implementing first of a given grammar using c
#include<stdio.h>
#include<ctype.h>
void FIRST(char[],char );
void addToResultSet(char[],char);
int numOfProductions;
char productionSet[10][10];
main()
{
int i;
char choice;
char c;
char result[20];
printf("How many number of productions ? :");
scanf(" %d",&numOfProductions);
for(i=0;i<numOfProductions;i++)//read production string eg: E=E+T
{
printf("Enter productions Number %d : ",i+1);
scanf(" %s",productionSet[i]);
}
do
{
printf("\n Find the FIRST of :");
scanf(" %c",&c);
FIRST(result,c); //Compute FIRST; Get Answer in 'result' array
printf("\n FIRST(%c)= { ",c);
for(i=0;result[i]!='\0';i++)
printf(" %c ",result[i]); //Display result
printf("}\n");
printf("press 'y' to continue : ");
scanf(" %c",&choice);
}
while(choice=='y'||choice =='Y');
}
/*
*Function FIRST:
*Compute the elements in FIRST(c) and write them
*in Result Array.
*/
void FIRST(char* Result,char c)
{
int i,j,k;
char subResult[20];
int foundEpsilon;
subResult[0]='\0';
Result[0]='\0';
//If X is terminal, FIRST(X) = {X}.
if(!(isupper(c)))
{
addToResultSet(Result,c);
return ;
}
//If X is non terminal
//Read each production
for(i=0;i<numOfProductions;i++)
{
//Find production with X as LHS
if(productionSet[i][0]==c)
{
//If X → ε is a production, then add ε to FIRST(X).
if(productionSet[i][2]=='$') addToResultSet(Result,'$');
//If X is a non-terminal, and X → Y1 Y2 … Yk
//is a production, then add a to FIRST(X)
//if for some i, a is in FIRST(Yi),
//and ε is in all of FIRST(Y1), …, FIRST(Yi-1).
else
{
j=2;
while(productionSet[i][j]!='\0')
{
foundEpsilon=0;
FIRST(subResult,productionSet[i][j]);
for(k=0;subResult[k]!='\0';k++)
addToResultSet(Result,subResult[k]);
for(k=0;subResult[k]!='\0';k++)
if(subResult[k]=='$')
{
foundEpsilon=1;
break;
}
//No ε found, no need to check next element
if(!foundEpsilon)
break;
j++;
}
}
}
}
return ;
}
/* addToResultSet adds the computed
*element to result set.
*This code avoids multiple inclusion of elements
*/
void addToResultSet(char Result[],char val)
{
int k;
for(k=0 ;Result[k]!='\0';k++)
if(Result[k]==val)
return;
Result[k]=val;
Result[k+1]='\0';
}
Output
#include<stdio.h>
#include<ctype.h>
void FIRST(char[],char );
void addToResultSet(char[],char);
int numOfProductions;
char productionSet[10][10];
main()
{
int i;
char choice;
char c;
char result[20];
printf("How many number of productions ? :");
scanf(" %d",&numOfProductions);
for(i=0;i<numOfProductions;i++)//read production string eg: E=E+T
{
printf("Enter productions Number %d : ",i+1);
scanf(" %s",productionSet[i]);
}
do
{
printf("\n Find the FIRST of :");
scanf(" %c",&c);
FIRST(result,c); //Compute FIRST; Get Answer in 'result' array
printf("\n FIRST(%c)= { ",c);
for(i=0;result[i]!='\0';i++)
printf(" %c ",result[i]); //Display result
printf("}\n");
printf("press 'y' to continue : ");
scanf(" %c",&choice);
}
while(choice=='y'||choice =='Y');
}
/*
*Function FIRST:
*Compute the elements in FIRST(c) and write them
*in Result Array.
*/
void FIRST(char* Result,char c)
{
int i,j,k;
char subResult[20];
int foundEpsilon;
subResult[0]='\0';
Result[0]='\0';
//If X is terminal, FIRST(X) = {X}.
if(!(isupper(c)))
{
addToResultSet(Result,c);
return ;
}
//If X is non terminal
//Read each production
for(i=0;i<numOfProductions;i++)
{
//Find production with X as LHS
if(productionSet[i][0]==c)
{
//If X → ε is a production, then add ε to FIRST(X).
if(productionSet[i][2]=='$') addToResultSet(Result,'$');
//If X is a non-terminal, and X → Y1 Y2 … Yk
//is a production, then add a to FIRST(X)
//if for some i, a is in FIRST(Yi),
//and ε is in all of FIRST(Y1), …, FIRST(Yi-1).
else
{
j=2;
while(productionSet[i][j]!='\0')
{
foundEpsilon=0;
FIRST(subResult,productionSet[i][j]);
for(k=0;subResult[k]!='\0';k++)
addToResultSet(Result,subResult[k]);
for(k=0;subResult[k]!='\0';k++)
if(subResult[k]=='$')
{
foundEpsilon=1;
break;
}
//No ε found, no need to check next element
if(!foundEpsilon)
break;
j++;
}
}
}
}
return ;
}
/* addToResultSet adds the computed
*element to result set.
*This code avoids multiple inclusion of elements
*/
void addToResultSet(char Result[],char val)
{
int k;
for(k=0 ;Result[k]!='\0';k++)
if(Result[k]==val)
return;
Result[k]=val;
Result[k+1]='\0';
}
Output
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