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bdghEX. NO: 1

DATE:

CONSTRUCTION OF NFA FROM A GIVEN

REGULAR EXPRESSION

AIM:erwhyerh

To write a C program to convert the regular expression to NFA.

ALGORITHM:

1. Start the program

2. Get the regular expression from the user

3. Build a finite automatgfon with one final state (and one start state)

4. Use the recursisgdfve definition of regular expressions to build a finite automaton for a

compound expressiofgfn from the finite automata for the base expressions

5. Repeat the procedure to construct the NFA for the given regular expression

6. Display the NFA structure.

7. Stop the program

PROGRAM:

#include

#include

#include

#include

#include

struct s1

{

int from,to;

};

typedef struct s1 s2;

struct ed

{

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sgfdint from[30],to[30],n;

char on[30];

};

typedef struct ed edg;

int top=0,start,end;

s2 s3[100];

edg edg1,edg2;

s2 pop()

{

top--;

return s3[top];

}

void push(s2 a)

{fgs3[top]=a;

top++;

}

int f(char a)

{

if(a=='/') return 1;if(a=='.') return 3;

if(isalnum(a)) return 5;

if(a=='(') return 7;

if(a==')') return 0;

return -1;

}

int g(cfdgshar a)

{

if(a=='/') return 2;

if(a=='.') return 4;

if(isalnfgdfgum(a)) return 6;

if(a=='(') return 0;

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if(a==')') return 8;

return -1;

}

char *postfix(char in1[])

{

int i=0,j=0,top=0,a,l,t=0;

g char s[100],*out,*in;

l=strlen(in1);

in=malloc(100);

for(;t

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if(s[toGNWTHSRp]=='*') a=1;

SS

else

{

out[j++]=s[top-a];

if(a) out[j++]=s[top];

top-=(a+1);

}

}

if(f(in[i])!=g(s[top]))

s[++top]=in[i];

elseHSDGH

{

toSDGHp--;

if(in[i+1]=='*')

{

out[j++]='*';

i++;

}}

i++;

}

if(toDHSFHp!=-1)

{

printf("\n ERROR");

exit(1QSDFH0)DFHSH;

}

out[j]='\0';

reDFHSZFHrhgasfturn out;

}

grcon(char *v)

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{

int i=0,l=0,k;

ssdfhszdfh2 a,b;

edgdsafhsh1.n=0;

whsdzghshile(v[i])

{

if(isasdfhlnum(v[i]))

{

edg1.fsfdhromdfhszfh[edg1.n]=++l;

edg1.to[dfhsdfhedg1.n]=++l;

edg1.osdfhdfhn[edg1.n]=v[i];

a.from=edg1.from[edg1.n];

a.tsfhsdfho=edg1.to[edg1.n];

edgdfhszdfh1.n++;

pdfshdfush(a);

}

elsdhfe fdhsif(v[i]=='*')

{

a=psdfhsop();edg1dhfsdf.from[edg1.n]=a.to;

edg1.dsfhdfhto[edg1.n]=a.from;

edgsdfhsdfh1.on[edg1.n]='E';

edg1.n++;

edg1.fshfdhsdfhdhrom[edg1.n]=++l;

edg1.to[edg1.n]=a.from;

edg1.on[edg1.n]='E';

edshfdhg1.n++;

eddshsdhg1.from[edg1.n]=a.to;

edhsdg1.to[edg1.n]=++l;

edgshdfh1.on[edg1.n]='E';

eddfhg1.n++;

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edsdfhg1.from[edg1.n]=l-1;

edsfdhg1.tsdho[edg1.n]=l;

edfsdhg1sdfhhfdhs.on[edg1.n]='E';

edsdfhg1.n++;

a.fdhsfdhfrom=l-1;

a.thssdfho=l;

pufdshsdfsh(a);

}

else if(v[i]=='.')

{

b=pop();

a=pop();

for(k=0;k

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edg1.to[edg1.n]=b.from;


edg1.n++;

edg1.from[edg1.n]=b.to;

edg1.to[edg1.n]=++l;


edg1.n++;

edg1.from[edg1.n]=a.to;

edg1.to[edg1.n]=l;


edg1.n++;

a.from=l-1;

a.to=l;

push(a);

}

i++;

}

a=pop();

start=a.from;end=a.to;

return 0;

}

void main()

{

int i;

char re[20];

clrscr();

printf(" Enter the Expression\t:");

scanf("%s",re);

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grcon(postfix(re));

printf("\n START =%d \n FINAL =%d ",start,end);

printf("\n\n Here E is considered as epsilon\n");

for(i=0;i

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1. Start the program.

