Compiler Design practical file

Practical File on Compiler Design

BACHELOR OF TECHNOLOGY

IN

COMPUTER SCIENCE & ENGINEERING

Submitted By: Submitted To:

Shahrukhane Alam Mr. Pankaj Sejwal

B.Tech 6th Sem. Faculty of Computer Science

Roll No.13017001009

DEPARTMENT OF COMPUTER SCIENCE & ENGINEERING

P.M. COLLEGE OF ENGINEERING , KAMI , SONEPAT

INDEX

S.no. Program Date Sign Remarks

1. Study of Lex & Yacc Tools

2. PROGRAM TO CHECK

WHEATHER A STRING

BELONGS TO A GRAMMAR OR

NOT.

3. PROGRAM IS TO CALCULATE

LEADING FOR ALL THE NON-

TERMINALS OF THE GIVEN

GRAMMAR

4. TO CALCULATE

TRAILING FOR ALL THE

NON-TERMINALS OF THE

GIVEN GRAMMMAR

5. PROGRAM FOR COMPUTATION

OF FIRST

6. PROGRAM TO FIND THE

NUMBER OF WHITESPACES AND NEWLINES CHARACTERS

7. TO IMPLEMENT STACK USING ARRAY

8. PROGRAM TO IMPLEMENT

STACK USING LINKED LIST

9. PROGRAM TO FIND OUT

WHETHER A GIVEN STRING IS

A IDENTIFIER OR NOT

10. PROGRAM TO FIND WHETHER STRING IS A KEYWORD OR

NOT

Practical =1. Study of Lex & Yacc Tools Lex - A Lexical Analyzer Generator ABSTRACT Lex helps write programs whose control flow is directed by instances of regular

expressions in the input stream. It is well suited for editor-script type

transformations and for segmenting input in preparation for a parsing routine. Lex source is a table of regular expressions and corresponding program

fragments. The table is translated to a program which reads an input stream,

copying it to an output stream and partitioning the input into strings which

match the given expressions. As each such string is recognized the

corresponding program fragment is executed. The recognition of the expressions

is performed by a deterministic finite automaton generated by Lex. The program

fragments written by the user are executed in the order in which the

corresponding regular expressions occur in the input stream. The lexical analysis programs written with Lex accept ambiguous

specifications and choose the longest match possible at each input point. If

necessary, substantial lookahead is performed on the input, but the input stream

will be backed up to the end of the current partition, so that the user has general

freedom to manipulate it. Lex can generate analyzers in either C or Ratfor, a language which can be

translated automatically to portable Fortran. It is available on the PDP-11

UNIX, Honeywell GCOS, and IBM OS systems. 1. Introduction.

Lex is a program generator designed for lexical processing of character input streams. It accepts a high-level, problem oriented specification for character

string matching, and produces a program in a general purpose language which

recognizes regular expressions. The regular expressions are specified by the user

in the source specifications given to Lex. The Lex written code recognizes these expressions in an input stream and partitions the input stream into strings

matching the expressions. At the boundaries between strings program sections

provided by the user are executed. The Lex source file associates the regular

expressions and the program fragments. As each expression appears in the input

to the program written by Lex, the corresponding fragment is executed. The user supplies the additional code beyond expression matching needed to

complete his tasks, possibly including code written by other generators. The

program that recognizes the expressions is generated in the general purpose

programming language employed for the user's program fragments. Thus, a

high level expression language is provided to write the string expressions to be

matched while the user's freedom to write actions is unimpaired. This avoids forcing the user who wishes to use a string manipulation language for input

analysis to write processing programs in the same and often inappropriate string

handling language. Lex is not a complete language, but rather a generator representing a new

language feature which can be added to different programming languages, called ``host languages.'' Just as general purpose languages can produce code to run on

different computer hardware, Lex can write code in different host languages.

The host language is used for the output code generated by Lex and also for the program fragments added by the user. Compatible run-time libraries for the

different host languages are also provided. This makes Lex adaptable to

different environments and different users. Each application may be directed to the combination of hardware and host language appropriate to the task, the user's

background, and the properties of local implementations. At present, the only supported host language is C, although Fortran (in the form of Ratfor [2] has

been available in the past. Lex itself exists on UNIX, GCOS, and OS/370; but

the code generated by Lex may be taken anywhere where appropriate compilers exist. Lex turns the user's expressions and actions (called source in this pic) into the

host general-purpose language; the generated program is named yylex. The

yylex program will recognize expressions in a stream (called input in this pic)

and perform the specified actions for each expression as it is detected.

