a function) to perform a special task. When one wants to read from the keyboard or disk or mouse, or write to the screen, one uses an interrupt.

77. Cont… Each interrupt though, has a number of sub-functions which select the individual task that the function has to do MOV AH,02  To select subfunction 2, move the appropriate number, 2, to AH char display interupt code MOV DL,"!"  In the interrupt list, it says that the character to output should be in register DL. So we move the character to DL.

78. Cont… MOV AH,01 To select subfunction 1, accept input MOV AH,09 To select subfunction 9, to display string We can also use Print “” statement to display strings We have to include “emu8086.inc” to do so

79. Cont… MOV AH,02 - Function to output a char MOV DL,"!” - Character to output INT 21h - Call the interrupt to output "!" MOV AH,04Ch - Select exit function MOV AL,00 - Return 0 INT 21h - Call the interrupt to exit

80. Cont … INT 21h   Finally when all the registers are set as required, we call the interrupt.

81. DATA TRANSFER INSTRUCTIONS(cont..) PUSH and POP instructions The stack is a collection of memory locations pointed to by the stack pointer register and the stack segment register( SS:SP ). PUSH and POP instructions are used to load to or receive data from the stack memory. Storing a CPU register in the stack is called a push. Loading the contents of the stack into the CPU register is called a pop. The SP points at the current memory location used as the top of the stack. No flags are affected by this instruction(PUSH and POP).

82. DATA TRANSFER INSTRUCTIONS(cont..) PUSH SOURCE When we wish to write data into the stack area, we use the PUSH instruction. The source of the word can be a general-purpose register , a segment register , or memory. As data is pushed onto the stack it is decremented by 2. As data is popped off the stack into the CPU, it is incremented by 2. When an instruction pushes or pops a general purpose register, it must be the entire 16-bit register. One must code "PUSH AX". There are no instructions such as "PUSH AL" or "PUSH AH".

83. DATA TRANSFER INSTRUCTIONS(cont..) As each PUSH is executed, the register contents are saved on the stack and SP is decremented by 2. EXAMPLES: PUSH BX ; Decrement SP by 2, copy BX to stack PUSH DS ; Decrement SP by 2, copy DS to stack PUSH AL ; Illegal, must push a word

84. DATA TRANSFER INSTRUCTIONS(cont..) EXAMPLES:- The stack segment register has been loaded with 4000H and the stack pointer register with FFFFH. If register CX contains 1234H and AX contains 4455H, what is the result of : PUSH AX PUSH CX Solution: The stack pointer points to a location referred to as the top of the stack. Whenever we push an item onto the stack, the SP is decremented by 2. This is necessary because all pushes involve 2 bytes of data Figure 3-3 shows the new contents of memory after PUSH AX and PUSH CX have executed.

85. Instruction Set of 8086

86. DATA TRANSFER INSTRUCTIONS(cont..) The data contained in memory locations 4FFFEH and 4FFFDH is replaced by the contents of register AX and data contained in memory locations 4FFFCH and 4FFFBH is replaced by the contents of register CX. Notice that the new stack pointer value is 4FFFBH. Remember that the stack builds toward 0. Also notice that the contents of registers CX and AX remain unchanged. When the SP register is pushed, the value written to the stack is the value of SP before the push.

87. DATA TRANSFER INSTRUCTIONS(cont..) P USHA ; Save all 16-bit registers onto the stack in the following order: AX, CX, DX, BX, SP, BP, SI, DI. The value of the SP is that before the PUSHA instruction. PUSHF; copies the contents of the flag register to the stack.

88. DATA TRANSFER INSTRUCTIONS(cont..) POP DESTINATION: The POP instruction is used to perform the reverse of a PUSH. Copies a word from the stack location pointed to by the stack pointer to a destination specified in the Instruction. The stack pointer is used to read 2 bytes of data and copy them into the location specified in the operand field. The destination can be : A general-purpose register, A segment register except CS and IP A memory location. The data in the stack is not changed.

89. DATA TRANSFER INSTRUCTIONS(cont..) After the word is copied to the specified destination, the stack pointer is automatically incremented by 2 to point to the next word on the stack. No flags are affected by the POP instruction. EXAMPLES: POP DX ; Copy a word from top of stack to DX ; Increment SP by 2 POP DS ; Copy a word from top of stack to DS ; Increment SP by 2 NOTE: POP CS Is illegal.

90. DATA TRANSFER INSTRUCTIONS(cont..) EXAMPLES: Assume the contents of the stack segment register and the stack pointer are 4000H and FFFBH, respectively. What is the result of POP DX followed by POP BX? USE FIGURE 3.4 NEXT PAGE. SOLUTION: The contents of location 4FFFBH (34H) are copied into the lower byte of DX, and the contents of location 4FFFCH (12H) are copied into the upper half of DX. Similarly, the contents of location 4FFFDH (55H) are copied into the lower byte of BX, and the contents of location 4FFFEH (44H) are copied into the upper half of BX. The stack pointer is then incremented a second time and points to 4FFFFH. Fig. 3-4 shows a snap shot of memory contents in the stack area.