2. Get the regular expression from user.

3. Construct the Postfix for the given expression.

4. Convert the regular expression to DFA using conversion algorithm.

5. Display the result.

6. Stop the Program.

PROGRAM:

#include

#include

#include

#include

#define N 30

int Pos[N],Posct,Lng, Rlnk[N], Llnk[N], Root, Nl[N];

char Str[N], Type[N], FoPos[N][N], FPos[N][N], LPos[N][N];

char Inp[N], Sym[N], DSts[N+N][N];

int DStct, DTran[N+N][N];

int IsTerm (char c){

return (isalpha (c) || c == '#');

}

int Val (char c)

{

return (int)c - (int)'0';

}

char State (int c)

{

if (c == -1) return '-';

return (char) ((int) 'A' + c);

}

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void Sort (char s[])

{

int i,l, c = 1;

char ch;

for ( l=0; s[l]!='\0'; l++);

while (c)

{

c=0;

for (i=1; i

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{

fnd = 1;

j = l1;

}

}

if (fnd == 0) s1[l1++] = s2[i];

}

s1[l1] = '\0';

Sort (s1);

}

void Find (int Nd)

{

if (Nd == -1) return;

if (IsTerm (Str[Nd]))

{

Nl[Nd] = 0;

FPos[Nd][0] = LPos[Nd][0] = (char) (Pos[Nd] + (int) '0');

FPos[Nd][1] = LPos[Nd][1] = '\0';

}else

{

int Left = Llnk[Nd], Right = Rlnk[Nd];

Find (Left);

Find (Right);

if (Str[Nd] == '/')

{

Nl[Nd] = (Nl[Left] || Nl[Right]) ? 1 : 0;

strcpy (FPos[Nd], FPos[Left]);

Add (FPos[Nd], FPos[Right]);

strcpy (LPos[Nd], LPos[Left]);

Add (LPos[Nd], LPos[Right]);

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}

else if (Str[Nd] == '*')

{

for ( i=0; LPos[Left][i]!='\0'; i++)

Add (FoPos[Val (LPos[Left][i])], FPos[Left]);

}

}

void ConvPostFix ()

{

char s[N], St[N];

int Top = -1, i, j, k;

strcpy (s, Str);

for (i=j=0; i

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while (Top >= 0 && St[Top] != '(')

Str[j++] = St[Top--];

St[++Top] = s[i];

}

}

while (Top >= 0)

Str[j++] = St[Top--];

Str[j] = '\0';

Lng = j;

}

void FormTree ()

{

char c[2];

int i;

int Asgn[N], Ct = 0;

c[1] = '\0';

for ( i=0; i

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for (i=0; i

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Str[0] = s[0];

ss[0] = s[0];

while (loop)

{

loop = 0;

for ( i=1,j=1; s[i]!='\0';i++)

{

ss[j++] = s[i];

if (s[i] ==')' && s[i-1] == '(')

{

loop++;

j -= 2;

}

}

ss[j] = '\0';

}

for ( i=1; s[i] != '\0'; i++)

{if ((IsTerm (s[i]) && (IsTerm (s[i-1]) || s[i-1] == '*'

|| s[i-1] == ')')) || (s[i] == '(' && s[i-1] != '('

&& s[i-1] != '/'))

Str[++Lng] = '.';

Str[++Lng] = s[i];

}

Str[++Lng] = '\0';

ConvPostFix ();

}

void ConDFA ()

{

char c[N];

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int fnd,i,j,k;

FormTree ();

FindN_F_L ();

strcpy(DSts[0], FPos[Root]);

DStct = 1;

for ( i=0; i

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}

}

}

}

void Print ()

{

int i,j;

printf(" \nPostFix Form : " );

printf( "%s",Str);

printf( "\n\nState\t");

for ( i=1; Inp[i]!='\0'; i++) printf("%c\t",Inp[i]);// printf("\t");

printf( "\n-");

for (i=1; Inp[i]!='\0'; i++) printf( "--------");

for (i=0; i

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GetRE (s);

ConDFA ();

Print ();

getch ();

}

OUTPUT:

Enter the Regular Expression: (a/b)*abb

PostFix Form: ab/*a.b.b.#.