+-------+ Source -> | Lex | -> yylex

+-------+

+-------+ Input -> | yylex | -> Output

+-------+

An overview of Lex

For a trivial example, consider a program to delete from the input all blanks or

tabs at the ends of lines. %% [ \t]+$ ;

is all that is required. The program contains a %% delimiter to mark the beginning of the rules, and one rule. This rule contains a regular expression

which matches one or more instances of the characters blank or tab (written \t for visibility, in accordance with the C language convention) just prior to the

end of a line. The brackets indicate the character class made of blank and tab;

the + indicates ``one or more ...''; and the $ indicates ``end of line,'' as in QED. No action is specified, so the program generated by Lex (yylex) will ignore

these characters. Everything else will be copied. To change any remaining string of blanks or tabs to a single blank, add another rule:

%% ;

[ \t]+$

[ \t]+ printf(" "); The finite automaton generated for this source will scan for both rules at once,

observing at the termination of the string of blanks or tabs whether or not there is a newline character, and executing the desired rule action. The first rule

matches all strings of blanks or tabs at the end of lines, and the second rule all

remaining strings of blanks or tabs. Lex can be used alone for simple transformations, or for analysis and statistics

gathering on a lexical level. Lex can also be used with a parser generator to perform the lexical analysis phase; it is particularly easy to interface Lex and

Yacc [3]. Lex programs recognize only regular expressions; Yacc writes parsers

that accept a large class of context free grammars, but require a lower level analyzer to recognize input tokens. Thus, a combination of Lex and Yacc is

often appropriate. When used as a preprocessor for a later parser generator, Lex

is used to partition the input stream, and the parser generator assigns structure to

the resulting pieces. The flow of control in such a case (which might be the first half of a compiler, for example) is shown in Figure 2. Additional programs,

written by other generators or by hand, can be added easily to programs written

by Lex.

lexical grammar rules rules

| |

+

v

+

v

+

Lex +---------

| | | Yacc |

+----

-----

| + +--------- +

|

+

v

+

v

+

yylex +---------

Input -> | | -> | yyparse | -> Parsed input +-

------

-- + +--------- +

Lex with Yacc

Yacc users will realize that the name yylex is what Yacc expects its lexical

analyzer to be named, so that the use of this name by Lex simplifies interfacing. Lex generates a deterministic finite automaton from the regular expressions in

the source. The automaton is interpreted, rather than compiled, in order to save

space. The result is still a fast analyzer. In particular, the time taken by a Lex program to recognize and partition an input stream is proportional to the length

of the input. The number of Lex rules or the complexity of the rules is not

important in determining speed, unless rules which include forward context

require a significant amount of rescanning. What does increase with the number

and complexity of rules is the size of the finite automaton, and therefore the size of the program generated by Lex. In the program written by Lex, the user's fragments (representing the actions

to be performed as each regular expression is found) are gathered as cases

of a switch. The automaton interpreter directs the control flow. Opportunity

is provided for the user to insert either declarations or additional statements

in the routine containing the actions, or to add subroutines outside this

action routine. Lex is not limited to source which can be interpreted on the basis of one

character lookahead. For example, if there are two rules, one looking for ab and

another for abcdefg, and the input stream is abcdefh, Lex will recognize ab and

leave the input pointer just before cd. . . Such backup is more costly than the

processing of simpler languages. 2. Lex Source. The general format of Lex source is:

{definitions} %%

{rules} %% {user subroutines}

where the definitions and the user subroutines are often omitted. The

second %% is optional, but the first is required to mark the beginning of the rules. The absolute minimum Lex program is thus