91. Cont …

92. DATA TRANSFER INSTRUCTIONS(cont..) POPA Destination (Pop All Registers). All general purpose registers are popped from the stack in the order indicated in Table 3-6. Note that the contents of the SP are not loaded with the data popped off the stack. This is necessary to prevent the stack from changing locations halfway through the execution.

93. DATA TRANSFER INSTRUCTIONS(cont..) Overflow and Underflow of Stack: PUSH instruction decrements SP by 2. At some point, if SP=0000H and if there is an attempt to PUSH data on the stack, Stack overflow will result. On the other hand, POP instruction increments SP by 2. At some point, if SP=FFFFH and if there is an attempt to POP data from the stack, Stack Underflow will result.

94. DATA TRANSFER INSTRUCTIONS(cont..) IN ACCUMULATOR, PORT Input byte or word from port to accumulator register(AL or AX). Data read from an input port always ends up in the accumulator. IN: COPY DATA FROM A PORT TO ACCUMULATOR. The input port is actually a hardware device connected to the processor's data bus. When executing the IN instruction, the processor will output the address of the input port on the address bus. The selected input port will then place its data onto the data bus to be read by the processor. The processor allows two different forms of the IN instruction: Direct and Indirect.

95. DATA TRANSFER INSTRUCTIONS(cont..) Direct: If the port number is between 00 and FFH, we would use: IN AL,80H or IN AX,80H. Using AL in the operand field causes 8 bits of data to be read. Two bytes can be input by using AX in the operand field. Example: IN AL, 0F8H ; Copy a byte from port 0F8H to AL IN AX, 95H ; Copy a word from port 95H to AX. Example: What is the result of IN AL, 80H if the data at input port 80H is 22H? Solution: The byte value 22H is copied into register AL (AL = 22H)

96. DATA TRANSFER INSTRUCTIONS(cont..) Indirect: If a full 16-bit port address must be specified, the port address is loaded(MOVED) into register DX, and IN AL, DX ; To copy a byte from 8-bits port DX to AL. IN AX, DX ; To copy a Word from 16-bits port DX to Ax. Example: MOV DX, 30F8H ; Load 16-bit address of the port in DX. IN AL, DX ; Copy a byte from 8bit port 30F8H to AL. IN AX, DX ; Copy a word from 16-bit port 30F8H to AX.

97. DATA TRANSFER INSTRUCTIONS(cont..) OUT PORT, ACCUMULATOR Output byte or word from accumulator to port. If the output port is 16-bit, then this port address loaded to DX and , OUT DX, AL or OUT DX, AX Example: MOV DX, 30F8H ; Load 16-bit address of the port in DX. OUT DX, AL ; Copy the contents of AL to pert 30F8H OUT DX, AX ; Copy the contents of AX to port 30F8H. When the port address is in range of 00H to FFH, then: OUT 80H, AL or OUT 80H, AX. Examples: OUT 0F8H, AL ; Copy content of AL to 8 bit port 0F8H. OUT 0F8H, AX ; Copy contents of AX to 16-bit port 0F8H.

98. DATA TRANSFER INSTRUCTIONS(cont..) Example: What happens during execution of OUT DX, AL if AL contains 7CH and DX contains 3000H? Solution: The port address stored in register DX is output on the address bus, along with the 7C from AL on the data bus. The output port circuitry must recognize address 3000H and store the data.

99. DATA TRANSFER INSTRUCTIONS(cont..) LEA DESTINATION, SOURCE (Load effective address): This instruction is used to load the offset of the source memory operand into one of the processor's registers. The memory operand may be specified by any number of addressing modes. The destination may not be a segment register. Determines the offset of the variable or memory location named as the source and loads this address in the specified 16-bit register. Flags are not affected by LEA instruction. Example: LEA CX, TOTAL ; Load CX with offset of TOTAL in DS. LEA AX, [BX] [DI] ; Load AX with EA = BX + DI LEA DX, 10 [SI] ; LOAD DX with EA = SI + 10

100. DATA TRANSFER INSTRUCTIONS(cont..) Example: What is the difference between: MOV AX, [40H] and LEA AX, [40H]? Solution: In MOV AX, [40H] ; Places 2 bytes of data from locations 40H ; and 41H into register AX. In LEA AX, [40H] ; Places 40H into register AX. Example: What does LEA AX, [SI] do? Solution: The value of SI at execution time is loaded into AX. Example: MOV BX, 35H MOV DI, 12H LEA SI, [BX+DI] ; SI = 35H + 12H = 47H

101. DATA TRANSFER INSTRUCTIONS(cont..) XCHG DESTINATION, SOURCE (Exchange data): Used to swap the contents of two 8-, or 16-bit operands. One operand must be a processor register (excluding the segment registers). The other operand may be a register or a memory location. If a memory location is used as an operand it is assumed to be within a data segment. Example: Registers AL and BL contain 30H and 40H, respectively. What is the result of XCHG AL, BL? Solution: After execution, AL = BL = 40H and BL =AL = 30H.

102. DATA TRANSFER INSTRUCTIONS(cont..) Individual Assignment: XLAT LDS LES LAHF and SAHF