State a b

---------------------------------------

A B A Start State

B B C

C B D

D B A Final State

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RESULT:

Thus the C program for construction of DFA from a Regular expression is executed and

verified.

EX. NO: 3

DATE:

IMPLEMENTATION OF LEXICAL ANALYSIS USING

LEX TOOL

AIM:

To write a C program to implement lexical analysis using LEX tool.

ALGORITHM:


2. Lex program consists of three parts.

Declaration %%

Translation rules %%

Auxilary procedure.

3. The declaration section includes declaration of variables, maintest, constants and regular

definitions.

4. Translation rule of lex program are statements of the form

P1 {action}

P2 {action}

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Pn {action}

5. Write a program in the vi editor and save it with .l extension.

6. Compile the lex program with lex compiler to produce output file as lex.yy.c.

Eg: $ lex filename.l

$ cc lex.yy.c -ll

7. Compile that file with C compiler and verify the output.

Eg: $ ./a.out inputfilename.c

8. Stop the program.

PROGRAM:

%{

%}

digit[0-9]+

identifier[a-zA-Z][a-zA-Z0-9]*

%%

#.* printf("%s is a preprocessor directive\n",yytext);

int |

float |

double |

if |

char |

goto printf("%s is a keyword\n",yytext);

{identifier} printf("%s is a function \n",yytext);

\{ printf("%s block begin\n",yytext);

\} printf("%s block end \n",yytext);

{identifier}(\[0-9]*\])? printf("%s is identifier\n",yytext);

\"+\" printf("%s is string\n",yytext);

-{digit} printf("%s is anegative number\n",yytext);

"+"?{digit} printf("%s is a positive number\n",yytext);

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\; |

$ |

$ printf("%s is a special operator\n",yytext);

\= printf("%s is a assignment operator\n",yytext);

\< |

\> |

\= printf("%s is a relation operator\n",yytext);

%%

int main(int argc,char *argv[])

{

if(argc>1)

{

FILE *fp;

fp=fopen(argv[1],"r");

if(!fp)

{

printf("cant open");exit(0);

}

yyin=fp;

yylex();

}

return(0);

}

INPUT:

#include

void main()

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{

int a,b;

a=10;

b=20;

}

OUTPUT:

[cse4@localhost ~]$ lex cse.l

[cse4@localhost ~]$ cc lex.yy.c -ll

[cse4@localhost ~]$ ./a.out input.c#include is a preprocessor directive

void is identifier

main() is a function

{ block begin

int is a keyword

a is identifier

,b is identifier

; is a special operator

a is identifier

= is a assignment operator

10 is a positive number


b is identifier

= is a assignment operator

20 is a positive number


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} block end

RESULT:

Thus the program for implementation of lexical analysis using lex tool is executed and

verified.

EX. NO: 4

DATE:

IMPLEMENTATION OF SYMBOL TABLE

AIM:

To write a C program to implement the symbol table.

ALGORITHM:

1. Start the program

2. Declare the variables

3. Display the menu for the user

1. Insert 2. Delete 3. Lookup 4. Exit

4. To insert a value in the symbol table, get the values of token, attribute and position from

the user.

5. Index and next position is generated by the values of token, attribute and position

6. To delete a value in the symbol table, get the value of token.

7. To search a token, get the value of token and search it in symbol table.


PROGRAM:

#include

#include

#include

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#include

#include

#define NULL 0

int size=0;

void insert();

void del();

int search(char lab[]);

void display();

void dis(char le[]);

struct symtab

{

char label[10];

char att[10];

int addr;

struct symtab *next;

};

struct symtab *first,*last;

void main()

{int op;

int y;

char la[10];

clrscr();

do

{

printf("\n SYMBOL TABLE IMPLEMENTATION \n") ;

printf("\n 1.Insert \n");

printf("\n 2.Delete \n");

printf("\n 3.Lookup \n");

printf("\n 4.Exit \n");

printf("\n Enter your choice: \n");

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scanf("%d",&op);

switch(op)

