Download UNIT-1 Introduction to System Programming

UNIT-1
Introduction to
System
Programming
 Introduction: Components of System
Software,
Language
Processing
Activities, Fundamentals of Language
Processing
Assemblers: Elements of Assembly
Language Programming, A simple
Assembly Scheme, Pass structure of
Assemblers, Design of Two Pass
Assembler, Single pass assembler
Macro Processor: Macro Definition and
call, Macro Expansion, Nested Macro
Calls and definition,Advanced Macro
Facilities, Design of Macro Processor
Unit-1




Introduction:
Components of System Software,
Language Processing Activities,
Fundamentals of Language Processing
Unit-1
Basic
 Software
 Types
of software
 Application software-Word, Excel etc.
 System Software- Compiler, Assembler,O.S.,
Linker etc.
Application
software
Operating
and
systems
software
Users
Hardware
Sytem Programming
Unit-1
Application Software
 Develop
using Programming Lang. and Tools
 Application software interact with systems
software;
 systems software then directs computer hardware
to perform the necessary tasks
Sytem Programming
Unit-1
System Software
 Definition:
The programs directly related to the
computer hardware and perform tasks associated
with controlling and utilizing computer hardware
 Controls operations of computer hardware
 System Software Consist of a variety of program
that support the operation of a computer.
Sytem Programming
Language processing Model
Source Program
C,C++,java etc
Target Assembly Program
Relocatable M/c Code
Lib+ relocatable obj file
Executable M/c Code
Unit-1
Components of System Software,
Example:
 Assembler
 Compiler
 Macro processor
 Loader
 Linker
 Interpreter
 Operating System
 Debugger
 Text editor
Sytem Programming
Programming Languages: Terminology
 Language
translator
 Systems
software that converts a programmer’s
source code into its equivalent in machine language
 Source
code
 High-level
 Object
program code written by the programmer
code
 Another
name for machine language code
Unit-1
Assembler
A
Translator
Assembly
Language
Program
Assembler
Machine
Language format
or object file
Fig.: Assembler
 m/c
and assembly lang. are low level lang.
 Assembly lang. a symbolic representation of machine
language.
 uses a mnemonic to represent each low-level
machine instruction or operation.
Unit-1
Assembler…
 Assembly
lang. is m/c dependent.
 Assemblers with different syntax for a particular CPU or
instruction set architecture.
 Example:- An instruction to add memory data to
a register
x86-family processor: add eax,[ebx],
whereas this would be written addl (%ebx),%eax
in the AT&T syntax used by the GNU Assembler.
Assembler…
 Assembler
make two passes over the assembly
file
 First Pass: reads each line and records labels in
a symbol table.
 Second Pass : use info in symbol table to
produce actual machine code for each line.
Sytem Programming
Unit-1
Compiler
A
compiler translates the high-level source programs
into target programs in machine languages for the
specific hardware.
Error Messages
Program in
Source
Language
Compiler
Target
code(Program in
machine code)
Fig.: A Compiler
 Once
the target program is generated, the user can
execute the program.
 compiler was written by Grace Hopper, in 1952
Unit-1
Compiler…
 Early
compilers were written in assembly language.
 A compiler verifies code syntax,
 generates efficient object code,
 performs run-time organization,
 and formats the output according to assembler
 and linker conventions
Unit-1
Macro Processor

It allows the programmer to write shorthand version of a
program .
 Macro
allows a sequence of source language code to be
defined once and then referred to by name each time it is to
be referred.
 Each time this name Occurs in a program, the sequence of
codes is substituted at that point.
Unit-1
Copy code -- Example
Source
STRG MACRO
STA
DATA1
STB
DATA2
STX
DATA3
MEND
.
STRG
.
STRG
.
.
Expanded source
.
.
.
STA
DATA1
STB
DATA2
STX
DATA3
.
STA
DATA1
STB
DATA2
STX
DATA3
.
{
{
Unit-1
Loader and Linker
A
program which accepts object program and
prepares them for execution.
 4 main functions:
i.
Allocation of space in main memory for the
programs.
ii.
Linking of object modules with each other.
iii. Adjust all address dependent locations.
iv. Physically loading the machine instructions and
data into the main memory.
Unit-1
Interpreter
A
Interpreter reads the source code one
instruction or line at a this line into machine
code or some intermediate form and executes
it.
Program
statement
Interpreter
Machine
language
statement
Statement
execution
Fig.: Interpreter
Unit-1
 Operating
System- Interface between users and
the hardware of a computer system.
 Debugger-
 Text
helps to detect error in the program.
editor- used to create and modify the
program.
Unit-1
Language Processing Activities
 It
Involves translation of a program written in a high
level language into machine code
 It
1.
2.
can be divided into two groups:
Program generation activities
Program execution activities
Unit-1
Language Processing Activities..
Language
processing
activities
Program
generation
Using compiler
Program
execution
Using
interpreter
Unit-1
1.Program generation activities
Error Messages
Program
specification
Program
generator
Program in target
PL
Fig.: Program generator

