assembly language programming with masm

8/10/2019 Assembly Language Programming With MASM

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ASSEMBLY LANGUAGE

PROGRAM DEVELOPMENT

WITH MASM

611 37100 Lecture 07-2

ASSEMBLY LANGUAGE PROGRAM

DEVELOPMENT WITH MASM

7.1 Statement Syntax for a Source Program

7.2 Assembler Directives

7.3 Creating a Source File with an Editor

7.4 Assembling and Linking Programs

7.5 Loading and Executing a Run Module

611 37100 Lecture 07-3

7.1 Statement Syntax for a Source

Program

Assembly language statement syntax

Directive statement syntax

Constants in a statement

Operand expressions using the arithmetic,relational, and logical operators

Value-returning and attribute operators

611 37100 Lecture 07-4


Program


EXAMPLES:

START: MOV CX, 10 ;Load a count of 10 into register CX

MOV CX, 10 ;Initialize the count in CX

CLC ;Clear the carry flag

LABEL: OPCODE OPERAND(S) ; COMMENT (S)

611 37100 Lecture 07-5


Program


Data register low byteDL

Data register high byteDH

Data registerDX

Counter register low byteCL

Counter register high byteCH

Counter registerCX

Base register low byteBL

Base register high byteBH

Base registerBX

Accumulator register low byteAL

Accumulator register high byteAH

Accumulator registerAX

RegisterSymbol

Extra segment registerES

Stack segment registerSS

Data segment registerDS

Code segment registerCS

Base pointer registerBP

Stack pointer registerSP

Destination index registerDI

Source index registerSI

RegisterSymbol

Symbols used for specifying register operands

611 37100 Lecture 07-6


ProgramAssembly language statement syntax

StackMOV [BP][DI]+DISP, AH

StackMOV [BP][SI]+DISP, AH

DataMOV [BX][DI]+DISP, AH

Data

Data

Data

Stack

Data

Data

Data

Stack

Data

Data

-

-

Segment

MOV [BX][SI]+DISP, AH

MOV AL, [DI]

MOV AL, [SI]

MOV [BP]+DISP, AL

MOV [BX]+DISP, AL

MOV AX, [BX]

MOV AX, [DI]

MOV AX, [BP]

MOV AX, [SI]

MOV VARIABLE, AX

MOV AL, 15H

MOV AX, VARIABLE

Example

DestinationBased indexed

SourceIndexed

DestinationBased

SourceRegister indirect

DestinationDirect

SourceImmediate

DestinationRegister

OperandAddressing mode


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611 37100 Lecture 07-7


Program

Directive statement syntax

EXAMPLES:

DB 0FFH ; Allocate a byte location initialized to FFH

DB 0FFH, 0FFH, 0FFH, 0FFH, 0FFH

(DB stands for Define Bytes)

LABEL: DIRECTIVE OPERAND(S) ; COMMENT (S)

611 37100 Lecture 07-8


Program

Constants in a statementConstants in an instruction or directives can be

expressed in any of many data types such binary,decimal, hexadecimal, octal, and characterdatatypes.

The first digit of a hexadecimal number must be on ofthe numbers in the range 0 through 9.

Typically, data and addresses are expressed inhexadecimal form, and the count for shift, rotate, andstring instructions in commonly expressed in decimalform.

2s complement of the number must be used fornegative numbers expressed in binary, hexadecimal,or octal form.

611 37100 Lecture 07-9


ProgramEXAMPLE

The repeat count in CX for a string instruction is to be equal todecimal 255. Assume that the instruction that is to load the counthas the form

MOV CX, XX

where XX stands for the count, an immediate operand that is to beloaded into CX. Show how the instruction should be written.

Solution:

MOV CX, 255D

or MOV CX, 255

or MOV CX, 0FFH

611 37100 Lecture 07-10


Program

EXAMPLEThe count in a move instruction that is to load CX is to be 10.

Write the instruction and express the immediate operand in binaryform.