{

case 1:

insert();

display();

break;

case 2:

del();

display();

break;

case 3:

printf("\n Enter the label to be searched:");

scanf("%s",&la);

y=search(la);

if(y==1)

{

printf("\n The label is present in the symbol table");

dis(la);}

else

printf("\n The lable is not found in the symbol table");

break;

case 4:

break;

}

}

while(op

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{

int n;

char l[10];

printf("\n Enter the token:");

scanf("%s",&l);

n=search(l);

if(n==1)

printf("\n The token is already exists duplicate can't be inserted \n");

else

{

struct symtab *p;

p=malloc(sizeof(struct symtab));

strcpy(p->label,l);

printf("\n Enter the position:");

scanf("%d",&p->addr);

printf("\n Enter the attribute:");

scanf("%s",&p->att);

p->next=NULL;

if(size==0){

first=p;

last=p;

}

else

{

last->next=p;

last=p;

}

size++;

}

}

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void display()

{

int i;

struct symtab *p;

p=first;

printf("\n Index \t Next \t Token \t Attribute \t Position \n");

for(i=0;ilabel,p->att,p->addr);

p=p->next;

}

}

int search(char lab[])

{

int i,flag=0;

struct symtab *p;

p=first;

for(i=0;ilabel,lab)==0)

{

flag=1;

}

p=p->next;

}

return flag;

}

void del()

{

int a;

char l[10];

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struct symtab *p,*q;

p=first;

printf("\n Enter the label to be deleted: \n");

scanf("%s",l);

a=search(l);

if(a==0)

printf("\n The label is not found");

else

{

if(strcmp(first->label,l)==0)

{

first=first->next;

}

else

if(strcmp(last->label,l)==0)

{

q=p->next;

while(strcmp(q->label,l)!=0)

{p=p->next;

q=q->next;

}

p->next=NULL;

last=p;

}

else

{

q=p->next;

while(strcmp(q->label,l)!=0)

{

p=p->next;

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q=q->next;

}

p->next=q->next;

}

size--;

}

}

void dis(char le[])

{

int i;

struct symtab *p;

p=first;

printf("\n Index \t Next \t Token \t Attribute \t Position \n");

for(i=0;ilabel,le)==0)

{

printf("\n %d \t %d \t %s \t %s \t %d \n",i,i+1,p->label,p->att,p->addr);

break;}

else

p=p->next;

}

}

OUTPUT:

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SYMBOL TABLE IMPLEMENTATION

1. Insert

2. Delete

3. Lookup

4. Exit

Enter your choice: 1

Enter the token: c

Enter the position: 4

Enter the attribute: identifier

Index Next Token Attribute Position

0 1 a identifier 1

1 2 = operator 2

2 3 b identifier 0

3 4 + operator 3

4 5 c identifier 0


1. Insert

2. Delete

3. Lookup

4. Exit


Enter the label to be deleted: c


0 1 a Identifier 1

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1 2 = operator 2

2 3 b identifier 0

3 4 + operator 3


1. Insert

2. Delete

3. Lookup

4. Exit


Enter the label to be searched: a

The label is present in the symbol table


0 1 a identifier 1


1. Insert

2. Delete

3. Lookup

4. Exit


RESULT:

Thus the program for implementation of symbol table is executed and verified.

EX. NO: 5

DATE:

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IMPLEMENTATION OF OPERATOR PRECENDENCE

PARSING

AIM:

To write a C program to implement operator precedence parsing algorithm.

ALGORITHM:

Input: An input string w and a table of precedence relations.

Output: If w is well formed, a skeletal parse tree, with a placeholder nonterminal E

labeling all interior nodes; otherwise, an error indication.

Method: Initially, the stack contains $ and the input buffer the string w$.

set ip to point to the first symbol of w$;

repeat forever

if$ is on top of the stack and ip points to $ then

return

else begin

let a be the topmost terminal symbol on the stack

and let b be the symbol pointed to by ip;

ifa b then

repeat

pop the stack

until the top stack terminal is related by

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PROGRAM:

#include

#include

#define msize 40

struct stack

{

int top;

char item[msize];

}s;

void main()

{

int i,j,re;

char *in;

char c1,c2;

s.top=-1;

clrscr();

printf("Enter the input language\n");

scanf("%s",in);

printf("\nStack \t\tInput\t\tActions\n\n");

push(&s,'$');

printf("$\t\t\t");

for(i=0;in[i];i++)

printf("%c",in[i]);

printf("$\tInitial\n");

for(i=0;in[i];i++)