The program generator is a software which accepts the
specification of program to be generated And generates
a program in target programming lang.
Sytem Programming
Unit-1
1.Program generation activities…
Fig.: Specification and Execution gap
• Execution gap is bridged by the compiler or interpreter.
Sytem Programming
Unit-1
1.Program generation activities…
•
Example- A Screen handling Program.
•
Employee name:
Address:
•
•
•
Married:
Age
Sex
Sytem Programming
Unit-1
2.Program Execution activities
2
Models for Program Execution:
1.Translation
2.Interpretion
Unit-1
1.Program Translation
Fig.: Program Translation Model
Sytem Programming
Unit-1
Program Interpretation
Sytem Programming
Unit-1
Fundamental of Language
Processing
 Language
Processing=Analysis of SP + Synthesis of TP
Fig.: Phases of language of language processing
System Programming
Unit-1
Analysis Phase
Three components:
1. Lexical rules -the formation of valid lexical units in the
source language.
Example:percent_profit = (profit * 100) / cost_price;
identifies =, * and / operators,
100 as constant,
and the remaining strings as identifiers.

2. Syntax rules the formation of valid statements in the source
language.
Example : percent_profit as the left hand side and (profit * 100) /
cost_price as the expression on the right hand side.
System Programming
Unit-1
Analysis Phase…
3. Semantic rules -associate meaning with valid
statements of the language.
example: percent_profit = (profit * 100) / cost_price;
Semantic analysis :
assignment of profit X 100 / cost_price to percent_profit
Sytem Programming
Unit-1
Synthesis Phase

1.
2.
It performs two main activities:
Creation of data structures in the target program
(memory allocation)
Generation of target code (code generation)
Example
MOVER AREG, PROFIT
MULT AREG, 100
DIV
AREG, COST_PRICE
MOVEM AREG, PERCENT_PROFIT
…
PERCENT_PROFIT DW 1
PROFIT DW 1
COST_PRICE DW 1
Sytem Programming
Unit-1






Assemblers:
Elements of Assembly Language Programming,
A simple Assembly Scheme,
Pass structure of Assemblers,
Design of Two Pass Assembler,
Single pass assembler
Unit-1
Assembler
A
Translator
Assembly
Language
Program
Assembler
Machine
Language format
or object file
Fig.: Assembler
 m/c
and assembly lang. are low level lang.
 Assembly lang. a symbolic representation of machine
language.
 uses a mnemonic to represent each low-level
machine instruction or operation.
Unit-1
Assembler…
 Assembly
lang. is m/c dependent.
 Assemblers with different syntax for a particular CPU or
instruction set architecture.
 Example:- An instruction to add memory data to
a register
x86-family processor: add eax,[ebx],
whereas this would be written addl (%ebx),%eax
in the AT&T syntax used by the GNU Assembler.
Unit-1
Assembler…
 The
1.
2.
3.
translated program contains 3 kinds of entities:
Relative entities- address of instructions and
variables.
Absolute entities- operation code , numeric and
string constant and fixed addresses.
The object program- addresses are relative
symbols are defined externally
Sytem Programming
Terminology
 Constant
 Literal
 Forward
reference
Sytem Programming
Unit-1
Literal and Constant
A
literal is an immediate operand
 A literal is an operand with constant value.
In the c-statement
int z=5;
x=y+5;
The constant value is ‘5’ known as literal.
 Literal can not be change during program
execution
 They are specified using immediate addressing.
Sytem Programming
Unit-1
Literal and Constant…
 Literal
in assembly language:
 Assembly instruction for 8086 with immediate
operand
 MOV
AX 15 (8086 instruction)
 But
Hypothetical machine does not support
immediate operand.
Sytem Programming
Unit-1
Literal and Constant…
 Handling
a literal by our machine is as follows:
ADD AREG,=‘6’
ADD AREG, X
-----------------------------------------------X DC ‘6’
Fig.: Handling of literal
Sytem Programming
Unit-1
Forward Reference
 It
is reference to the entity which precedes its
declaration.
 Examples
x=y+5
int x,y;
 The
compiler will not be able to generate the m/c
code for the statement “x=y+5” until it has been
declaration of two variables x & y
Sytem Programming
Unit-1
Forward Reference…