Solution:The binary form of 1010 is 010102. Forming the 2s complement

by complementing each bit and adding 1, we get

10101

+110110

Therefore, the instruction is written as

MOV CX, 10110B

611 37100 Lecture 07-11


Program


Subtract B from A and makes the operand

equal to the difference

Adds A to B and makes the operand equal to

the sum

Shifts the value in A right by n bits position

and assigns this shifted value to the operand

Shifts the value in A left by n bits position and

assigns this shifted value to the operand

Divides A by B and assigns the remainder to

the operand

Divides A by B and makes the operand equal

to the quotient

Multiplies A with B and makes the operand

equal to the product

Function

A-B

A+B

A SHR n

A SHL n

A MOD B

A/B

A*B

Example

-

+

SHR

SHL

MOD

/

*Arithmetic

OperatorType

611 37100 Lecture 07-12


Program Operand expressions using the arithmetic,

relational, and logical operators

Compares value of A to that of B. If A is

greater than or equal to B, the operand is set

to FFFFH and if it is less than it is set to 0H

Compares value of A to that of B. If A is less

than or equal to B, the operand is set to

FFFFH and if it is greater than it is set to 0H

Compares value of A to that of B. If A is

greater than B, the operand is set to FFFFH

and if it is equal or less than it is set to 0H

Compares value of A to that of B. If A is less

than B, the operand is set to FFFFH and if it is

equal or greater than it is set to 0H

Compares value of A to that of B. If A is not

equal to B, the operand is set to FFFFH and if

they are equal it is set to 0H

Compares value of A to that of B. If A equals B,

the operand is set to FFFFH and if they are not

equal it is set to 0H

Function

A GE B

A LE B

A GT B

A LT B

A NE B

A EQ B

Example

GE

LE

GT

LT

NE

EQRelational

OperatorType


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611 37100 Lecture 07-13


Program


A is XORed with B and makes the value that

results equal to the operand

A is ORed with B and makes the value that


A is ANDed with B and makes the value that


Takes the logical NOT of A and makes the

value that results equal to the operand

Function

A XOR B

A OR B

A AND B

NOT A

Example

XOR

OR

AND

NOTLogical

OperatorType

611 37100 Lecture 07-14


Program

EXAMPLEFind the value the assembler assigns to the source operand for

the instruction

MOV BH, ( A * 4 2 ) / ( B 3 )

for A=8 and B=5

Solution:The solution is calculated as

(8*4-2)/(5-3) = (32 2)/(2)

= (30)/(2)

= 15

And using hexadecimal notation, we get the instructionMOV BH, 0FH

611 37100 Lecture 07-15


ProgramEXAMPLE

What value is used for the source operand in the expression

MOV AX, A LE (B C)

if A=234, B=345, and C=111?

Solution:

Substituting into the expression, we get

234 LE (345 111)

234 LE 234

since the relational operator is satisfied, the instruction isequivalent to

MOV AX, 0FFFFH

611 37100 Lecture 07-16


Program


Returns the number of units (as specified by

TYPE) allocated for the variable A to the

operand

Returns the byte count of variable A to theoperand

Returns to the operand a number

representing the type of A; 1 for a byte

variable and 2 for a word variable; NEAR or

FAR for the label

Assigns the offset of the location A in its

corresponding segment to the operand

Assigns the contents held in the segment

register corresponding to the segment in

which A resides to the operand

Function

LENGTH A

SIZE A

TYPE A

OFFSET A

SEG A

Example

LENGTH

SIZE

TYPE

OFFSET

SEGValue-returning

OperatorType

611 37100 Lecture 07-17


Program


Returns to the operand A the low byte of

the word of A

LOW ALOW

Returns to the operand A the high byte of

the word of A

Assigns to operand A a distance or type

attribute: BYTE, WORD, NEAR, or FAR, and

the corresponding segment attribute

Assigns to operand A an attribute that

indicates that it is within +127 or 128

bytes of the next instruction. This lets the

instruction be encoded with the minimum

number of bytes

Overrides the normal segment for operand

A and assigns a new segment to A

Overrides the current type of label operand

A and assigns a new pointer type: BYTE,

WORD, NEAR, or FAR to A

Function

HIGH A

THIS BYTE A

JMP SHORT A

ES: A

NEAR PTR A

Example

HIGH

THIS

SHORT

DS:, ES:, SS:

PTRAttribute

OperatorType

611 37100 Lecture 07-18


Data directives

Segment-control directives

Modular programming directives

Directives for memory usage control

Directives for program listing control


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611 37100 Lecture 07-19


.TFCOND%OUT

.XLISTSUBTTL.LIST

.XCREF.SFCOND.LFCOND

.XALL.SALL.LALL

TITLEPAGE.CREFListing

MACROIRP

REPTLOCALEXITM

PURGEIRPCENDMMacro

IF2IFIDNIFB

IF1IFEIF

IFNDEFIFDIFENDIF

IFNBIFDEFELSEConditional

ENDP

STRUCLABELEND

SEGMENTINCLUDEDW OR WORD

RECORDGROUPDT or TBYTE

.RADIXEXTRNDQ or GWORD

PUBLIC

PROC

ORG

NAME

EVEN

= (EQUAL SIGN)

EQU

ENDS

DD or DWORD

DB or BYTE

COMMENT

ASSUMEData

DirectivesType

611 37100 Lecture 07-20


Data directives

Define or initialize double word size (4

byte) variable or locations

Define or initialized word size (2 byte)

variables or locations

Define or initialize byte size variables

or locations

Set or redefine the value of a symbol

Assign a permanent value to a symbol

Function

Define

double word

DD or DWORD

Define wordDW or WORD

Define byteDB or BYTE

Equal to=

EquateEQU

MeaningDirective

Commonly used data directives

611 37100 Lecture 07-21


Data directives

EXAMPLES:

AA EQU 0100H

BB EQU AA+5H

AA = 0100H

BB = AA+5H

CC DB 7

EE DB ?

MESSAGE DB JASBIR

TABLE_A DB 10 DUP(?), 5 DUP(7)

TABLE_B DB 0,1,2,3,4,5,6,7,8,9

611 37100 Lecture 07-22



Three kinds of memory segment: Code segment

Data segment

Stack segment

The segment-control directives partitions andassigns a source program to a specific memorysegment.

The beginning of a segment is identified by thesegment (SEGMENT) directive and its end ismarked by the end of segment (ENDS) directive.

611 37100 Lecture 07-23



Specifies the segment address register for a

given segment

ASSUME

Specifies the end of a segmentENDS

Defines the beginning of a segment and specifies

its kind, at what type of address boundary it is to

be stored in memory, and how it is to be

positioned with respect to other similar segments

in memory

SEGMENT

FunctionDirective

Segment directives

611 37100 Lecture 07-24



EXAMPLES:

SEGA SEGMENT PARA PUBLIC CODE

ASSUME CS:SEGA

MOV AX, BX

.

.

.

SEGA ENDS


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611 37100 Lecture 07-25



Segment begins on a 256 byte address boundary

in memory (8 LSB of the address are equal to 0)

PAGE

Segment begins on a word (2 byte) address

boundary in memory (LSB of the address is 0)

WORD

Segment begins anywhere in memoryBYTE

Segment begins on a 16 byte address boundary

in memory (4 LSB of the address are equal to 0)

PARA

FunctionAttribute

Align-type attributes

611 37100 Lecture 07-26



Locates the segment at an address above all

other segments

MEMORY

The segment is part of the run-time stack

segment

STACK

Locates the segment at the 16-bit paragraph

number evaluated from the expression

AT [expression]

Overlaps from the beginning segments with

the same name

COMMON

Concatenates segments with the same namePUBLIC

FunctionAttribute

Combine-type attributes

611 37100 Lecture 07-27



Specifies the extra segmentEXTRA

Specifies the stack segmentSTACK

Specifies the data segmentDATA

Specifies the code segmentCODE

FunctionAttribute

Class attributes

611 37100 Lecture 07-28



The specified symbols are defined in other

modules and are to be used in this module

EXTRN name:type[.]

