compiler lab manuals

8/13/2019 Compiler Lab Manuals

1/40

PROGRAM - 1

Objective WAP a program to check a given word is akeyword or not .

Theory Keywords are the reserve words of thelanguage. In C , these C keywords have special meaning to

the C compiler , so the pro-grammar must e careful not to

use these words as identifiers such as variale names.

Every C word is classified as either a keyword or an identifier.All keywords have fixed meanins and these meanings cannote changed. Keywords serve as uilding locks for program

statements.

Algorithm 1. Ass!me an array "a #-dimensional$ in which we store the

keywords as A% & % &.#. 'cans the in(!t word which we want to check in )% &.

*. +ow com(are the strin stored in )% & with ivenkeywords of A% & % &.

,. f strin matchedhen Print / t is a keyword 0Else Print / t is not a eyword 0.

Program 2incl!de3stdio.h4

2incl!de3conio.h42incl!de3strin.h4void main"$5 clrscr"$6char a%7&%8&95:;ein:


2/40

gets+*"

puts+*"

for+i&"i/"i00*

1

c&strcmp+a#i%,*"

if"c99>$5 (rintf":?n ' K23W456)*" reak"

!

c&'"

!

if+c&&'*printf+)(n748 A K23W456)*"

getch+*"

!

@9 :ere some words are used , further we enhance itaccording to our re;uirement 9

Output

Enter the strin =f ' EBOR

Enter the strin =)e

+O A EBOR


3/40

PROGRAM 2

Objective - WAP 84 I I6278I?I25 A76 C47>8A78>

Theory dentifiers refer to the names of varia;les< f!nctions and arrays.hese are !ser-defined names and consist of a seD!ence of lettersand diits< with a letter as a first character. )oth !((ercase andlower case letters are (ermitted< altho!h lowercase letters arecommonly !sed. he !nderscore character is also (ermitted in

identifiers. t is !s!ally !sed as a link ;etween two words in lonidentifiers. Constants in C refer to fixed val!es that do not chaned!rin the exec!tion of a (roram. C s!((orts several ty(es ofconstants as = nteer constants< Real constants< 'inle characterconstants< 'trin constants .

Algorithm 1. oad the iven strin in an array " arr%1>& $ .#. Check the first word of the strin &6


4/40

int i9>


5/40

Output

Enter strin = Bas

Constant= #*dentifier== a;s


6/40

PROGRAM 3

Objective-A program for acktrack parser for the grammar-

> -D cAdA -Daa

Theory n o(-down (arser< there is a well-known draw;ack


7/40

Else ret!rn falseEnd

Program

2incl!de3conio.h42incl!de3stdio.h42incl!de3iostream.h42incl!de3strin.h4int (t9>


8/40

5(tr6if"A"$991$

5

if"str%(tr&99d$5(tr6H

Helse

5co!t33:error@n:6H

Helse

5co!t33:error@n:6H

H

int A"$5int (6

(9(tr6if"str%(tr&99a$

5(tr6if"str%(tr&99;$

5(tr6ret!rn 16H

else5

(tr9(6if"str%(tr&99a$

5(tr6ret!rn 16H

else

5


9/40

ret!rn #6H

HH

else5ret!rn #6H

H

Output

Enter strin==cadcadN s!ccessf!lly Acce(ted

enter strin== ca;dca;dN s!ccessf!lly Acce(ted


10/40

PROGRAM 4

Objective -A program to check that the entered numeris Integer, 5eal,and 2Eponential

with the help of numer grammer-

digit -D 'B...F

sign -D 0-G

Integer -D +digit*0

>ignHInteger -D sign+digit*0

5eal -D >ignHInteger.Integer

2Eponential -D 5eal 2 >ignHInteger

Theory n the first (hase of a com(iler < lexical analyer takes in(!t as are!lar ex(ression and ives o!t(!t < a series of tokens< and then inf!rther (hases