START 100
MOVER AREG, X
-----------------------------------------------X DC ‘1’
Fig.: An Example of forward reference
START 100
BC ANY,L1
X DC ‘1’
-----------------------------------------------L1: MOVER AREG, X
Fig.: An Example of backword reference
Sytem Programming
Unit-1
Assembler
Elements of assembly language programs:
 Basic features
 Statement format
 Operation code
Sytem Programming
Unit-1
Basic features
• Assembly lang. Provides 3 basic features:
1. Mnemonic Operation Codes(Opcodes)
Ex: MOVER or MOVEM
2. Symbolic Operand:
Ex: DS – Declare as storage
DC – Declare as Constant
3. Data Declaration:
Ex: X DC ‘-10.5’
Sytem Programming
Unit-1
Statement Format
Statement Format for instruction:
[Label] <opcode> <operand1> [ <operand2>..]

Label- Optional
Opcode-it contain symbolic operation code
Operand- Operand can also be a CPU register: AREG,
BREG,CREG.
ExampleLOOP :
MOVER AREG, ‘=5’
Unit-1
Machine Instruction Format
Sign
Opcode(2)
Reg.operand(1)
Memory
operand(3)
Fig.: Machine Instruction format for assembly lang.
• Sign is not part of the instruction
•Reg.operand: AREG, BREG,CREG,DREG.
• Memory operand: Refers memory word using symbolic name
Sytem Programming
Unit-1
Machine supports 11 Different Operations
Symbolic opcode
Remark
STOP
Stop Execution
ADD
Operand  Oper1+Oper2
SUB
Operand  Oper1- Oper2
MULT
Operand Oper1*Oper2
MOVER
CPU Register  Memory move
MOVEM
Memory operand  CPU register
COMP
BC
Set condition code
Comparison instruction sets a condition code
The condition code can be tested by BC
Branch on condition
Format for BC : BC <condition code spec.>, <Memory address>
1. LT-Less than
2. LE-Less or equal
3. EQ-Equal
4. GT-Greater than
5. GE-Greater or equal
6. ANY-Implies unconditional transfer of control
Unit-1
Machine supports 11 Different Operations
Symbolic
opcode
Remark
DIV
Operand Oper1/Oper2
READ
Operand2 input value
PRINT
Outputoperand2
First operand is always a CPU register
Second operand is always a memory operand
Unit-1
Assembly language statements

1.
2.
3.
3 types of Statement in Assembly Language
Imperative
Declaration
Assembler directives
Unit-1
1.Imperative Statement
 Executable
statement
 Indicates an action to be taken.
 translates into a machine instruction.
 Ex.: class IS
1. MOVER BREG, X
2. STOP
3. READ X
4. PRINT Y
5. BC NE,L1
Unit-1
2.Declaration statements
 Reserves
memory for variables
 Initial value of a variable can also be specified.
[label] DS <const specifying size of memory to be reserved>
[label] DC <Initial value of variable>
DS  declare storage,
DC  declare constant
Class-DL
Unit-1
2.Declaration statements…
Reserve memory
area of 1 word for
variable X
X
Y
DS
DS
1
5
Unit-1
2.Declaration Statements…
ONE
DC
‘1’
Reserve one word of
memory for variable
ONE
Memory is initialized
with ‘1’
Unit-1
MOT for declaration statements
Mnemonic opcode
Class
Opcode
DS
DL
01
DC
DL
02
Unit-1
3.Assembler Directive
 Instructs
the assembler to perform certain actions
during assembly of a program.






