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error: invalid operands to binary

Posted on 2009-03-29
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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 !
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Question by:unknown_
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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);
}
 
 
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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);

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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.
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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.
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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!
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