11/40


12/40

Helse if"c99N$5

t916

Helse if"c99eTTc99E$5

e9e16(t9(t16if"e99#$5

t9>6Helse5

"$6H

Helse if"c99,8SSe99>$5

(9(16(t9(t16if"(99#$

5t9>6

(t9l6

Helse5

"$6H

Helse5

t9>6H

H

Output

Enter a n!m;er to check inteer< real < ex(onential=


13/40

,.1*,.1*N

+!m;er is Real

Enter a n!m;er to check inteer < real < ex(onential =,U,UN

+!m;er is nteer


14/40

Program -5

Objective A program for 5ecursive descent parser forthe grammar-

2 -D 82

2-D 082 G

8 -D ?8

8-D 9?8 G

? -D +2* id

where symol G is used to represent null

Theory A (arser that !ses a set of rec!rsive (roced!res to reconie itsin(!t with no ;ack trackin is called a rec!rsive descent (arser.Rec!rsive (roced!res are easy to write and are efficient if written ina (ro(er lan!ae. Be can constr!ct the rec!rsive descent (arserfor the rammar of exam(le.. after removal of ;acktrackin !sinleft rec!rsion removal ;y means of m!t!ally rec!rsive (roced!resas follows.

Algorithm Proced!re E"$

)ein

"$ 5 E EK HEPRME"$

End

Proced!re EPRME"$

)ein f in(!t sym;ol 9 JK then)ein

AVA+CE"$6 5 EK EK H"$6EPRME"$6End

End

Proced!re "$)ein


15/40

L"$6 5 LK HPME"$

End

Proced!re PRME"$)einf in(!t sym;ol 9 JIK then)ein

AVA+CE"$6 5 K ILK HL"$6PRME"$62nd

Procedure ?+*

Jegin

If input symol & idL then

A6MA7C2+*"

2lse if input symol & + then

Jegin

A6MA7C2+*

2+*

If input symol & *L then

AVA+CE"$6

ElseERROR"$6

EndElse ERROR"$6end

Program -

2incl!de3conio.h4

2incl!de3stdio.h42incl!de3iostreame.h42incl!de3strin.h4int (t9>&6void E"$6void EPRME"$6void "$6void PRME"$6

void L"$6


16/40

void main"$5co!t33:enter strin==:6cin44str6

len9strlen"str$6str%len&9N6co!t33str6E"$6if"str%(t&99N$5co!t33:s!ccess:6Helse5co!t33:error:6H@@k(=etch"$6Hvoid E"$

5co!t33:e?n:6"$6

EPRME"$6@@co!t33:e?n:6H

void EPRME"$5co!t33:e(?n: 6if"str%(t&99$

5(t9(t16

"$6EPRME"$6H

Hvoid "$

5co!t33:t?n:6L"$6PRME"$6

H


17/40

void PRME"$5co!t33:t(?n:6if"str%(t&99I$

5(t9(t16L"$6PRME"$6H

Hvoid L"$

5co!t33:f?n:6if"str%(t&99d$

5(t9(t16H

else if"str%(t&99"$5

(t9(t16E"$6if"str%(t&99$$

5

(t9(t16H

elseco!t33:error?n:6

Helseco!t33:error?n:6

H

Output Enter strin == dIddIdNe

t


18/40

ft(ft(

e(s!ccess


19/40

Program

Objective WAP to remove a left recursion of a givenproduction.

Theory it is (ossi;le for a rec!rsive descent (arser to loo(forever. A (ro;lem arises with a / left rec!rsive0 (rod!ctions like

Ex(r eEpr 0 termWhere the left-most symol of the ody is the same as the non-

terminal at the head of the production.

A left-recursive production can e eliminated y rewriting the

offending production. Consider a non-terminal A with twoproductions

A Aa N Bhere Q and S are seD!ences of terminals and non-terminals thatdo not start with A.