A directive is a direction for the assembler
A directive is also known as pseudo instruction
machine code is not generated for AD.
Example: START, END
Advanced assembler directive: LTORG,EQU,ORIGIN
Class: AD
Unit-1
3.Assembler Directive…
 START
<Constant>
 It indicates that the first word of the m/c code
should be placed in the memory word with the
address <CONSTANT>
Unit-1
3.Assembler Directive…
 END
[<OPERAND SPECIFICATION>]
 Optional,
indicates address of the instruction
where the address of program should begin.
 By default, execution begins from the first
instruction.
 It indicates the end of the source program.
 Class:AD
Advanced Assembler Directives
 These
1.
2.
3.
directive include:
ORIGIN
EQU
LTORG
ORIGIN
 Useful
when m/c code is not stored in consecutive memory
location.
 ORIGIN <address specification>
it containing an Operand or constant or expression.
 Sets
LC to the address given by <address specification>
 LC processing in a relative rather than absolute manner
ORGIN…..Example
Sr. Assembly Program
No
LC
1
START
100
2
MOVER BREG,’=2’
100
3
LOOP MOVER AREG,N
101
4
ADD BREG,’=1’
102
5
ORGIN LOOP+5
6
NEXT BC ANY,LOOP
7
LTORG
8
ORGIN NEXT+2
9
LAST STOP
108
10
N DC ‘5’
109
11
END
106
Remark
ORGIN LOOP+5,
Set LC to the value 106
(101+5=106)
Here, LOOP associated with
address 101
ORGIN NEXT+2
Sets LC to the value 108
(106+2=108)
Here, NEXT associated with
address 106
EQU
Syntax:
<symbol> EQU <address specification>
Where,
<address specification> :can be operand specification or a
constant
<symbol>: EQU Associates symbol with the <address
specification>
Ex. BACK EQU LOOP
The symbol BACK is set to the address of LOOP
LTORG
 Permits
a programmer to specify where literal should
be placed.
 If the LTORG statement not present, literal are present
at the END statement
 At every LTORG Statement, memory is allocated to
the literal of the current pool of literals.
 The pool contains all literal used in the program since
the start of the program or since the last LTORG
statement.
Sytem Programming
Class Field Indicate
Sr.No.
Type of Mnemonic Symbol
Value of class Field
1
Imperative Statement(IS)
1
2
Declaration Statement(DL)
2
3
Assembler directive(AD)
3
4
CPU Register(RG)
4
5
Condition code(CC)
5
Sytem Programming
Unit-1
Mnemonic Opcode Table(MOT)
Mnemonic
opcode
m/c code for
opcode
Class
Size of
instructions
STOP
00
IS
1
ADD
01
IS
1
SUB
02
IS
1
MULT
03
IS
1
MOVER
04
IS
1
MOVEM
05
IS
1
COMP
06
IS
1
BC
07
IS
1
DIV
08
IS
1
READ
09
IS
1
PRINT
10
IS
1
Unit-1
Mnemonic Opcode Table(MOT)…
Mnemonic
opcode
m/c code for
opcode
Class
Size of
instructions
START
01
AD
-----
END
02
AD
-------
ORIGIN
03
AD
-----
EQU
04
AD
-------
LTROG
05
AD
-----
DS
01
DL
-------
DC
02
DL
1
AREG
01
RG
-----
BREG
02
RG
-------
CREG
03
RG
-----
Unit-1
Mnemonic Opcode Table(MOT)…
Mnemonic
opcode
m/c code for
opcode
Class
Size of
instructions
EQ
01
CC
-----
LT
02
CC
-------
GT
03
CC
-----
LE
04
CC
-------
GE
05
CC
-----
NE
06
CC
-------
ANY
07
CC
------
Unit-1
Types of Assembler
3
1.
2.
3.
Types of Assemblers
Load and Go-Assembler
One-pass Assembler
Two-pass Assembler
Sytem Programming
Unit-1
Load and Go-Assembler
 Simplest
form of assembler
 It produces machine language as output which
are loaded directly in main memory and executed
 The ability to design code and test the different
program components in parallel
Sytem Programming
Unit-1
One Pass Assembler
A
One-pass assembler requires 1 scan of the
source program to generate machine code.
 Normally , it does not allow forward referencing.
 An assembler cannot generate m/c code for an
assembly instruction with FR.
 Machine code is generated ,after the address of
variable used in the instruction is known.
 Symbol table is used to record the address of the
variables.
One Pass Assembler with
Forward Reference
 FR
can be tackled using technique Backpatching
1. Operand field of instruction containing FR is left blank initially.
2. A table of instruction containing forward reference is made
separately.
(<Instruction address> ,<Symbol making a forward reference>)
This table can be used o fill-up address in incomplete instruction.
An example of backpatching:
Assign addresses to all statements in the program
Sytem Programming
L1
START 100
MOVER AREG, X………….
ADD BREG, ONE ………….
COMP BREG, TEN…………
BC EQ, LAST
ADD AREG,ONE
BC ANY, L1………….
100
101
102
103
104
105
LAST STOP…………………………… 106
X
DC ‘5’………………… 107
ONE DC ‘1’……………………108
TEN DC ’10 …………………..109
END
Step1: Generation of M/C code and FR left as blank
Assembly instruction
Machine Instruction
START 100
L1
LAST
MOVER AREG, X
100
04
1
ADD BREG, ONE
101
01
2
-----
COMP BREG, TEN
102
06
2
------
BC EQ, LAST
103
07
1
-----
ADD AREG,ONE
104
01
1
-----
BC ANY, L1
105
07
7
101
STOP
106
00
00
000
X
-----
DC
‘5’
107
00
00
005
ONE DC
‘1’
108
00
00
001
109
00
TEN DC ‘10’
00
Sytem Programming
010
Backpatch list:
Instruction Address
Symbol making a
forward reference
000
X
101
ONE
102
TEN
103
LAST
104
ONE
Step2: Backpatch list can be used to fill up address in blank field of instruction
Assembly instruction
Machine Instruction
START 100
L1
LAST
MOVER AREG, X
100
04
1
107
ADD BREG, ONE
101
01
2
108
COMP BREG, TEN
102
06
2
109
BC EQ, LAST
103
07
1
106
ADD AREG,ONE
104
01
1
108
BC ANY, L1
105
07
7
101
STOP
106
00
00
000
X
DC
‘5’
107
00
00
005
ONE DC
‘1’
108
00
00
001
109
00
00
010
TEN DC ‘10’
Sytem Programming
Machine Code Meaning
100
04
Memory Address
Opcode
1
107
Stands for 1-AREG Address of
2-BREG Symbolic operand
3-CREG
4-DREG
Start 100 – is assembler directive,100 LC(location count) 0f first instruction
STOP –M/c opcode-00,absense of register,
108 X
DS
1 : Memory is reserved but no code is generated.
Two Pass Assembler
Two passes(stages),
Pass I
1. Separate the symbols, mnemonic op-code and operand
fields.
2. Determine the storage requirement for every assembly
language statement and update the location counter.
3. Build the symbol table.
Pass II
Generate machine code.