%% (no definitions, no rules) which translates into a program which copies the

input to the output unchanged. In the outline of Lex programs shown above, the rules represent the user's

control decisions; they are a table, in which the left column contains regular

expressions and the right column contains actions, program fragments to be

executed when the expressions are recognized. Thus an individual rule might

appear

integer printf("found keyword INT"); to look for the string integer in the input stream and print the message ``found keyword INT'' whenever it appears. In this example the host procedural language is C and the C library function printf is used to print the string. The

end of the expression is indicated by the first blank or tab character. If the action is merely a single C expression, it can just be given on the right side of the line;

if it is compound, or takes more than a line, it should be enclosed in braces. As a slightly more useful example, suppose it is desired to change a number of words from British to American spelling. Lex rules such as colour

printf("color") mechaniseprintf("mechanize");

petrolprintf("gas"); would be a start. These rules are not quite enough, since the word

petroleum would become gaseum; a way of dealing with this will be a bit

more compl

Practical=2

PROGRAM TO CHECK WHEATHER A STRING BELONGS TO A

GRAMMAR OR NOT. #include<stdio.h>

#include<conio.h> #include<ctype.h>

#include<string.h> void main() {

int a=0,b=0,c,d; char str[20],tok[11]; clrscr();

printf("Input the expression = "); gets(str); while(str[a]!='\0')

{ if((str[a]=='(')||(str[a]=='{'))

{ tok[b]='4';

b++; }

if((str[a]==')')||(str[a]=='}')) { tok[b]='5';

b++; }

if(isdigit(str[a])) {

while(isdigit(str[a])) {

a++; }

a--; tok[b]='6'; b++; }

if(str[a]=='+') {

tok[b]='2'; b++; }

if(str[a]=='*')

{ tok[b]='3';

b++; }

a++;

} tok[b]='\0';

puts(tok); b=0;

while(tok[b]!='\0') {

if(((tok[b]=='6')&&(tok[b+1]=='2')&&(tok[b+2]=='6'))||((tok[b]=='6')&&(tok[b+1

]=='3')&&(tok[b+2]=='6'))||((tok[b]=='4')&&(tok[b+1]=='6')&&(tok[b+2]=='5'))/*||((tok[b

]!=6)&&(tok[b+1]!='\0'))*/) {

tok[b]='6'; c=b+1;

while(tok[c]!='\0') {

tok[c]=tok[c+2]; c++;

} tok[c]='\0'; puts(tok);

b=0; }

else {

b++;

puts(tok); }

} d=strcmp(tok,"6"); if(d==0)

{ printf("It is in the grammar.");

} else

{ printf("It is not in the grammar.");

} getch();

}

OUTPUT Input the expression = (23+) 4625 4625 4625 4625 4625

It is not in the grammar. Input the expression = (2+(3+4)+5) 46246265265 46246265265 46246265265

46246265265 46246265265 462465265 462465265 462465265 462465265 4626265 4626265 46265 46265 465 6 6

It is in the grammar.

Practical=3 TO CALCULATE LEADING OF NON-TERMINALS

#include<conio.h>

#include<stdio.h>

char arr[18][3] =

{

{'E','+','F'},{'E','*','F'},{'E','(','F'},{'E',')','F'},{'E','i','F'},{'E','$','F'}, {'F','+','F'},{'F','*','F'},{'F','(','F'},{'F',')','F'},{'F','i','F'},{'F','$','F'},

{'T','+','F'},{'T','*','F'},{'T','(','F'},{'T',')','F'},{'T','i','F'},{'T','$','F'},

}; char prod[6] = "EETTFF"; char res[6][3]=

{

{'E','+','T'},{'T','\0'}, {'T','*','F'},{'F','\0'},

{'(','E',')'},{'i','\0'}, };

char stack [5][2]; int top = -1; void install(char pro,char re)

{ int i; for(i=0;i<18;++i)

{ if(arr[i][0]==pro && arr[i][1]==re) {

arr[i][2] = 'T'; break; }

} ++top;

stack[top][0]=pro; stack[top][1]=re;

} void main()

{

int i=0,j; char pro,re,pri=' '; clrscr();

for(i=0;i<6;++i)

{

for(j=0;j<3 && res[i][j]!='\0';++j) {

if(res[i][j] =='+'||res[i][j]=='*'||res[i][j]=='('||res[i][j]==')'||res[i][j]=='i'||res[i][j]=='$')

{ install(prod[i],res[i][j]);

break; }

} } while(top>=0)

