CSE 331/ETE 332/EEE 332Microprocessor and Assembly

Language Programming

Presented ByDr. Shazzad Hosain

Asst. Prof. EECS, NSU

Books

• Fundamentals of Digital Logic and Microcomputer Design – M. Rafiaquzzaman

• The Intel Microprocessors 8086/8088 … , and Pentium Pro Processors – Barry B. Brey

• Assembly Language Programming and Organization of the IBM PC – Ytha Yu, Charles Marut

Mark DistributionItems Percentage

Attendance 5%

Quizzes 25%

Mid Term 25%

Lab 10%

Final 35%

Total 100%

http://www.northsouth.edu/php/faculty/shazzad/teaching.html and follow the corresponding link.

There will be no makeup for quizzes or exams. For details please visit

Best 3 out of 5

Assembly Language Programming

What is Microcomputer?

The Microprocessor and Its Architecture

• ALU (Arithmetic and Logic Unit)• The Control Unit• Registers

HDD

RAM

Cache Memory

Microprocessor

R1 R1

R3

The Programming Model

• Program visible registers– We can access via programs

• Program invisible registers– We can not access via programs

Registers

16 bit Segment registers

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 occured

• Signed interpretation– FFFFh = -1, 0001h = 1, summation is -1+1 = 0– Singned 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)

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

Registers

16 bit Segment registers

Segment Registers

16 bit Segment registers

***

Segment 1

Segment 2

Segment n

0000hCS

8000hDS

A000hSS

An Assembly Program#include <stdio.h>void main (){ 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

Charls Marut