Sytem Programming
Design of Two Pass Assembler
 First
pass has to fix address of variables ,
 the second pass can generate the machine code.
Sytem Programming
Mnemonic
Opcode
length
ADD
01
1
MOVER
04
1
Source
Program
Assembly
PASS II
PASS I
Symbol
Table
X
110
ONE
112
Fig.: Data Structures for the Assembler
Target
Program
Data Access
Control Transfer
Working of pass I
 Data
structures required:
MOT-use to search the opcode
Symbol table-use to search the symbol
Literal table –add literal
Pool table : starting literal number of each pool.
Symbol Table
 It
1.
2.
3.
4.
contains:
Name of variable or a label
Its address
Its size in number of words
ExampleSYMBOL
Index
TABLE
Symbol
address
0
X
214
1
L1
202
2
NXT
207
3
BACK
202
Sytem Programming
Literal Table
 It
contains:
1. Value of the literal
2. Address of the memory location associated with
the literal.
LITERAL TABLE
ExampleLiteral
address
0
5
205
1
2
206
2
1
210
3
2
211
4
4
215
Sytem Programming
Index
POOL Table
 This
table contains the literal number of the starting
literal of each literal pool.
Index
POOL TABLE
0
0
1
2
2
4
Sytem Programming
Intermediate Code
 Is
equivalent representation of source program.
 Pass-I of the assembler involve scanning of the source file.
 Every
opcode is searched in MOT
 Every
operand is searched in symbol table.
 It
helps in avoiding:
1. Scanning of source file in PASS-II
2. Searching MOT and ST in PASS-II
Sytem Programming
Format of Intermediate Code

