Memory Address Segment-offset address Base location (segment) + logical location (offset) Example: For 32-bits segment-offset address, 08F1:0100 represents a 20-bits absolute address F 1 (0)

Floating-Point Unit 80-bit Registers ST(0) ST(1) ST(7) FPU Instruction Pointer FPU Data Pointer Tag Register Control Register Status Register 48-bit Pointer Register 16-bit Control Register Opcode Register

DUP Operator db 20 dup(0) ; 20 bytes, all equal to zero db 20 dup (?) ; 20 un-initialized bytes db 4 dup(“ABC”) ; 12 bytes: “ABCABCABCABC” db 4096 dup(0);

Assembly Language Basic elements 1. Constant and expression E+07

Integer Constants 26 decimal 1Ah hexadecimal 1101b binary 36q octal 2BH hexadecimal 43Q octal 36D decimal 48d decimal

Constant Expression, Symbolic constants and Character or String Constants Constant expression *20 -3*4/ E+04 A symbolic constant rows = 5 columns = 10 Number = rows * columns Character or string constant ‘ABC’ ‘X’ “ This is a test” “The dog is black”

Assembly Language Statement Consists of : 1. A name, an instruction mnemonic, operands, and a comment. It has the following form: [name] [mnemonic] [operands] [;comment] Example: call sub1 ; transfer of control

Names A name identifies a label, variable, symbolic, or a keyword. What can be used as a name: 1. A… Z a… z ? 4. _ Underscore 6. $ Length limit----maximum of 247 characters (in MASM)

Variables and Labels 1. A variable is a location in a program’s data area that has been assigned a name. Such as: count1 db 50 ; message db “ Hello world!”, odh, 0ah 2. Label. If a name appears in the code area of a program, it is called a label. Example: LabelA: mov ax, 0 mov bx, ffh

Hello World Program title Hello World Program (hello.asm) ; This program displays "Hello, world!".model small.stack 100h.data message db "Hello, world!",0dh,0ah,'$'.code main proc mov mov ds,ax mov ah,9 mov dx,offset message int 21h mov ax,4C00h int 21h main endp end main

Analyze the Program.model small ----specify the program’s memory model For that model we can have: Tiny--- code + data less than 64k Samll--- code <=64k, data,<=64k One code segment and one data segment Medium---Data <=64k, code any size. Multiple code segment, one data segment Compact---Code <= 64k, data any size. Multiple data segment, one code segment Large---Code >64k, data >64k both multiple segment Huge---Same as the large model, except that individual variables such as arrays may be large than 64k Flat--- No segments. 32-bits addresses are used for both code and data. ***Protected mode only

Standard Assembler Directives end---end of program assembly endp--- end of procedure page---Set a page format for the listing file proc---Begin procedure title---Title of the listing file.code--- Mark the beginning of the code segment.data---.model---.stack---Set the size of the stack segment

Data Allocation Directives DB byte DW DD DF, DP define far pointer DQ define Quadword DT define 10 bytes

Examples Char1 db ‘A’ ; ASCII char. Char2 db “A’ –10 ; expression Smallest db –128 Max_unsign db 255 Myval db ? Only reserve memory, no initializeers List db 10,20,30,40,50

Characters and Integers are one and the same Char db ‘A’ Hex db 41h Dec db 65 Bin db Oct db 101q

Type mix is allowed List db 10, 20, 41h, b List2 db 0ah, 29h, ‘a’, 22

Strings Cstring db “Good morning”,0 Pstring db 12, “Good morning” Longstring db “This is a long string, that” db “ clearly is going to take” db “ several lines to store”,0

Symbolic Constants Equate directives They allow constants and literals to be given symbolic names. 1. Equal-Sign Directive (redefinable) The syntax is: name = expression Examples: string = ‘XY’ count = 500 minInt = 8000h

To assign a 32 bits integer is required.386 maxLong = 7FFFFFFFh ; maximum 32-bit signed value minLong = h ; Minimum 32-bit signed value “=“ defined symbols can be redefined any number of times.

