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lex program to count the number of Positive and negative integer


/* decleration section in this sections we will decleared the different  value and include the header file which we are using in this program to run this program */

%{
#include<stdio.h>
int p_count=0;
int n_count=0;
%}


/* this sections is known as defined section in which we defined the rule and regulation of regular expression which will going to accept or not */

 %%
[+]?([0-9])+  {

 /* this is for increment and print the positive integer eg: 96, +36 */

p_count++; printf("Positive Integer:%s\n",yytext);}
[-]([0-9])+   {

/* this is for increment and print the negative integer eg: -96, -36 */

n_count++; printf("negative Integer:%s\n",yytext);}

%%

 /* main function for calling the function */

void main(int argc[],char *argv[])
{
yylex();
printf("the total number of Postive Integer are: %d\n" ,p_count);
printf("the total number of Negative  Integer are:%d\n",n_count);
}

                                download the program file

How to run this program

 1 step- save as .l extension and run "  flex  ass1.3_a.l  "
 2 step- there would be a c program generated by flex analyser which may be " lex.yy.c "   run this file as this command "gcc lex.yy.c -o first -ll " or "cc lex.yy.c -o first -ll"

 3rd step- for output "./a.out " and enter the input after complete input press ctrl + D so that input sections will complete  ,it will display the result
 


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