Each Mnemonic opcode field is represented as:
( Statement class
IS
AD
DL
,
Machine code )
MOT entry of opcode
Ex.: MOVER  (IS,04)
LTORG  (AD, 05)
START(AD,01)
DC  (DL,02)
Operand
(
Operand Class,
C: constant
S: symbol
L: literal(varible)
RG: register
CC:condition code
reference )
for a symbol or
literal , reference
field contains the
index of the operand’s
entry in the symbol
table or literal table
START 200
MOVER AREG, ‘=5’…………. 200
MOVEM AREG, X ………… 201
L1
MOVER BREG, ‘=2’…………. 202
202+3=205
ORIGIN L1+3
LTORG
………… 205,206
NEXT ADD AREG, ‘= 1’
………… 207
SUB BREG,’=2’
………… 208
BC LT, BACK
…………. 209
LTORG
210, 211
BACK EQU L1
………… 202
ORIGIN NEXT+5
MULT AREG,’=4’
…………212
STOP……………………………213
X
DS
‘5’………………………. 214
END
220
SYMBOL TABLE
Symbol
address
0
X
214
1
L1
202
2
NEXT
207
3
BACK
202
Index
LITERAL TABLE
Literal
address
0
5
205
1
2
206
2
1
210
3
2
211
4
4
220
Index
Index
POOL TABLE
0
0
1
2
2
4
Assembly Program
LC
Intermediate Code
………..(AD,01) (C,200)
200 ….(IS,04) (RG,01) (L,0)
201….(IS,05) (RG,01) (S,0)
L1
202 ….(IS,04) (RG,02) (L,1)
(AD,03) (C,205)
205……(DL,02)
(C,5)
206 ……(DL,02) (C,2)
NEXT ADD AREG, ‘= 1’……. 207 ….(IS,01) (RG,01) (L,2)
SUB BREG,’=2’ ……… 208 ….(IS,02) (RG,02) (L,3)
BC LT, BACK…………. 209 ….(IS,07) (CC,02) (S,3)
LTORG ………..
210, ……(DL,02) (C,1)
211 ……(DL,02) (C,2)
BACK EQU L1 ………..202 ………..(AD,01) (C,202)
ORIGIN NEXT+5
………..(AD,01) (C,212)
MULT AREG,’=4’…….. 212 ….(IS,03) (RG,03) (L,4)
STOP…………………… 213 ….(IS,00)
X
DS
‘5’……………. 214 ……(DL,01) (C,1)
END
-------(AD,02)
220 ………(DL,02) (C,4)
START 200
MOVER AREG, ‘=5’…..
MOVEM AREG, X……
MOVER BREG, ‘=2’…..
ORIGIN L1+3
LTORG
START 100
MOVER AREG, ‘=5’
ADD CREG, ‘= 1’
A DS ‘3’
L1 MOVER AREG, B
ADD AREG,C
MOVEM AREG,D
LTORG
EQU A+1
L2 PRINT D
ORIGIN A-1
SUB AREG,’=1’
MULT CREG,B
C DC ‘5’
ORIGIN L2+1
STOP
B DC 19
END
SYMBOL TABLE
Symbol
address
0
A
102
1
L1
105
2
B
112
3
C
103
4
D
103
POOL TABLE
LITERAL TABLE
Literal
address
0
5
108
1
1
109
2
1
113
0
0
1
2
START 100
(AD,01) (C,100)
MOVER AREG, ‘=5’
100(IS,04) (RG,01) (L,0)
ADD CREG, ‘= 1’
101(IS,01) (RG,03) (L,1)
A
DS
‘3’
102(DL,01) (C,3)
L1: MOVER AREG, B
105(IS,04) (RG,01) (S,2)
ADD AREG,C
106(IS,01) (RG,01) (S,3)
MOVEM AREG,D
107(IS,05) (RG,01) (S,4)
LTORG
108 (DL,02)(C,5)
109 (DL,02)(C,1)
D EQU A+1
L2 PRINT D
ORIGIN A-1
SUB AREG,’=1’
MULT CREG,B
C DC ‘5’
ORIGIN L2+1
STOP
B
DC 19
END
(AD,04) (C,103)
110 (IS,10) (S,4)
(AD,03) (C,101)
101(IS,02) (RG,01) (L,2)
102(IS,03) (RG,03) (S,2)
103(DL,02)(C,5)
(AD,03) (C,111)
111(IS,00)
112(DL,02) (C,19)
(AD,02)
113(DL,02) (C,1)
(AD,01) (C,100)
100 (IS,04) (RG,01) (L,0)
100) 04 01
101) 01 03
102)
108
109
105 (IS,04) (RG,01) (S,2)
105)
04
01
112
106 (IS,01) (RG,01) (S,3)
106)
01
01
103
107(IS,05) (RG,01) (S,4)
107)
05
01
103
108)
109)
00
00
00
00
005
001
101 (IS,01) (RG,03) (L,1)
102 (DL,01) (C,3)
108
109
110
101
102
103
111
112
113
(AD,04) (C,103)
(IS,10) (S,4)
(AD,03) (C,101)
(IS,02)(RG,01) (L,2)
(IS,03) (RG,03) (S,2)
(DL,02)(C,5)
(AD,03) (C,111)
(IS,00)
(DL,02) (C,19)
(AD,02)
(DL,02) (C,1)
110)
10
00
103
101)
102)
103)
02
03
00
01
03
00
113
112
005
111)
112)
00
00
00
00
000
019
113)
00
00
001
 Macro
Processor:
 Macro Definition and call,
 Macro Expansion,
 Nested Macro Calls and definition,
 Advanced Macro Facilities,
 Design of Macro Processor
Sytem Programming
94
Introduction
A
macro instruction is simply a
notational convenience for the
programmer.
 A macro represents a commonly used
group of statements in the source
programming language
 Each
time this name Occurs in a program, the sequence
of codes is substituted at that point.

