Thread: Lex and yacc program

Hi, i have few questions in the program which is given bellow ,
there are two file first one is lex and second is yacc file.
Here is the OUTPUT:
student@localhoststudent]$ yacc -d cal2.y
[student@localhoststudent]$ lex cal1.l
[student@localhoststudent]$ cc lex.yy.c y.tab.c -ll -lm
[student@localhoststudent]$ ./a.out
3+2
5.000000

Code:

//Lex program starts here
%{
/* LEX FILE*/
/*TO RECOGNIZE INFIX EXPRESSION AND EVALUATE IT.*/
#include"y.tab.h"
#include<stdio.h>
void yyerror(char *);
%}


%%
(([0-9]+)|([0-9]*\.[0-9]+))   {
    yylval.dval=atof(yytext);                
    return NUM;
}


[+-/*\n,~()]  {return (*yytext);}


sin   return SIN;
cos   return COS;
tan   return TAN;
sqrt  return SQRT;
exit  exit(0);
[ \t]  /*ignore*/
%%




int yywrap(void)
{
    return 1;
}

//End of the program

(([0-9]+)|([0-9]*\.[0-9]+))   {  
    yylval.dval=atof(yytext);                
    return NUM;
}
//What happens in abouve block.
// "\." why this slash is given before "."  


[+-/*\n,~()]  {return (*yytext);}
// what this line is doing?

//Yacc program stars here
%{
/* YACC FILE */
/*LEX AND YACC PROGRAM*/
/*TO RECOGNIZE INFIX EXPRESSION AND EVALUATE IT.*/


#include<stdio.h>
#include<math.h>
void yyerror(char *);
%}


%union 
{
    double dval;
}


%token <dval> NUM
%token SIN COS TAN SQRT
%right '~'
%left '+' '-'
%left '*' '/'
%type <dval> expression


%%
program: program statement'\n'
    | 
    ;


statement: 
      expression  { printf("%lf\n", $1); }
      ;


expression:  
        NUM
        | expression '+' expression  {$$ = $1 + $3;}
        | expression '-' expression  {$$ = $1 - $3;}
        | expression '*' expression  {$$ = $1 * $3;}
        | expression '/' expression  {$$ = $1 / $3;}             
        | '~' expression  {$$ = -(1) * $2;}
        | SQRT'('expression')'  {$$ = sqrt( $3 );}
        | SIN'('expression')'   {$$ = sin ($3*3.142/180);}
        | COS'('expression')'   {$$ = cos ($3*3.142/180);}
        | TAN'('expression')'   {$$ = tan ($3*3.142/180);}
        ;
%%


 main()
{
    yyparse();
}


void yyerror(char *s)
{
    fprintf(stderr,"%s\n",s);
}


//End of the program

// how is the execution of following block
%%
program: program statement'\n'
    | 
    ;


statement: 
      expression  { printf("%lf\n", $1); }
      ;


expression:  
        NUM
        | expression '+' expression  {$$ = $1 + $3;}
        | expression '-' expression  {$$ = $1 - $3;}
        | expression '*' expression  {$$ = $1 * $3;}
        | expression '/' expression  {$$ = $1 / $3;}             
        | '~' expression  {$$ = -(1) * $2;}
        | SQRT'('expression')'  {$$ = sqrt( $3 );}
        | SIN'('expression')'   {$$ = sin ($3*3.142/180);}
        | COS'('expression')'   {$$ = cos ($3*3.142/180);}
        | TAN'('expression')'   {$$ = tan ($3*3.142/180);}
        ;
%%

Please any suggestions are highly appreciated! Thanks a lot for all your time and support!

I haven't actually made use of lex or yacc before (shame), but:

Originally Posted by Lanpan

Hi, i have few questions in the program which is given bellow ,

Code:

(([0-9]+)|([0-9]*\.[0-9]+))   {
    yylval.dval=atof(yytext);                
    return NUM;
}
//What happens in abouve block.
// "\." why this slash is given before "."

The bold red bit is a regular expression. I presume these are fall-thrus used by the lexer in matching elements. What that will match is

([0-9]+)
1 or more (+) digits

|
OR

([0-9]*\.[0-9]+)

Zero or more (*) digits, then a literal . (\ is an escape used since . by itself has a special meaning (any single character)), then 1 or more digits.

Ie., it is matching numbers in any of these forms:
123
5.6
.777