Solved

error: invalid operands to binary

Posted on 2009-03-29
6
2,699 Views
Last Modified: 2012-05-06
Hello,

I try to compile my yacc grammar but i get the following errors :

error: invalid operands to binary *
error: invalid operands to binary /


Thanks in advance for any help !
0
Comment
Question by:unknown_
6 Comments
 

Author Comment

by:unknown_
ID: 24015024
%{
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

      
#define YYDEBUG 1

extern int  yyparse();
extern FILE *yyin;      

int yywrap()
 {
    return 1;
 }

void yyerror(const char *s)
 {
    fflush(stdout);
    printf("\n%*s\n%*s\n", s);
 }



enum treetype {operator_node, number_node, variable_node};
 typedef struct tree {
   enum treetype nodetype;
   union {
     struct {struct tree *left, *right; char operator;} an_operator;
     int ival;
     char sval;
   } body;
 } tree;
 static tree *make_operator (tree *l, char o, tree *r) {
   tree *result= (tree*) malloc (sizeof(tree));
   result->nodetype= operator_node;
   result->body.an_operator.left= l;
   result->body.an_operator.operator= o;
   result->body.an_operator.right= r;
   return result;
 }
 static tree *make_number (int n) {
   tree *result= (tree*) malloc (sizeof(tree));
   result->nodetype= number_node;
   result->body.ival= n;
   return result;
 }
 static tree *make_variable (char v) {
   tree *result= (tree*) malloc (sizeof(tree));
   result->nodetype= variable_node;
   result->body.sval= v;
   return result;
 }
 static void printtree (tree *t, int level) {
 #define step 4
   if (t)
     switch (t->nodetype)
     {
       case operator_node:
        printtree (t->body.an_operator.right, level+step);
        printf ("%*c%c\n", level, ' ', t->body.an_operator.operator);
        printtree (t->body.an_operator.left, level+step);
        break;
       case number_node:
        printf ("%*c%d\n", level, ' ', t->body.ival);
        break;
       case variable_node:
        printf ("%*c%c\n", level, ' ', t->body.sval);
     }
 }



%}
 
%start goal

%union {
      int ival;
      char * sval;
        tree* btree;
}
 
%token <sval> IDENTIFIER
%token <sval> VARIABLE
%token <ival> INTEGER
 /*%token <ival> FLOAT*/
%token <sval> STRING
%token <sval> LITERAL
%token <sval> UNKNOWN
%token <sval> PLUS
%token <sval> MINUS
%token <sval> TIMES
%token <sval> SLASH
%token <sval> LPAREN
%token <sval> RPAREN
%token <sval> SEMICOLON
%token <sval> COMMA
%token <sval> EQL
%token <sval> OR
%token <sval> OR2
%token <sval> AND
%token <sval> AND2
/* keywords */
%token <sval> IF
%token <sval> ELSE
%token <sval> statement
%token <sval> DO
%token <sval> INT
%token <sval> RETURN
%token <sval> VOID
%token <sval> FLOAT
%token <sval> WHILE
%type <btree> goal
%type <ival> expression
%type <ival> term
%type <sval> factor
%type <sval> ifstatement
%type <sval> whilestatement
%type <sval> dostatement
%type <sval> variable_list
%type <sval> variable



%%

goal                  :
variable '=' expression SEMICOLON {printf("rule : VAR = Expression\n");}
|      ifstatement SEMICOLON
|      whilestatement SEMICOLON
|       dostatement SEMICOLON
|      INT variable_list SEMICOLON {printf("rule: INT_VARIABLE\n");}      
|      FLOAT variable_list SEMICOLON   {printf("rule: FLOAT_VARIABLE\n");} {printtree ($1, 1);} ;

expression            :      term '+' term {printf("rule 1\n");} {$$ = make_operator ($1, '+', $3);}
|      term '-' term {printf("rule 2\n");} {$$ = make_operator ($1, '-', $3);}
|      term {printf("rule 3\n");}  {$$ = $1;}   ;

term                  :      factor '*' factor  {printf("rule 4\n");} {$$ = $1 * $3;}
|      factor '/' factor  {printf("rule 5\n");} {$$ = $1 / $3;}
|      factor {printf("rule 6\n");}  {$$ = $1;}  ;