{

re=check(s.item[s.top],in[i]);

if(re==1)

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{

push(&s,in[i]);

for(j=0;j

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for(j=0;j

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printf("\t\t\t");

printf("$");

printf("\t\tpop(%c>$)\n",s.item[s.top+1]);

}

}

if(s.top==0)

printf("\nACCEPTED");

else

printf("\nERROR");

getch();

}

int check(char st,char ip)

{

char id[8]={'+','-','*','i','$','/','(',')'};

char pre[8][8]={{'=','>','','=','','>','=','','>',''},

{'>','>','>','=','>','>','0','>'},{'','>','0','>'}};

int i,t1,t2;

for(i=0;i

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if(pre[t1][t2]=='>')

return 0;

else

return 1;

}

push(ps,x)

struct stack *ps;

char x;

{

ps->item[++(ps->top)]=x;

return;

}

pop(ps)

struct stack *ps;

{

ps->item[--(ps->top)];

return;

}

OUTPUT:

Enter the input language

i*i+(i/i)

Stack Input Actions

$ i*i+(i/i)$ Initial

$i *i+(i/i)$ push($*)

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$* i+(i/i)$ push($+)

$+ (i/i)$ push($))

$) $ push($$)

ACCEPTED

RESULT:

Thus the C program for implementation of operator precedence parsing is executed and

verified.

EX. NO: 6

DATE:

IMPLEMENTATION OF SYNTAX ANALYSIS USING

YACC TOOLAIM:

Write a program to implement a syntax analysis using Yacc tool.

ALGORITHM:

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2. Yacc program consists of three parts namely

Declarations

%%

Transition Rule

%%

Supporting C routines.

3. Declaration part consists of two sections, first section contains only include statements

and the second statements contains declaration of the grammar tokens.

4. Each rule in set of transition rules consists of grammar production and semantic action.

The set of productions are of the form

: {semantic action 1}

| {semantic action 2}

..

| {semantic action n}

;

5. In the third part, error recovery routines are added.

6. The program is typed using vi editor, and saved with .y extension.

7. It is first compiled with the yacc compiler to produce the C code for C compiler (yaccfilename.y).

8. After that compile that program with C compiler (cc y.tab.c ly ).

9. See the output using ./a.out.


PROGRAM:

%{

#include

#include

#define YYSTYPE double

%}

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%token number

%left '+''-'

%left '*''/'

%right uminus

%%

lines: expr'\n'{printf("The Result is : \t %g\n",$$);}

expr:expr'+'expr{$$=$1+$3;}

|expr'-'expr{$$=$1-$3;}

|expr'*'expr{$$=$1*$3;}

|expr'/'expr{$$=$1/$3;}

|'('expr')'{$$=$2;}

|'-'expr%prec uminus{$$=-$2;}

|number

;

%%

int yylex()

{

char c;

while((c=getchar())=='\0');if((c=='.')||(isdigit(c)))

{

ungetc(c,stdin);

scanf("%lf",&yylval);

return number;

}

return c;

}

main()

{

printf("Enter The Expression : \t");

yyparse();

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}

void yyerror()

{

printf("\n Error....Invalid expression");

return;

}

OUTPUT:

[cse4@localhost ~]$ yacc cse1.y

[cse4@localhost ~]$ cc y.tab.c

[cse4@localhost ~]$ ./a.out

Enter The Expression : 2+4*7-4/2

The Result is : 28

RESULT:

Thus the program for syntax analysis using YACC tool is executed and verified.

EX. NO: 7

DATE:

IMPLEMENTATION OF SHIFT REDUCE PARSING

AIM:

To write a C program to implement the shift reduce parsing.

ALGORITHM:


2. Get the input string from the user.

3. Push $ onto top of the stack.

4. Set ip to point to the first input symbol.

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5. If there is any production which can be used to reduce the input symbol reduce the

string otherwise push it to the top of the stack.

6. Set ip to point to next input symbol.

7. Repeat the above steps until the top of the stack contains the $ and the starting symbol.

If so, then the string is valid, otherwise the string is invalid, return an error message.


