C program for Syntax Analyzer
#include<stdio.h> #include<conio.h> #include<string.h> #include<ctype.h> void main() { int i,j,k=0,count,inc=0,n; char name[30],open[30],ch,chh,o[30]; char op[20]={'=','+','-','*','/','%','^','&','|'}; clrscr(); textcolor(3); cprintf("--Syntax Analyser--"); printf("\n"); printf("\n Enter Syntax"); printf("\n"); scanf("%s",name); n=strlen(name); for(i=0;i<n;i++) { ch=tolower(name[i]); for(j=0;j<9;j++) { if(ch==op[j]) { open[k]=i; o[k]=ch; k++; } } } for(i=0;i<k;i++) { count=open[i]; ch=tolower(name[count-1]); chh=tolower(name[count+1]); if(isalpha(ch)&&isalpha(chh)||isdigit(chh)) ++inc; } if(k==inc) printf("\n %s is a valid syntax",name); else printf("\n %s is an invalid syntax",name); getch(); }
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bekar ka program no explanation is given and only print syntax is valid chahe wrong sentence hi kyu na likha ho