Algorithm 1. Arrane the non-terminals of G in some order A1


20/40

int i"

clrscr+*"

for+""*

1

if+&&3NN&&y* 1

printf+)(n 2nter the production to remove left recursion)*"

gets+ar*"

gets+str*"

if+ar#%&&ar#B%strQ&7R==*

1

for+i&"str#B0i%Q&7R=="i00*

1

ar'#%&ar#%"

ar'#'%&ar#'%"

ar'#B0i%&str#B0i%"

!

ar'#B0i%&S"

for+i&"ar#0i%Q&7R=="i00*

1

str#%&S"

str#'%&& "

str#B0i%&ar#0i%"

! str#B0i%&S"

i00"

str#B0i%&N"

i00"

str#B0i%&"

!

else

1

printf+)Aove prodection is not left 5ecursive)*" for+i&"ar#i%Q&7R=="i00*

ar'#i%&ar#i%"

!

printf+)(n(n ur resulting productions are given elowT

(n(n)*"

puts+ar'*"

printf+)(n)*"

puts+str*"


21/40

printf+)(n(nif u want to agin then press y)*"

&getchar+*"

!

else 1

reak"

!

!

!


22/40

Program X

Objective WAP to make a 7?A from a regulareEpression .

Theory he re!lar ex(ression is the notation of choice fordescri;in lexical analyers and other (attern-(rocessinsoftware6 im(lementation of that software reD!ires thesim!lation of a LA or (erha(s the sim!lation of +LA .

;eca!se an +LA often has a choice of move on an in(!tsym;ol < or even a choice of makin a transaction on Y

"n!ll$ or on a real in(!t sym;ol< its sim!lation is lessstraihtforward than for a LA . th!s often it is im(ortantto convert an +LA to a LA that acce(ts the samelan!ae.

Algorithm (Thompson algorithm)Method = - ;ein ;y (arsin r into its constituent su-eEpressions. 8he rules for constructing an 7?A consist of

asis rules for handling sueEpressions with no operators ,and inductive rules for constructing larger 7?ALs from the

7?ALs for the immediate su-eEpressions of a given

eEpression .

Induction T- >uppose 7+s* and 7+t* are 7?ALs for regular

eEpression s and t , respectively.

a$ >uppose r & sNt. 8hen 7+r * ,the 7?A for r , isconstructed .

Zere < i and f are new states < the start and acce(tin states of+" r$


23/40

the only acce(tin state of +" r$ . the acce(tin state of +"s$and the start state of 7+t* are merged into a single state ,will all the transitions in or out of either state. A path

from I to f must go first through 7+s*, and therefore its

lael will egin with some string in =+s*.c$ >uppose r & s9. 8hen for r we construct the 7?A 7+r *.:ere , I and f are new states , the start state and lone

accepting state of 7+r *. 8o get from I to f , we can either

follow the introduced path laeled e , which takes care of

the one string in =+s*' , or we can go to the start state of

7+s* , through that 7?A , then from its accepting state

ack to its start state Uero or more times. 8hese options

allow 7+r * to accept all the strings in =+s*', =+s*B, and so

on , so the entire set of strings accepted y 7+r * is =+s*9.

d$ Linally< s!((ose r 9 "s$ . then "r $ 9 "s$ < and we can !sethe +LA +"s$ as +"r $.

Program

2incl!de3stdio.h42incl!de3conio.h42incl!de3stdli;.h4

int n&&&6char t%#>&6char I(tr 9 : :6char Istr 9 : :6

void nfa"void$6

@@ MA+ L[+CO+

void main"$5

clrscr"$6(rintf":?n Enter Re!lar Ex(ression = :$6scanf":\s:


24/40

void nfa"void$5

int i$5

if"I(tr 49 X SS I(tr 39 1##$5

st%F& 9 i16end%F& 9 i#6t%F& 9 I(tr6i6

F6(tr6

H

if"I(tr 99 " SS I(tr ]9 $ SS I"(tr #$ 99 @$5

if"I"(tr U$ 99 I$5st%F& 9 i6end%F& 9 i 16t%F& 9 e6

F6

st%F& 9 i 86end%F& 9 i16

t%F& 9 e6F6st%F& 9 i 86end%F& 9 i X6t%F& 9e6

F6

st%F& 9 i6end%F& 9 i X6t%F& 9 e6

F6


25/40

i6H

st%F& 9 i6end%F& 9 i16

t%F& 9 e6F6

st%F& 9 i6end%F& 9 i*6t%F& 9 e6

F6i6

(tr6

st%F& 9 i6end%F& 9 i16t%F& 9 I(tr6

F6i6

(tr6

st%F& 9 i6end%F& 9i*6

t%F& 9 e6F6i6

(tr6st%F& 9 i6end%F& 9 i16t%F& 9 I(tr6

F6i6

(tr6

st%F& 9 i6end%F& 9 i16t%F& 9 e6

F6i6str 9 (tr6if"I"str 1$ 99 I SS I"str #$ 99?>$

;reak6

if"I"str 1$ 99 I$


26/40

5(tr6i6while"I(tr ]9 ?>$

5 st%F& 9 i6end%F& 9 i16

(tr6t%F& 9 I(tr6

F6i6

(tr6H

;reak6H

(tr6

Hif"I(tr 99 "$5

n9i16x9i#6st%F&9i16

end%F&9i#6t%F& 9 e6

F6i6

(tr6H

if"I"(tr 1$ 99 I$5

st%F& 9 i16end%F& 9 i#6t%F& 9 e6

F6i6

(tr6H

if"I(tr 99I$5st%F& 9 i6


27/40

end%F& 9 x6t%F& 9 e6

F6st%F& 9 n6

end%F& 9 i16t%F& 9 e6F6(tr6H

H(rintf":?n ransition ta;le for +LA = :$6(rintf":?n?n 'AR i@o 'ym;ol E+ ?n:$6(rintf": 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 99 9 ?n?n:$6n 9 F6for"r9>6r3F6r$

(rintf":?n \d\Uc\8d:


28/40

Program /

Objective A program for stack implementation of shift

reduce parser for the grammar-9 2 -D 8:8 -D ?R

: -D 08: e where e & 7R==

R -D 9?R e

? -D i where i& any alphaetic terminal word.

Theory his (arsin method is ;ottom-!( ;eca!se it attem(ts to constr!ct a

(arse tree for an in(!t strin ;einnin at the leaves "the ;ottom $

and workin !( towards the root " the to( $ .here are two (ro;lems that m!st ;e solved if we are to a!tomate

(arsin ;y handle (r!nin. he first is how to locate a handle in ariht-sentential from < and the second is what (rod!ction to choosein case there is more than one (rod!ction with the same riht side.Bhole the (rimary o(erations of the (arser are shift and red!ce6i--6Helse 5state91X6startid9i-#6 i--6H

;reak6@@'tate for whilecase *=if"str%i&99h$ state9,6else5state91X6startid9i-16i--6H

;reak6case ,=if"str%i&99i$ state9U6else5state91X6startid9i-#6i--6H

;reak6case U=if"str%i&99l$ state986else5state91X6startid9i-*6i--6H

;reak6case 8= if"str%i&99e$ state9X6else5state91X6startid9i-,6i--6H

;reak6case X= if"str%i&99"TTstr%i&99+[$

5(rintf":while = eyword?n?n:$6state9>6i--6Helse5state91X6startid9i-U6i--6H

;reak6@@state for docase 7=if"str%i&99o$ state96

else5state91X6startid9i-16i--6H;reak6case = if"str%i&995TTstr%i&99 TTstr%i&99+[TTstr%i&99"$5