PROGRAM:

#include

#include

void pus(char);

void reduce();

char inp1[5][10]={"E+E","E*E","(E)","a"},stk[50],inpt1[10],stk1[50];

int ct=0;

void main()

{

int k,len;

char inpt[10];

clrscr();

printf("Enter the input string\n\n");

gets(inpt);

printf("\n\n");

printf("STACK\tINPUT \tACTION");

printf("\n\n");

len=strlen(inpt);

inpt[len++]='$';inpt[len]='\0';

strcpy(inpt1,inpt);

len=strlen(inpt);

inpt[len++]='$';

inpt[len++]='\0';

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pus('$');

pus(inpt[0]);

for(k=1;k0)

{

for(i=0;i

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}

void reduce()

{

int j=0,ct1,i,ct2,t,ct3,true=0;

char temp[10],tmp;

strcpy(stk1,stk);

for(i=0;i

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t=ct2;

while(ct2

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$EE) +a$ Shift

$E +a$ Reduce

$E+ a$ Shift

$EE a$ Reduce

$EEa $ Shift

$EEE $ Reduce

$E $ Reduce

$E $ Reduce

$E $ Reduce

String Accepted

RESULT:

Thus the program for implementations of shift reduce parsing is executed and verified.

EX. NO: 8

DATE:

IMPLEMENTATION OF LR PARSING

AIM:

To write a C program to implement LR parsing algorithm.

ALGORITHM:

Input: An input string w and an LR parsing table with functions action and goto for a

grammar G.

Output: If w is in L(G), a bottom up parse for w; otherwise an error indication.

Method: Initially, the parser has s0 on its stack, s0 is the initial state, and w$ in the input

buffer. The parser then executes the program until accept or error action is encountered.

set ip to point to the first symbol of w$;

repeat forever begin

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let s be the state on the top of the stack and

a the symbol pointed to by ip;

ifaction [s, a] = shift sthen begin

push a then s on the top of the stack;advance ip to the next input symbol

end

else ifaction [s, a] = reduce A then begin

pop 2*|| symbols off the stack;

let s be the state now on top of the stack;

push A then goto [s, A] on top of the stack;

output the production A

end

else ifaction [s, a] = accept then

return

else error()

end

PROGRAM:

char stk[10],inp[10],pat[20][20][20],prod[10][10],ipsymb[10];

int sp,ip,tp;

char c,v;

int i,k,t;

void gettable()

{

int i,j,k,n;

char c;

strcpy(pat[0][0],"s5");strcpy(pat[0][3],"s4");strcpy(pat[0][6],"1");

strcpy(pat[0][7],"2");strcpy(pat[0][8],"3");

strcpy(pat[1][5],"A"); strcpy(pat[1][1],"s6");

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strcpy(pat[2][1],"r2");strcpy(pat[2][2],"s7");strcpy(pat[2][4],"r2");

strcpy(pat[2][5],"r2");

strcpy(pat[3][1],"r4");strcpy(pat[3][2],"r4");strcpy(pat[3][4],"r4"); strcpy(pat[3]

[5],"r4");


strcpy(pat[4][7],"2");strcpy(pat[4][8],"3");

strcpy(pat[5][2],"r6");strcpy(pat[5][1],"r6");strcpy(pat[5][4],"r6");



strcpy(pat[6][8],"3");

strcpy(pat[7][0],"s5");strcpy(pat[7][3],"s4");strcpy(pat[7][8],"a");

strcpy(pat[8][1],"s6");strcpy(pat[8][4],"sb");

strcpy(pat[9][1],"r1");strcpy(pat[9][2],"s7");strcpy(pat[9][4],"r1");






ipsymb[0]='i';ipsymb[1]='+';ipsymb[2]='*';ipsymb[3]='(';ipsymb[4]=')';ipsymb[5]='$';ipsymb[6]='E';ipsymb[7]='T';ipsymb[8]='F';

strcpy(prod[0],"E'->E");strcpy(prod[1],"E->E+T");strcpy(prod[2],"E->T");

strcpy(prod[3],"T->T*F");strcpy(prod[4],"T->F");

strcpy(prod[5],"F->(E)");strcpy(prod[6],"F->i");

}

int ipnum(char c)

{

int i;

for(i=0;i

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}

return i;

}

int stknum(char c)