{ pro = stack[top][0]; re = stack[top][1]; --top; for(i=0;i<6;++i)

{

if(res[i][0]==pro && res[i][0]!=prod[i]) {

install(prod[i],re); }

} }

for(i=0;i<18;++i) {

printf("\n\t"); for(j=0;j<3;++j)

printf("%c\t",arr[i][j]); }

getch(); clrscr();

printf("\n\n"); for(i=0;i<18;++i)

{ if(pri!=arr[i][0])

{ pri=arr[i][0]; printf("\n\t%c -> ",pri);

} if(arr[i][2] =='T')

printf("%c ",arr[i][1]); }

getch();}

OUTPUT

E + T E * T

E ( T E ) F

E I T E $ F

F + F F * F F ( T

F ) F F I T

F $ F T + F

T * T T ( T

T ) F T I T

T $ F

PRACTICAL =4

TO CALCULATE TRAILING FOR ALL THE NON-TERMINALS

OF THE GIVEN GRAMMMAR #include<conio.h>

#include<stdio.h>

char arr[18][3] =

{ {'E','+','F'},{'E','*','F'},{'E','(','F'},{'E',')','F'},{'E','i','F'},{'E','$','F'},

{'F','+','F'},{'F','*','F'},{'F','(','F'},{'F',')','F'},{'F','i','F'},{'F','$','F'}, {'T','+','F'},{'T','*','F'},{'T','(','F'},{'T',')','F'},{'T','i','F'},{'T','$','F'},

}; char prod[6] = "EETTFF"; char res[6][3]=

{ {'E','+','T'},{'T','\0','\0'},

{'T','*','F'},{'F','\0','\0'}, {'(','E',')'},{'i','\0','\0'},

}; char stack [5][2]; int top = -1;

void install(char pro,char re)

{

int i; for(i=0;i<18;++i) {

if(arr[i][0]==pro && arr[i][1]==re) { arr[i][2] = 'T'; break;

} }

++top; stack[top][0]=pro;

stack[top][1]=re;

}

void main()

{ int i=0,j;

char pro,re,pri=' ';

clrscr();

for(i=0;i<6;++i)

{ for(j=2;j>=0;--j)

{ if(res[i][j]=='+'||res[i][j]=='*'||res[i][j]=='('||res[i][j]==')'||res[i][j]=='i'||res[i][j]=='

$')

{ install(prod[i],res[i][j]);

break; }

else if(res[i][j]=='E' || res[i][j]=='F' || res[i][j]=='T') {

if(res[i][j-1]=='+'||res[i][j-1]=='*'||res[i][j-1]=='('||res[i][j-1]==')'||res[i][j-1]=='i'||res[i][j-1]=='$')

{ install(prod[i],res[i][j-1]); break; }

} }

}

while(top>=0)

{ pro = stack[top][0]; re = stack[top][1]; --top; for(i=0;i<6;++i)

{

for(j=2;j>=0;--j) {

if(res[i][0]==pro && res[i][0]!=prod[i]) {

install(prod[i],re); break;

} else if(res[i][0]!='\0') break;

} }}

for(i=0;i<18;++i) {

printf("\n\t");

for(j=0;j<3;++j) printf("%c\t",arr[i][j]); }

getch(); clrscr();

printf("\n\n"); for(i=0;i<18;++i)

{ if(pri!=arr[i][0])

{ pri=arr[i][0]; printf("\n\t%c -> ",pri);

} if(arr[i][2] =='T')

printf("%c ",arr[i][1]); } getch();

}

OUTPUT

E + T

E * T E ( F

E ) T E i T E $ F F + F

F * F F ( F F ) T

F i T F $ F

T + F T * T

T ( F T ) T

T i T T $ F

E -> + * ) i F -> ) i

T -> * )

PRACTICAL= 5

PROGRAM FOR COMPUTATION OF FIRST

#include<stdio.h>

#include<conio.h> #include<string.h>

void main() {

char t[5],nt[10],p[5][5],first[5][5],temp; int i,j,not,nont,k=0,f=0; clrscr();

printf("\nEnter the no. of Non-terminals in the grammer:"); scanf("%d",&nont); printf("\nEnter the Non-terminals in the grammer:\n"); for(i=0;i<nont;i++)

