lecture 1 assembly language programming
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Lecture 1 Assembly Language Programming. Presented By Dr. Rajesh Palit Asst. Professor, EECS, NSU Originally Prepared By Dr. Shazzad Hosain , EECS, NSU. What is Microcomputer?. Interface Circuitry. Central Processing Unit. BUS Interface Unit Execution Unit Control Unit. - PowerPoint PPT PresentationTRANSCRIPT
Lecture 1
Assembly Language Programming
Presented ByDr. Rajesh Palit
Asst. Professor, EECS, NSUOriginally Prepared By
Dr. Shazzad Hosain, EECS, NSU
What is Microcomputer?
Basic Blocks of a microcomputer
BUS Interface Unit
Execution UnitControl Unit
Interface Circuitry
Central Processing Unit
Von-Neumann Concept/Model
Von-Neumann cycle• The Von-Neumann concept is the basic concept for
universal microprocessors. The ix86 architecture is based on that concept.
• It means a stored-program computer in which an instruction fetch and a data operation cannot occur at the same time because they share a common bus.
• It comprises a Control unit, an Execution unit, Memory, I/O unit
• Instructions are treated in 5 cycles – (1) Fetch, (2) Decode, (3) Fetch Operands, (4) Execute, and (5) Update instruction Counter
Simplified Block Diagram of CPU
Registers• General Purpose Registers– AX, BX, CX, DX– Base Pointer (BP), Stack Pointer (SP)– Source Index (SI), Destination Index (DI)
• Segment Registers– Code Segment (CS), Data Segment (DS), Stack
Segment (SS), Extra Segment (ES)• Status Flag Register (FLAGs)• Instruction Pointer (IP)
Registers
16 bit Segment registers
Decimal, Binary and Hexadecimal Numbers
• 23910= 11 1101 01112 = 3D716 • 239d = 1111010111b = 3D7h (d,b,h upper or
lower case)• If no suffix is given, decimal is assumed• If a hexadecimal starts with a letter, an extra
zero must be put in the beginning• Signed Numbers – 2’s complements form
Example Data
• If AX = 20A2H then AH = 20H, AL = A2H• In other words, if AH = 1CH and AL = A2H then AX = 1CA2H
0010 0000 1010 0010
AH AL
AX
The FLAGS register• FLAGS indicate the condition of the MP• Also control the operations• FLAGS are upward compatible from 8086/8088
to Pentium/Pentium Pro
Figure 2.2: The EFLAG and FLAG registers
The FLAGs
• Carry Flag – C – C = 1 if there is a carry out from the msb on addition– Or, there is a borrow into the msb on subtraction– Otherwise C = 0– C flag is also affected by shift and rotate instructions
1010101011101010
111010100 C = 1, in this case
The FLAGs
• Parity Flag – P – P = 1 for even parity, if number contains even
number of ones– P = 0 for odd parity, if odd number of ones
10101010 10101011
P = 1 P = 0Even number of ones Odd number of ones
Definition changes from microprocessor to microprocessor
The FLAGs
• Zero Flag – Z– Z = 1 for zero result– Z = 0 for non-zero result
• Sign Flag – S– S = 1 if MSB of a result is 1, means negative number– S = 0 if MSB of a result is 0, means positive number
The FLAGs
• Trap Flag – T– Enables trapping through an on-chip debugging feature– T = 1 MP interrupts the flow of a program, i.e. debug mode is
enabled– T = 0 debug mode is disabled
• Direction Flag – D– Selects increment/decrement mode of SI and/or DI registers
during string instructions– D = 1, decrement mode, STD (set direction) instruction used– D = 0, increment mode, CLD (clear direction) instruction used
The FLAGs
• Overflow Flag – O– O = 1 if signed overflow occurred– O = 0 otherwise– Overflow is associated with the fact of range of
numbers represented in a computer• 8 bit unsigned number range (0 to 255)• 8 bit signed number range (-128 to 127)• 16 bit unsigned number range (0 to 65535)• 16 bit signed number range (-32768 to 32767)
Signed vs. Unsigned Overflow
• Let, AX = FFFFh, BX = 0001h and execute • ADD AX, BX
1111 1111 1111 1111+ 0000 0000 0000 00011 0000 0000 0000 0000
AXBX
• Unsigned interpretation– Correct answer is 10000h = 65536– But this is out of range.– 1 is carried out of MSB, AX = 0000h, which is wrong– Unsigned overflow occurred
• Signed interpretation– FFFFh = -1, 0001h = 1, summation is -1+1 = 0– Signed overflow did not occur
How instructions affect the flags?
• Every time the processor executes a instruction, the flags are altered to reflect the result
• Let us take the following flags and instructions
• Sign Flag – S• Parity Flag – P• Zero Flag – Z• Carry Flag – C
• MOV/XCHG• ADD/SUB• INC/DEC• NEG
NoneAllAll except CAll (C = 1 unless result is 0, O = 1,
80H, or 8000H)
Example 1
• Let AX = FFFFh, BX = FFFFh and execute ADD AX, BX FFFFh+ FFFFh1 FFFEh
The result stored in AX is FFFEh = 1111 1111 1111 1110
SPZC
= 1 because the MSB is 1= 0 because the are 15 of 1 bits, odd parity= 0 because the result is non-zero= 1 because there is a carry out of the MSB on addition
Example 2
• Let AX = 8000h, BX = 0001h and execute SUB AX, BX 8000h- 0001h 7FFFh
The result stored in AX is 7FFFh = 0111 1111 1111 1111
SPZC
= 0 because the MSB is 0= 0 because the are 15 of 1 bits, odd parity= 0 because the result is non-zero= 0 because there is no carry
Title Example
.model small
.data
list db 10,17,11,25,13 large db 0 count db 5
.code main proc mov ax, @data mov ds, ax mov al, 5 call fact
An Assembly Program mov ax,4C00h int 21h main endp
fact proc push bx mov bl, al mov ax, 1 again: mul bl dec bl jnz again pop bx ret fact endp
end main
An Assembly Program#include <stdio.h>int main (void){ int i, j ; ********* // comment *********}
Example 3-5 of Barry B. Brey’s book
An Assembly Program Cont.
• What is the content of BX?
00h 10hAX
AH AL
10h 00h 00h 00h AAh AAhDATA1 DATA2 DATA3 DATA4
AAh AAhBX
BH BL
Assembly Language Structure
An Assembly Program
• SMALL model allows one data segment and one code segment
• TINY model directs the assembler to assemble the program into a single segment
• DB for Define Byte (one single byte)• DW for Define Word (two consecutive bytes)
10h 00h 00h 00h AAh AAhDATA1 DATA2 DATA3 DATA4
Another Example
References
• Ch 6, Digital Logic and Microcomputer Design – by M. Rafiquzzaman
• Ch 2, Intel Microprocessors – by Brey• Ch 5, Assembly Language Programming – by
Charles Marut