{

char t[10];

int i;

if(c

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sp=sp-b;

t=stk[sp];

prev=stknum(t);

stk[++sp]=pr[0];

z=ipnum(pr[0]);

stk[++sp]=pat[prev][z][0];

stk[sp+1]='\0';

}

void main()

{

int q;

clrscr();

printf("ENTER THE INPUT...");

scanf("%s",inp);

t=strlen(inp);

inp[t]='$';

inp[t+1]='\0';

stk[0]='0';

gettable();printf("\n\n\nSTACK\t\tINPUT\t\tOPERATION\n");

while(1)

{

c=inp[ip];

v=stk[sp];

k=ipnum(c);

i=stknum(v);

if(pat[i][k][0]=='s')

shift();

else if(pat[i][k][0]=='r')

reduce();

else if(pat[i][k][0]=='A')

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{

printf("\n\nVALID...");

getch();

exit(0);

}

else

{

printf("\n\nINVALID...");

getch();

exit(0);

}

printf("%s\t\t",stk);

q=ip;

while(inp[q]!='\0')

printf("%c",inp[q++]);

if(pat[i][k][0]=='s')

printf("\t\tShift\n");

else if(pat[i][k][0]=='r')

printf("\t\tReduced by %s\n",prod[pat[i][k][1]-48]);}

}

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OUTPUT:

ENTER THE INPUT: (i*i)+i

STACK INPUT OPERATION

0(4 i*i)+i$ Shift

0(4i5 *i)+i$ Shift

0(4F3 *i)+i$ Reduced by F->i

0(4T2 *i)+i$ Reduced by T->F

0(4T2*7 i)+i$ Shift0(4T2*7i5 )+i$ Shift

0(4T2*7Fa )+i$ Reduced by F->i

0(4T2 )+i$ Reduced by T->T*F

0(4E8 )+i$ Reduced by E->T

0(4E8)bFa +i$ Shift

0F3 +i$ Reduced by F->(E)

0T2 +i$ Reduced by T->F

0E1 +i$ Reduced by E->T

0E1+6)bFa i$ Shift

0E1+6i5Fa $ Shift

0E1+6F3 $ Reduced by F->i

0E1+6T9 $ Reduced by T->F

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0E1 $ Reduced by E->E+T

VALID...

RESULT:

Thus the program for construction of LR parsing is executed and verified.

EX. NO: 9

DATE:

IMPLEMENTATION OF CODE GENERATION

AIM:

To write a C program to implement the code generation algorithm.

ALGORITHM:

1. The code generation algorithm takes as input a sequence of three address statements

constituting a basic block.2. For each three address statement of the form x := y op z we perform the following

actions:

Invoke a function getreg to determine the location L where the result of the computation

y op z should be stored. L will usually be a register, but it could also be a memory

location. We shall describe getreg shortly.

Consult the address descriptor for y to determine y, (one of) the current location(s) of y.

prefer the register for y if the value of y is currently both in memory and a register. Ifthe value of y is not already in L, generate the instruction MOV y , L to place a copy of

y in L.

Generate the instruction OP z, L where z is a current location of z. Again, prefer a

register to a memory location if z is in both. Update the address descriptor of x to

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indicate that x is in location L. If L is a register, update its descriptor to indicate that it

contains the value of x, and remove x from all other register descriptors.

If the current values of y and/or z have no next users, are not live on exit from the block,

and are in register descriptor to indicate that, after execution of x := y op z, thoseregisters no longer will contain y and/or z, respectively.


PROGRAM:

#include

#include

#include

char exp[10][10],*ope;

int i,j,n,s[10],flag2,flag3,r1,r2;

void check();

void oper();

void main()

{

clrscr();

printf("\nEnter no of Expression:\n");

scanf("%d",&n);

printf("\nEnter the expression(In Three Address code):\n");

for(i=0;i

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void oper()

{

for(i=0;i

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}

}

if(flag2==1 && flag3==1)

{

s[i]=0;

printf("%sR%d,R%d \t(Result In R%d)\n",ope,r2,r1,r1);

}

else if(flag2==1&&flag3!=1)

{

s[i]=r1;

printf("%s %c,R%d \t(Result In R%d)\n",ope,exp[i][4],r1,r1);

}

else if(flag2!=1&&flag3==1)

{

s[i]=r2;

printf("%s %c,R%d \t(Result In R%d)\n",ope,exp[i][2],r2,r2);