{ scanf("\n%c",&nt[i]);

} printf("\nEnter the no. of Terminals in the grammer: ( Enter e for absiline ) "); scanf("%d",&not);

printf("\nEnter the Terminals in the grammer:\n"); for(i=0;i<not||t[i]=='$';i++) {

scanf("\n%c",&t[i]); }

for(i=0;i<nont;i++) {

p[i][0]=nt[i]; first[i][0]=nt[i];

} printf("\nEnter the productions :\n"); for(i=0;i<nont;i++)

{ scanf("%c",&temp);

printf("\nEnter the production for %c ( End the production with '$' sign ) :",p[i][0]);

for(j=0;p[i][j]!='$';) {

j+=1; scanf("%c",&p[i][j]);

}

} for(i=0;i<nont;i++)

{

printf("\nThe production for %c -> ",p[i][0]); for(j=1;p[i][j]!='$';j++) {

printf("%c",p[i][j]);

} }

for(i=0;i<nont;i++) {

f=0; for(j=1;p[i][j]!='$';j++) {

for(k=0;k<not;k++) { if(f==1)

break;

if(p[i][j]==t[k])

{ first[i][j]=t[k]; first[i][j+1]='$'; f=1;

break;

} else if(p[i][j]==nt[k])

{ first[i][j]=first[k][j]; if(first[i][j]=='e')

continue; first[i][j+1]='$'; f=1;

break; }

} }

} for(i=0;i<nont;i++)

{ printf("\n\nThe first of %c -> ",first[i][0]); for(j=1;first[i][j]!='$';j++)

{ printf("%c\t",first[i][j]);

} }

getch(); }

OUTPUT Enter the no. of Non-terminals in the grammer:3 Enter the Non-terminals in the grammer: ERT Enter the no. of Terminals in the grammer: ( Enter e for absiline ) 5 Enter the Terminals in the grammer: ase*+ Enter the productions : Enter the production for E ( End the production with '$' sign ) :a+s$ Enter the

production for R ( End the production with '$' sign ) :e$ Enter the production for

T ( End the production with '$' sign ) :Rs$

The production for E -> a+s The production

for R -> e The production for T -> Rs The first of E -> a The first of R -> e The

first of T -> e s

PRACTICAL-6

PROGRAM TO FIND THE NUMBER OF WHITESPACES AND

NEWLINES CHARACTERS

#include<stdio.h> #include<conio.h> #include<string.h>

void main() {

char str[200],ch;

int a=0,space=0,newline=0; clrscr();

printf("\n enter a string(press escape to quit entering):"); ch=getche();

while((ch!=27) && (a<199)) {

str[a]=ch; if(str[a]==' ')

{ space++;

} if(str[a]==13) {

newline++; printf("\n");

} a++;

ch=getche(); }

printf("\n the number of lines used : %d",newline+1); printf("\n the number of spaces

used is : %d",space); getch(); }

OUTPUT enter a string(press escape to quit entering):hello! how r u? Do you like prog. in compiler? the number of lines used : 4 the number of spaces used is : 7

PRACTICAL-7 TO IMPLEMENT STACK USING ARRAY #include<stdio.h>

#include<conio.h> #include<string.h>

void main() {

char a[20]={NULL},inp; int ans=0,pos,i;

clrscr(); while(ans<4)

{ pos=0;

while(a[pos]!=NULL && pos<=20) pos++;

printf("\n\n####\tstack=%s, pos=%d",a,pos); printf("\n\n\t\t-- Main Menu --\n\n1. Push\n2. Pop\n3. View Stack\n4.Exit\nYour Choice: ");

scanf("%d",&ans); switch(ans)

{ case 1:

{ if(pos==20)

printf("\nStack is already Full."); else

{ printf("\nEnter input character: ");

scanf("%s",&a[pos]); printf("\nPush Operation Successful."); }

break; }

case 2: {

if(pos==0) printf("\nStack already empty.");

else

{ a[pos-1]=NULL;

printf("\nPop operation Successful."); }

break; }

case 3: {

if(pos==0) printf("\nEmpty Stack.");

else {

printf("\nStack Content:--\n"); for(i=pos-1;i>=0;i--)

{ printf("\n %c",a[i]); if(i==pos-1)

printf(" (Top of the Stack.)"); }

} break;

} case 4:

{ exit();

break; }

default: { printf("\nInvalid Input.");

getch(); exit();

} }

getch(); }

}

OUTPUT

#### stack=, pos=0

-- Main Menu --

1. Push 2. Pop

3. View Stack 4.Exit

Your Choice: 1 Enter input character: karan

Push Operation Successful.