The defined symbols can be referenced

from other modules

PUBLIC Symbol[..]

Defines the end of a procedureproc-name ENDP

Defines the beginning of a far-proc

procedure

proc-name PROC FAR

Defines the beginning of a near-proc

procedure

proc-name PROC [NEAR]

FunctionDirective

611 37100 Lecture 07-29



Two kinds of procedures: Near procedure - IP in stack when called

Far procedure - IP and CS in stack when called

NEAR is assumed as the default attribute of aprocedure.

If a procedure can be called from other modules,its name must be made public by using thePUBLIC directive.

If a procedure in another module is to be called

from the current module, its name must bedeclared external by using the EXTRN directive.

611 37100 Lecture 07-30



PUBLIC SUB

CSEG1 SEGMENT

.

.

SUB PROC FAR

MOV AX, BX

.

.

RET

SUB ENDP

.

.

.

CSEG1 ENDS

EXTRN SUB:FAR

CSEG2 SEGMENT

.

.

CALL SUB

.

.

.

.

.

CSEG2 ENDS

Module 1 Module 2


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611 37100 Lecture 07-31


Directives for memory usage control

EXAMPLE:

ORG 100H

ORG 100H

ORG $+200H(memory locations 10016 to 30016 are skipped and the machine

code of the program can start at address 30116)

Specifies the end of the source programEND [expression]

Specifies the memory address starting from

which the machine code must be placed

ORG [expression]

FunctionDirective

611 37100 Lecture 07-32


Directive for program listing control

EXAMPLE:

PAGE 50 100PAGE +

Prints text on the third line of each

page of the listing

SUBTTL text

Prints text on the second line of

each page of the listing

TITLE text

Selects the number of lines printed

per page and the maximum number

of characters printed per line in the

listing

PAGE operand_1 operand_2

FunctionDirective

611 37100 Lecture 07-33


Examples of a source program using directives

611 37100 Lecture 07-34


Examples of a source program using directives

611 37100 Lecture 07-35

7.3 Creating a Source File with an

EditorC:\_

EDIT A:BLOCK.SRC []

Type the program line by line.

MOV AX, DATASEGADDRMOV DS, AX

.etc.

Menu to save or print the fileNEWOPEN

.etc.

Alt-F

[]Loaded file not saved. Save it now

611 37100 Lecture 07-36


SourceProgram

AssemblerProgram

ObjectModule

SourceListing

Assembling a source program


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611 37100 Lecture 07-37

7.4 Assembling and Linking Programs Source listing

Programstartingaddress

Originalsource code

MachineCode

611 37100 Lecture 07-38

7.4 Assembling and Linking Programs Source listing

Symboltable

611 37100 Lecture 07-39

7.4 Assembling and Linking Programs Source listing example of listing syntax error

Comment outthe variable N

List syntaxerror A2008

611 37100 Lecture 07-40

7.4 Assembling and Linking Programs Source listing example of listing syntax error

Spelling errorof DEC

List syntaxerror A2006

611 37100 Lecture 07-41


Link program and modular programming

Object module 1(MOD1.OBJ)

LinkProgram

Runmodule

Linkmap

Linking object modules



611 37100 Lecture 07-42


Benefits of modular programming

Programmers can be working on a softwareproject simultaneously.

Smaller sizes of the modules require less time toedit and assemble.

Modular programming makes it easier to reuse oldsoftware.


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611 37100 Lecture 07-43


Initiating the assembly and linking processes

ML [OPTIONS] file name [[OPTIONS] file name] [/link link-options]

ML /Fl /Fm BLOCK.ASM ()

Display sequences for initiating assembly of a program

611 37100 Lecture 07-44


Initiating the assembly and linking processesLINK ()

Display sequences to initiate linking of object files

Link map file

611 37100 Lecture 07-45

7.5 Loading and Executing a Run

Module

611 37100 Lecture 07-46

7.5 Loading and Executing a Run

Module

assembly language programming with masm

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