}

else

{printf("MOV %c,R%d \t(%c Moved To R%d)\n",exp[i][2],s[i],exp[i][2],s[i]);

printf("%s %c,R%d \t(Result In R%d)\n",ope,exp[i][4],s[i],s[i]);

}

}

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OUTPUT:

Enter no of Expression:

4

Enter the expression (In Three Address code):

A=b+c

B=d-e

C=f*g

D=h/i

Generated code is:

MOV b,R0 (b Moved To R0)

ADD c,R0 (Result In R0)

MOV R0,A (Result Moved To A)MOV d,R1 (d Moved To R1)

SUB e,R1 (Result In R1)

MOV R1,B (Result Moved To B)

MOV f,R2 (f Moved To R2)

MUL g,R2 (Result In R2)

MOV R2,C (Result Moved To C)

MOV h,R3 (h Moved To R3)

DIV i,R3 (Result In R3)

MOV R3,D (Result Moved To D)

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RESULT:

Thus the program for implementation of code optimization is executed and verified.

EX. NO: 10

DATE:

IMPLEMENTATION OF CODE OPTIMIZATION TECHNIQUE

AIM:

To write a C program to implement code optimization technique.

ALGORITHM:


2. Enter the expression for which intermediate code is to be generated.

3. If the length of the string is greater than 3, then call the procedure to return the

precedence among the operands.

4. Assign the operand to exp array and operators to the array.

5. Create the three address code using quadruple structure.

6. Reduce the no of temporary variables.

7. Continue this process until we get an output.


PROGRAM:

#include

#include

#include

char s[20],o[20];void main()

{

int i=0,j=0,k,f1=1,f=1,k1=0;

void part();

clrscr();

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printf("\n Enter the input string\n");

scanf("%s",o);

strlen(o);

while(o[k1]!='\0')

{

I f((o[k1]=='=')==1)

{

break;

}

k1++;

}

for(j=k1+1;j3)

{

while(s[j]!='\0')

{

if((s[j]=='*')==1||(s[j]=='/')==1)

{

k=j;

if(f1!=0)

{

printf("t1=%c\n",s[k+1]);

printf("t2=%c%ct1",s[k-1],s[k]);

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}

else

{

if(k>3)

{

printf("t2=t1%c%c\n",s[k],s[k+1]);

}

else

{

printf("\t2=t1%c%c\n",s[k],s[k-1]);

}

}

f=0;

break;

}

j++;

}

j=0;

while(s[j]!='\0'){

if((s[j]=='+')==1||(s[j]=='-')==1)

{

k=j;

if(f==0)

{

if(k

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printf("\nt3=t2%c%c\n",s[k],s[k+1]);

}

}

else

{

printf("t1=%c%c%c\n",s[k-1],s[k],s[k+1]);

}

f1=0;

}

j++;

}

printf("%c=t3",o[0]);

}

else

{

printf("t1=%s\n",s);


}

part();getch();

}

void part()

{

int i=0,j=0,k,f1=1,f=1,k1=0;

while(o[k1]!='\0')

{

if((o[k1]=='=')==1)

{

break;

}

k1++;

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}

for(j=k1+1;j3)

{

while(s[j]!='\0')

{

if((s[j]=='*')==1||(s[j]=='/')==1)

{

k=j;

if(f1!=0)

{printf("t1=%c%c%c\n",s[k-1],s[k],s[k+1]);

}

else

{

if(k>3)

{

printf("t2=t1%c%c\n",s[k],s[k+1]);

}

else

{

printf("\t2=t1%c%c\n",s[k],s[k-1]);

}

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}

f=0;

break;

}

j++;

}

j=0;

while(s[j]!='\0')

{

if((s[j]=='+')==1||(s[j]=='-')==1)

{

k=j;

if(f==0)

{

if(k

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}

else

{

printf("t1=%s\n",s);


}

}

OUTPUT:

Enter the input stringa=b+c*d

Three address code is

t1=d

t2=c*t1

t3=t2+b

a=t3

OPTIMIZED CODE

t1=c*d

t2=t1+b

a=t2

RESULT:

Thus the program fogegsfadgsdfggr implementation of code

optimizegerfgdfgdfgdfation is executed and verified.

85708439 pcd lab record

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