Example of Using Directives count =5 mov al, count mov dl, al count =10 mov cx, count count = 2000 mov ax, count

EQU and TEXTEQU directives EQU is not redefinable TEXTEQU is redefinable Examples: maxint equ 32767; Numeric float1 equ ; String TEXTEQU syntax: name TEXREQU name TEXTEQU textmacro

Intel chip Instructions 1. MOV---Data transfer instruction Instruction format: MOV reg, reg MOV reg, immed MOV mem, reg MOV mem, immed MOV reg, mem MOV mem16, segreg MOV reg16, segreg MOV segreg, immed16 MOV segreg, reg16 *** Flag bits are not affected by MOV instruction

Examples.data count db 10 total dw 4126h bigVal dd h.code mov al, bl mov bl, count mov count, 26 mov bl, 1

Examples—cont… mov dx, cx mov bx, 8FE2h mov total, 1000h mov eax, ebx mov edx, bigVal ; 32-bit memory to reg ***Pay attention the the fact that operands size must match. Otherwise, the assembler will return an error for any size mismatch.

Example.data count dw 20h ; 20h---here really mean 0020h.code mov al, count; error:

XCHG Instruction XCHG reg, reg xchg ax, bx xchg eax, ebx XCHG reg, mem xchg ax, value1 XCHG mem, reg xchg value2, cx

Program example Title Exchange Two Variables (Exchange.asm).model small.stack 100h.data Value1 db 0ah Value2 db 14h.code Main proc Mov ; initialize DS register Mov ds, ax Mov al, value1 Xchg value2, al Mov value1, al Mov ax, 4c00h Int 21h Main endp End main; Question: Why don’t we use xchg value1, value2?

Arithmetic Instructions ADD and SUB instructions ADD reg, reg ADD mem, reg ADD reg, mem ADD reg, immed ADD mem, immed ADD accum, immed Flag O D I S Z A P C SUB has the similar instruction format * * *

Examples Mov ax, 10 Sub ax, 10 ; AX=0, ZF =1 Mov bx, 1 Sub bx, 2 ; BC = FFFF, SF = 1 Mov ax, 0FFh Add al, 1 ; AL = 00, CF = 1

Basic Operand Types There are three basic types of operands: 1. Immediate----a constant 2. Register one of the CPU’s register 3. Memory---a reference to a mem.Location There are six different types of memory operands: Direct, direct-offset, register-indirect Indexed, base-indexed, and base-indexed with displacement

Example mov ax, [si] --- this is called register-indirect Here si register hold the beginning address for a list of numbers. The first number in the address will be moved to al and the second number will be moved into ah.

Immediate operands, Direct Operands, and Direct-Offset operands 1. Immediate operands: Mov al, 10 ; ----a number Mov eax, h ; ---a 4-byte number Mov dl, ‘X’ ; --- a character Mov ax, (40 *50) ; arithmetic expression

Direct Operands.data Count db 20 Wordlist dw 1000h, 2000h LongVal dd h.code Mov al, Count Mov bx, Wordlist +2 Mov edx, LongVal

Direct Offset Operands.data Array db 0ah, 0bh, 0ch, 0dh.code Mov al, array ; 0ah-  al Mov bl, array +1 ; 0bh  bl Mov cl, array + 2 ; 0ch  cl Mov dl, array +3 ; 0dh  dl

Another Example

IA – 32 Processor Architecture IA – 32 processor have three basic mode of operations: 1. Protected Mode 2. Real-address Mode 3. System Management Mode

Protected Mode IA-32 processors have three basic modes of operation: Protected mode, Real-address mode and System Management mode. In addition, the Virtual-8086 mode is a special case of Protected mode.

1. Protected Mode Protected mode is the native state of the processor, in which all instructions and features are available. Programs are given separate memory areas (called segment), and the processor detects any attempt by a program to reference memory outside its assigned segment. Each program can address up to 4GB memory.

Virtual – 8086 Mode While in Protected mode, the processor can directly execute Real-address mode software such as MS-DOS programs in a safe multitasking environment. In other word, even if an MS-DOS program crashes, it will not affect other program running at the same time.

Real-address mode It offers us a few more features, such as the ability to switch into other two modes. Only 1 MB of memory can be addressed: 00000h-FFFFFh The processor can only run one program a time.

System Management Mode SSM-mode It provides an operating system with a mechanism for implementing such functions as power management and system security. These functions are usually implemented by computer manufacturers who want to customize the processor for a particular system setup.

Homework6 Write and run an assembly program under debug that will: 1) Add numbers from 1 to Note: you would need these following instructions: A) jz – jump if zero B) mov cx, number of times (1000 in Hex.) C) create the labels that tells program where to go when certain conditions are met. Due: 11/07 Turn in: Your code and the running result.

Note to the debug usage You can assemble your assembly language program with other software and load the code into the memory by using the debug as follows: Under debug: (Click start, click Run, type cmd and type enter) N a:hello.exe L This will load your code from a: to the memory address beginning at CS:0100.