Enter the string (a,(a,a)) Input (a,(a,a)) a, (a,a)) ,(a,a)) Action S →(L) L -> SL' S →a ,(a,a)) L' ->,SL' (a,a)) S →(L) a,a)) L -> SL' ,a)) ,a)) )) S →a L' ->,SL' S →a L' -> ε L' -> ε String is successfully parsed Enter the string (aa,a) Input Action (aa,a) S →(L) aa,a) L -> SL' a,a) S →a a,a) L' -> € Error in parsing string
Use the logic in that
for the following grammar:
S →aAB
A →Abc | b
B →d
Make sure that you choose at least two strings that pass the grammar and one
string that does not pass the grammar just like I did in the model program. Output should follow same layout for new C program.
/* Recursive Descent Parser for the Expression Grammar:
S → (L) |a
L' →,SL'|ε
L → SL'
Valid inputs: (a,(a,a)) and (a,(a,(a,a),(a,a)))
Invalid inputs:(aa,a)
*/
#include <stdio.h>
#include <string.h>
int S(), Ldash(), L();
char *ip;
char string[50];
int main()
{
printf("Enter the string\n");
scanf("%s",string);
ip = string;
printf("\n\nInput\t\tAction\n ------------------------------\n");
if (S())
{
printf("\n------------------------------\n");
printf("\n String is successfully parsed\n");
}
else
{
printf("\n------------------------------\n");
printf("Error in parsing string\n");
}
}
int L()
{
printf("%s\t\tL -> SL'\n", ip);
if (S())
{
if (Ldash())
{
return 1;
}
else
return 0;
}
else
return 0;
}
int Ldash()
{
if (*ip == ',')
{
printf("%s\t\tL' -> ,SL'\n", ip);
ip++;
if (S())
{
if (Ldash())
{
return 1;
}
else
return 0;
}
else
return 0;
}
else
{
printf("%s\t\tL' -> ε\n", ip);
return 1;
}
}
int S()
{
if (*ip == '(')
{
printf("%s\t\tS →(L) \n", ip);
ip++;
if (L())
{
if (*ip == ')')
{
ip++;
return 1;
}
else
return 0;
}
else
return 0;
}
else if (*ip == 'a')
{
ip++;
printf("%s\t\tS →a \n", ip);
return 1;
}
else
return 0;
}
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