system dp mukesh 2003

8/8/2019 System Dp Mukesh 2003

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DESIGN PROBLEM-1

OF

SYSTEM PROGRAMMING

Submitted to: Submitted by:Submitted to: Submitted by:

RUCHI VINAYAK MukeshRUCHI VINAYAK Mukesh

SamtaSamta

B.techB.tech

(H)-CSE(H)-CSE

Section:Section:

A1802A1802


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Roll No.:Roll No.:

RA1802B29RA1802B29

Ques:-Implement all the steps of a compiler for the following

program module:

while (AD OR E>F)

{

If (A==1)

Call ADD (A+B*-C, -D- -F*-G)

Else

W [I, J] = X [I, J] + Y[W[k, L]] + Z[I+J]

U [I, J, K] = V [I, J, K, L]

}

Expectations:

1. Find out all the tokens generated by lexical analyzer.

2. Perform the syntax tree for last two operations (elsestatements) used in program.

3. Perform the semantic analysis over the syntax.

4. Evaluate intermediate code (3-address code) for the given loop.

5. Optimize the code in terms of memory allocation.


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Solution of the design problem

1. Find out all the tokens generated by lexical analyzer.

1.lexical analyzer is basically used for the analysis of keywords,identifiers using a

table called symbol maintenance table. Its classification depends on the source

language that we used for the implementation of all the items produced by the

lexical analyzer.

Analyzer is basically used for the creation or development of the descriptor i.e.

lexical. Token basically consist of two parts i.e. class code and number in class.

Class code is used to address the class to which a lexical unit belongs. Number in

the class is the entry number of the lexical unit in the relevant table.

According to the ques. All elements are divided into following categories:-

a) Identifier-it basically uses id as a symbol or token class code for theidentifier. Numbers can be used as per our requirements.

so ,here are all the identifiers and all the corresponding tokens used for them.


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SR NO TOKEN

I.A id#1

II.B id#2

III.C id#3

IV.D id#4

V .E id#5

VI.F id#6

VII. G id#7

VIII .W id#8

IX .X id#9

X. Y id#10

XII . Z id#11

XIII .U id#12

XIV .V id#13

XV . I id#14

XVI .J id#15

XVII.K id#16

XVII.L id#17


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b).keywords-it is a name that has a special meaning and functionality. It can be

also used to represent symbol or as an operator. In this the class code is a key.

SR NO TOKEN

I .if key#1

II .while key#2

III . ADD key#3

IV. else key#4

3.operator-their main purpose is to compare between the various variables and

identifiers .they are represented by the symbol (op)

SR NO TOKEN

I . < op#1

II. > op#2

III. AND op#3

IV. OR op#4

V. + op#5

VI. - op#6

VII. * op#7

VIII.== op#8

4.special symbol-these are special notations like array parameter etcthe class

code is (ss)


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SR NO TOKEN

I.( ss#1

II.) ss#2

III .[ ss#3

IV.] ss#4

V., ss#5

Expectation2:-Perform the syntax tree for last two operations (else

statements) used in program.


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Solution 2. A)


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B)

Expectation3:- Perform the semantic analysis over the syntax.

Solution of 3:- it is an perception that semantic analysis is always depends on thesyntactic but it is not so it also provides the info. About the syntax provided. Main

purpose of semantic is to do the generation of the intermediate code from the

source language. And it is also used to define an action that has to be implemented

on the source statement. Semantic analysis not only determines the meaning of the

parse tree that has been created by the help of syntax analysis but it also adds


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info. To the the tables .these tables and sequences results in the production of

their in the analysis phase.