It allows the programmer to write shorthand version of a
program .
Example
|
|
|
ADD 1,DATA
ADD 2,DATA
ADD 3,DATA
|
|
|
ADD 1,DATA
ADD 2,DATA
ADD 3,DATA
|
|
DATA DC 5
96
Basic Macro Processor
Functions
A program with
Macro definitions and
Macro invocations
Expanded program
Macro
Processor
A program without
Macro definitions
Assembler
Object program
Macro
3
main step of macro
1.Deifine the macro name
2. Write its definition
3.Macro call
Macro call : macro processor replaces each
macro call with macro instruction .
98
Macro Instruction definition
 Macro

Definition
Two new assembler directives
 MACRO
 MEND
Syntax
Start of definition………………………..
Macro name……………………………
Sequence of instruction…………..
MACRO
{
End of definition……………………………...MEND
 MACRO:
is the first line of the definition &
identifies the following line as the macro
instruction name.
 The definition is terminated by with MEND
statement
Example
Source program
|
MACRO
INCR
ADD 1,DATA
ADD 2,DATA
ADD 3,DATA
MEND
|
INCR
|
INCR
|
|
|
DATA DC 5
Expanded Source
ADD 1,DATA
ADD 2,DATA
ADD 2,DATA
ADD 1,DATA
ADD 2,DATA
ADD 2,DATA
Difference between macro and
subroutine:
Difference between macro and subroutine:
1. a macro call is an instruction to replace the macro
name with its body, whereas subroutine call is an
instruction to transfer the program’s control to the
subroutine’s definition with all parameters, if required.
2. A macro call results in macro expansion, whereas
subroutine call results in execution.
3. Macro expansion increases the size of the program
but subroutine execution doesn’t affect the size of the
program
4. Macro expansion doesn’t affect the execution speed
of the program much in comparison to subroutines
affect the execution speed of the program

102
Macro Instruction with
parameter
Syntax
Start of definition……………………….. MACRO
Macro name with parameter……………………………
Sequence of instruction…………..
{
End of definition……………………………...MEND
Example:
MACRO
INCR &ARG
ADD AREG,&ARG
ADD AREG,&ARG
ADD AREG,&ARG
MEND
 Macro
Processor:
 Macro Definition and call,
 Macro Expansion,
 Nested Macro Calls and definition,
 Advanced Macro Facilities,
 Design of Macro Processor
Sytem Programming
Unit-1
Macro Processor
 Macro
allows a sequence of source language code to be
defined once and then referred to by name each time it is to
be referred.
 Each time this name Occurs in a program, the sequence of
codes is substituted at that point.

It allows the programmer to write shorthand version of a program .