Practical- 8

PROGRAM TO IMPLEMENT STACK USING LINKED LIST

#include<stdio.h> #include<conio.h> struct stack

{

int no; struct stack *next;

} *start=NULL;

typedef struct stack st; void push(); int pop(); void display(); void main()

{

char ch; int choice,item; do

{ clrscr();

printf("\n 1: push"); printf("\n 2: pop"); printf("\n 3: display"); printf("\n Enter your choice"); scanf("%d",&choice);

switch (choice)

{ case 1: push(); break;

case 2: item=pop(); printf("The delete element in %d",item); break;

case 3: display(); break;

default : printf("\n Wrong choice"); }; printf("\n do you want to continue(Y/N)"); fflush(stdin);

scanf("%c",&ch); }

while (ch=='Y'||ch=='y'); }

void push()

{ st *node;

node=(st *)malloc(sizeof(st)); printf("\n Enter the number to be insert"); scanf("%d",&node->no);

node->next=start; start=node;

}

int pop()

{

st *temp; temp=start; if(start==NULL)

{ printf("stack is already empty"); getch();

exit(); }

else {

start=start->next; free(temp); }

return(temp->no); }

void display() {

st *temp; temp=start; while(temp->next!=NULL)

{ printf("\nno=%d",temp->no); temp=temp->next;

} printf("\nno=%d",temp->no);

}

OUTPUT

1: push

2: pop 3: display

Enter your choice3

no=234 do you want to continue(Y/N)

Practical -9

THIS PROGRAM IS TO FIND OUT WHETHER A GIVEN STRING IS A

IDENTIFIER OR NOT

#include<stdio.h> #include<conio.h> int isiden(char*);

int second(char*);

int third(); void main() { char *str; int

i = -1; clrscr(); printf("\n\n\t\tEnter the desired String: "); do

{

++i; str[i] = getch();

if(str[i]!=10 && str[i]!=13)

printf("%c",str[i]); if(str[i] == '\b')

{

--i; printf(" \b"); }

}while(str[i] != 10 && str[i] != 13); if(isident(str))

printf("\n\n\t\tThe given strig is an identifier"); else

printf("\n\n\t\tThe given string is not an identifier");

getch(); } //To Check whether the given string is

identifier or not //This function acts like first

stage of dfa int isident(char *str) { if((str[0]>='a' && str[0]<='z') || (str[0]>='A' && str[0]<='Z'))

{

return(second(str+1)); }

else return 0;

} second(char *str)

{ if((str[0]>='0' && str[0]<='9') || (str[0]>='a' && str[0]<='z') || (str[0]>='A' && str[0]<='Z'))

{

return(second(str+1));

} else

{ if(str[0] == 10 || str[0] == 13)

{ return(third(str));

} else

{ return 0; }

}

} int third() {

return 1; }

OUTPUT:

Enter the desired String: a123

The given strig is an identifier ________________________________________________________________

______________

Enter the desired String: shailesh

The given strig is an identifier ________________________________________________________________

_______________

Enter the desired String: 1asd

The given string is not an identifier ________________________________________________________________

________________

Enter the desired String: as-*l

The given string is not an identifier ________________________________________________________________

_____

Practical -10

PROGRAM TO FIND WHETHER STRING IS A KEYWORD OR

NOT

#include<stdio.h> #include<conio.h> #include<string.h>

void main() {

int i,flag=0,m; char s[5][10]={"if","else","goto","continue","return"},st[10];

clrscr(); printf("\n enter the string :");

gets(st); for(i=0;i<5;i++)

{ m=strcmp(st,s[i]);

if(m==0) flag=1;

} if(flag==0)

printf("\n it is not a keyword"); else

printf("\n it is a keyword");

getch(); }

OUTPUT enter the string :return it is a keyword

enter the string :hello it is not a keyword