Node no. Semantic action

L.place = i

(1) E.place = i

(2) Elist.array = X

Elist.place= i

Elist.dim=1

(3) L.place = j

(4) E.place= j

(5) M=2 , t1 (newtemp) t1=i*d2

Elist.place = t1 t1=t1+j

Elist.dim =m

Elist.array = X

(6) L.place = t2 (newtemp) t2= C(X)L.offset = t3 (newtemp) t3=t1*w

(8) L.place = i

(9) E.place = i

(10) L.place=j

(11) E.place = j

(12) t4 (newtemp ) t4 = i+j

(13) Elist.array = Y

Elist.place = t4

Elist.dim =1


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(14) L.place = k

(15) E.place = k

(16) m=2 , t5 (newtemp) t5=t4*d4

Elist.array = Y

Elist.place = t5 t5=t5 + k

Elist.dim=2

(17) t6 (newtemp) t6= C(Y)

t7 (newtemp) t7= t5*w

(18) t8 (newtemp) t8=t6[t7]

(19) L.place = z

L.offset= NULL

(20) L.place= z

(21) t9 (newtemp) t9 = t8 + z

(22) t2[t3] := t9

Expectation4:- Evaluate intermediate code (3-address code) for the given

loop.

Solution of 4. it is an activity performed by the analyzer for the generation of the

code .but to d this first we have to understand the architecture of the target

language. It helps in understanding of the complete knowledge of all the

instructions and addressing modes that has been used in the target computer.

We have three steps to generate the code:-

1.allocation of the intermediate result in the memory which leads to aseparate process.

2. specification of the instruction that has to be used for type conversion

of operations.


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3.specification of the all the specification modes that has been used for the

variable accessing.

Code generation is as follows:-

1) Compare ( (id # 3)

2) Compare (>) (id # 4) to (id # 5) ( id # 6)

3) We are giving comparison , again them, we get ( id # 2)

4) Assignment operator (=) , with (id # 1) to (id # 20) which -( id # 10)

5) ( - ) -minus sign ; negative (id # 2),which is giving another id ( id # 2)

6) Now comes , multiplication, (id # 11) with (id # 4)(id #11)

7) Next step is of addition (id # 1) with (id # 11) which will give (id # 11)

8) Again there is unary operation (negation) from (id # 12) after (id # 18)

9) After this , it is repetition of unary operation (negation) from (id # 14) to

(id # 13)

10) After this we multiply (id # 12) with (id # 13) giving (id # 13)

11) Unary operation negation (id # 15) (id # 12)

12) After doing all operations , we substrate (id # 12) from (id # 13)

13) Addition of (id # 21 ) with (id # 22) gives (id # 17)

14) Addition of (id # 25) gives (id # 16) (id # 17)

15) Store the result in (id # 19)

16) Store (id # 19) in (id # 15) which is the result.

Expectation 5:-.optimize the code in terms of memory allocation.

Solution 5:-Its main concern is about proper usage or proper management of the

memory by the analyzer and to divide memory in order to get memory optimization.


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Memory optimization can be done in different ways but before doing that we must

know how to allocate the memory and use it in our code.

So first we will concern about the memory allocation and after that steps to

reduce the memory used by us.

Code optimization is as follows:

Symbol type address

A real 2001

C real 2002

temp1 real 2003

B real 2004

D real 2005

temp2 real 2006

E real 2007

F real 2008

temp3 real 2009

temp4 real 2010

temp5 real 2011

temp6 real 2012

temp7 real 2013

b1 real 2014

w[i,j] array 2015

x[i,j] array 2025

temp9 real 2026


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w[i,j] array 2027

x[i,j] array 2037

above we can see that we have used memory for variable, constant or any

operator. There are several Steps to reduce the memory that has been used for

temp allocation but we concern only about the result only.

Steps are as follows:-

1.first,we have to remove all the memory that has been used for temp allocation in

between the calculations.

2.secondly,memeory optimization has to be done for all the operators we have used

like :,and,or,etc.

3.thirdly,identifying and collecting garbage value due to program execution and

compilation.

4.fourthly,by using dynamic memory instead of the of the static memory for the

allocations of variables.

system dp mukesh 2003

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