views:

158

answers:

3

I made a program that is supposed to recognize a simple grammar. When I input what I think is supposed to be a valid statement, I get an error. Specifically, if I start out with an identifier, I automatically get a syntax error. However, I noticed that using an identifier won't generate an error if it is preceded by 'int'. If a is an identifier, then if I type 'int a;' this is ok. But if I type 'a = 3' I get an error. Just typing a by itself will generate an error.

The lex file:

%{
#include <stdlib.h>
#include <ctype.h>
#include <string.h>

#include "y.tab.h"

%}

else else
if if
int int
return return
void void
while while
id [a-zA-Z]+
num [0-9]+
lte <=
gte >=
equal ==
notequal !=


%%

{else}  {   return ELSE; }
{if}    {   return IF; }
{int}   { return INT; }
{return} {  return RETURN; }
{void} {    return VOID; }
{while} {   return WHILE; }
{id} {   return ID; }
{num} {  return NUM; }
{lte} {  return LTE; }
{gte} {  return GTE; }
{equal} {   return EQUAL; }
{notequal} {    return NOTEQUAL; }


[\[\];] { return yytext[0];}


%%

The yacc file

/* C-Minus BNF Grammar */

%token ELSE
%token IF
%token INT
%token RETURN
%token VOID
%token WHILE

%token ID
%token NUM

%token LTE
%token GTE
%token EQUAL
%token NOTEQUAL
%%

program : declaration_list ;

declaration_list : declaration_list declaration | declaration ;

declaration : var_declaration | fun_declaration ;

var_declaration : type_specifier ID ';'
                | type_specifier ID '[' NUM ']' ';' ;

type_specifier : INT | VOID ;

fun_declaration : type_specifier ID '(' params ')' compound_stmt ;

params : param_list | VOID ;

param_list : param_list ',' param
           | param ;

param : type_specifier ID | type_specifier ID '[' ']' ;

compound_stmt : '{' local_declarations statement_list '}' ;

local_declarations : local_declarations var_declaration
                   | /* empty */ ;

statement_list : statement_list statement
               | /* empty */ ;

statement : expression_stmt
          | compound_stmt
          | selection_stmt
          | iteration_stmt
          | return_stmt ;

expression_stmt : expression ';'
                | ';' ;

selection_stmt : IF '(' expression ')' statement
               | IF '(' expression ')' statement ELSE statement ;

iteration_stmt : WHILE '(' expression ')' statement ;

return_stmt : RETURN ';' | RETURN expression ';' ;

expression : var '=' expression | simple_expression ;

var : ID | ID '[' expression ']' ;

simple_expression : additive_expression relop additive_expression
                  | additive_expression ;

relop : LTE | '<' | '>' | GTE | EQUAL | NOTEQUAL ;

additive_expression : additive_expression addop term | term ;

addop : '+' | '-' ;

term : term mulop factor | factor ;

mulop : '*' | '/' ;

factor : '(' expression ')' | var | call | NUM ;

call : ID '(' args ')' ;

args : arg_list | /* empty */ ;

arg_list : arg_list ',' expression | expression ;
+1  A: 

Are you chewing up whitespace?, that is an important factor, since you stated a[space]=[space]3.

tommieb75
It doesn't have to do with whitespace because if I just type a and press enter then I still get an error.
Phenom
Ok. But I would consider to chew up whitespace at the lexer level before handling the parser.' ' { ; }Do you not think so?
tommieb75
That doesn't work. Wouldn't just . work?
Phenom
+1  A: 

program expands to declaration_list, and declaration_list only allows a series of declarations. You never actually allow statements/expressions/etc in your grammar (except inside a function definition).

Cory Petosky
A: 

Such a time waste yaar

nikhil gupta