factor                  :      variable {printf("rule 7\n");} {$$ = make_variable ($1);}
|      INTEGER {printf("rule 8\n");} {$$ = make_number ($1);}
|      FLOAT {printf("rule 9\n");} {$$ = make_number ($1);}
|      STRING {printf("rule 10\n");}
|      '(' expression ')' {printf("rule 11\n");} {$$= make_block ($2);};

ifstatement            :      IF '(' expression ')' goal ELSE goal SEMICOLON {printf("rule: IF_ELSE_STATEMENT\n");}
|      IF '(' expression ')' goal {printf("rule: IF_STATEMENT\n");};

whilestatement      :      WHILE '(' expression ')' goal {printf("rule: WHILE_STATEMENT\n");} {$$= make_while ($3, $5);} ;      
            
dostatement : DO expression WHILE '(' expression ')'   {printf("rule DO_STATEMENT\n");} {$$= make_do ($2, $5);} ;


variable : VARIABLE {printf("rule var2\n");}  {$$= list ($1); /* [$1] */} ;

      
variable_list : VARIABLE {printf("rule 15\n");} {$$= list ($1); /* [$1] */}
|        variable ',' variable_list {printf("rule 16\n");}  ;

                     
%%

#include "lex.yy.c"

main(int argc,char *argv[])
{
 if(argc<1)
  {
   printf("Please specify the input file\n");
   exit(0);
  }
 FILE *fp=fopen(argv[1],"r");
 if(!fp)
 {
  printf("couldn't open file for reading\n");
  exit(0);
 }
  yyin=fp;
  yyparse();
  fclose(fp);
}
 
 
0
 
LVL 40

Accepted Solution

by:
mrjoltcola earned 500 total points
ID: 24015077
You have the %type factor declared as sval:

%type <sval> factor


But you are using it as numeric in the "term" rule:

term                  :      factor '*' factor  {printf("rule 4\n");} {$$ = $1 * $3;}
|      factor '/' factor  {printf("rule 5\n");} {$$ = $1 / $3;}
|      factor {printf("rule 6\n");}  {$$ = $1;}  ;


$1  *  $3 is the same as if you wrote:   yylval1.sval  *  yylval2.sval

You can't multiple 2 strings, and your intention here is to treat them has numbers, so define them as such in your %type declarations

0
 
LVL 7

Expert Comment

by:HalfAsleep
ID: 24018943
Is it just me, or does the printf statement in yyerror look a bit suspect?

printf("\n%*s\n%*s\n", s);

I don't think I have ever seen a "%*s %*s", s statement before.

I can understand something like the code I have supplied, but what on earth does that printf statement do?
 const char* msg = "Hello printf";
 int string_size = strlen (msg);
 printf("msg: %.*s", string_size, msg);

Open in new window

0
ScreenConnect 6.0 Free Trial

Want empowering updates? You're in the right place! Discover new features in ScreenConnect 6.0, based on partner feedback, to keep you business operating smoothly and optimally (the way it should be). Explore all of the extras and enhancements for yourself!

 
LVL 53

Expert Comment

by:Infinity08
ID: 24019034
HalfAsleep, %*s also works - it provides a minimum width to print the string. However, that printf is missing three parameters :) The output will likely show garbage.
0
 
LVL 7

Expert Comment

by:HalfAsleep
ID: 24019074
That is what I meant, I could not understand the single string parameter.  For that printf, I would expect 4.
0
 

Author Closing Comment

by:unknown_
ID: 31564127
Thank you very much !
 Can you please have a look to the print parse tree question:
http://www.experts-exchange.com/Programming/Languages/C/Q_24279119.html

Im not sure how to work with the printout of the parse tree.


Thanks!
0

Featured Post

Netscaler Common Configuration How To guides

If you use NetScaler you will want to see these guides. The NetScaler How To Guides show administrators how to get NetScaler up and configured by providing instructions for common scenarios and some not so common ones.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Preface I don't like visual development tools that are supposed to write a program for me. Even if it is Xcode and I can use Interface Builder. Yes, it is a perfect tool and has helped me a lot, mainly, in the beginning, when my programs were small…
This tutorial is posted by Aaron Wojnowski, administrator at SDKExpert.net.  To view more iPhone tutorials, visit www.sdkexpert.net. This is a very simple tutorial on finding the user's current location easily. In this tutorial, you will learn ho…
Video by: Grant
The goal of this video is to provide viewers with basic examples to understand and use while-loops in the C programming language.
The goal of this video is to provide viewers with basic examples to understand and use switch statements in the C programming language.

820 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question