(rintf":do = eyword?n?n:$6state9>6i--6H

;reak6

@@state for else


36/40

case 1>=if"str%i&99l$ state9116else5state91X6startid9i-16i--6H

;reak6case 11=if"str%i99s&$ state91#6

else5state91X6startid9i-#6i--6H;reak6case 1#=if"str%i&99e$ state91*6else5state91X6startid9i-*6i--6H

;reak6case 1*=if"str%i&995TTstr%i&99+[$5

(rintf":else = eyword?n?n:$6state9>6i--6Helse5state91X6startid9i-,6i--6H

;reak6@@state for forcase 1,=if"str%i&99o$ state91U6else5state91X6startid9i-16i--6H

;reak6case 1U=if"str%i&99r$ state9186else5state91X6startid9i-#6i--6H

;reak6case 18=if"str%i&99"TTstr%i&99+[$5

(rintf":for =eyword?n?n:$6state9>6i--6Helse5state91X6startid9i-*6i--6H

;reak6

@@'tates for identifierscase 1X=if"isaln!m"str%i&$TTstr%i&99-$5state9176i6Helse if"str%i&99+[TTstr%i&993TTstr%i&994TTstr%i&99"TTstr%i&99$TTstr%i&996TTstr%i&999TTstr%i&99TTstr%i&99-$ state9176i--6

;reak6


37/40

case17=if"str%i&99+[TTstr%i&993TTstr%i&994TTstr%i&99"TTstr%i&99$TTstr%i&996TTstr%i&999TTstr%i&99TTstr%i&99-$5endid9i-16

(rintf"::$6for"F9startid6F39endid6F$(rintf":\c:6else if"isaln!m"str%i&$TTstr%i&99$5

(rintf":3 = Relational o(erator?n?n:$6i--6state9>6H

;reak6case #>= if"isaln!m"str%i&$TTstr%i&99$5

(rintf":39 = Relational o(erator?n?n:$6i--6state9>6H

;reak6@@states for relational o(erator 4S 49case #1= if"str%i&999$ state9##6else if"isaln!m"str%i&$TTstr%i&99$5

(rintf":4 = relational o(erator?n?n:$6i--6state9>6H

;reak6case ##= if"isaln!m"str%i&$TTstr%i&99$5

(rintf":49 =relational o(erator?n?n:$6i--6

state9>6


38/40

H;reak6@@states for relational o(erator 99 S assinment o( 9case #*= if"str%i&999$ state9#,6

else5(rintf": = Assinment o(erator?n?n:$6i--6state9>6H

;reak6case #,= if"isaln!m"str%i&$$5

(rintf":99 =Relational o(erator?n?n:$6state9>6i--6H

;reak6@@states for constantscase #U= if"isal(ha"str%i&$$5

(rintf": =Error=?n?n:$6(!ts"str$6

for"F9>6F3i6F$(rintf"::$6 (rintf":Y:$6 (rintf":Error at (osition \d Al(ha;et can not follow diit?n?n:6i--6H

;reak6

@@states for s(ecial character "


39/40

case #8= (rintf":" ='(ecial character?n?n:$6startid9i6state9>6i--6

;reak6@@state for s(ecial character $case #X= (rintf":$ = '(ecial character?n?n:$6state9>6i--6

;reak6@@state for 6case #7= (rintf":6 ='(ecial character?n?n:$6state9>6i--6

;reak6@@state for case #= (rintf": =O(erator?n?n:$6state9>6i--6

;reak6case *>= (rintf":- =O(erator?n?n:$6state9>6i--6

;reak6@@error state@@case = oto E+6Hi6H

(rintf":End of the (roram:$6@@E+6etch"$6

H

@I his (roram works as a model of lexical analyser" or shows workin $ < which se(erates the syntax andenerates token I@


40/40

Output

IIIIIIIIProram for lexical analysisIIIIIIII

Enter the strin = ifAnalysis

f = eywordEnd of the (roram.

IIIIIIIIProram for lexical analysisIIIIIIII

Enter the strin = Analysis

= O(erator End of the (roram.

compiler lab manuals

Documents