lex and yacc to parse trignometric expression - yacc

I have the following code for lex and yacc. I am getting kind of extra values in the printed statement can anyone tell. whats wrong with the code?
Lex code:
%{
#include <stdio.h>
#include "y.tab.h"
%}
%%
[ \t] ;
[+-] { yylval=yytext; return Sym;}
(s|c|t)..x { yylval=yytext; return Str;}
[a-zA-Z]+ { printf("Invalid");}
%%
int yywrap()
{
return 1;
}
yacc code:
%{
#include<stdio.h>
%}
%start exps
%token Sym Str
%%
exps: exps exp
| exp
;
exp : Str Sym Str {printf("%s",$1); printf("%s",$2); printf("%s",$3);}
;
%%
int main (void)
{
while(1){
return yyparse();
}
}
yyerror(char *err) {
fprintf(stderr, "%s\n",err);
}
Input:
sinx+cosx
output:
sinx+cosx+cosxcosx
look at the output of the code!!!

yytext is a pointer into flex's internal scanning buffer, so its contents will be modified when the next token is read. If you want to return it to the parser, you need to make a copy:
[+-] { yylval=strdup(yytext); return Sym;}
(s|c|t)..x { yylval=strdup(yytext); return Str;}
Where symbols are a single character, it might make more sense to return that character directly in the scanner:
[-+] { return *yytext; }
in which case, your yacc rules should use the character directly in '-single quotes:
exp : Str '+' Str {printf("%s + %s",$1, $3); free($1); free($3); }
| Str '-' Str {printf("%s - %s",$1, $3); free($1); free($3); }

Related

Yacc %define parse.error verbose generates error

When I try to get more out of my "syntax error", I seem to use the way described on so many websites, but all seem to create their own errors, for some reason.
I was getting standard "syntax error" on line 5 of the Input file... so I wanted to add better error handling so I can see what exactly is the issue. But
%define parse.error verbose
However, it gives me this;
error: %define variable 'parse.error' is not used
Below are my files, as long as you keep it constructive, feel free to comment on more then just the error parts, any help is welcome :)
(As long as the errors get fixed as well :P )
Thanks in advance!
lex file;
%option nounput yylineno
%{
#include "yaccTest.tab.h"
void InvalidToken();
void extern yyerror (char *s);
%}
whitespace [ \t\r\v\f]
linefeed \n
%%
";" {return SEMICOLON;}
"=" {return EQ;}
"+" {return PLUS;}
"-" {return MINUS;}
"*" {return MULTIPLY;}
"/" {return DEVIDE;}
"(" {return BO;}
")" {return BC;}
"^" {return POWER;}
"print" {return PRINT;}
[a-zA-Z][a-zA-Z0-9]* {yylval.charValue = yytext[0]; return IDENTIFIER;}
[0-9]+ {yylval.intValue = atoi(yytext); return NUMBER;}
{whitespace} {;}
. {InvalidToken();}
%%
void yyerror(char *s) {
fprintf(stderr, "\nERROR ON LINE %d : \n %s\n", yylineno, s);
exit(0);
}
void InvalidToken(){
printf("ERROR ON LINE %d : \n Invalid Token %s\n", yylineno,yytext);
exit(0);
}
int yywrap (void) {return 1;}
yacc file;
%{
#include <stdio.h>
#include <stdlib.h>
int getVariableValue(char varID);
extern int yylineno;
int varIDs[52] = {0};
int varValues[52] = {0};
%}
%define parse.lac full
%define parse.error verbose
%union YYSTYPE {int intValue; char charValue;}
%token COLON SEMICOLON ST SE EQ GE GT PLUS MINUS MULTIPLY DEVIDE BO BC CBO CBC POWER LOOP PRINT
%token <intValue> NUMBER
%token <charValue> IDENTIFIER CHAR
%type <charValue> declaration expression
%type <intValue> numval
%right EQ
%left PLUS MINUS
%left MULTIPLY DEVIDE
%left POWER
%%
declaration : IDENTIFIER EQ expression
| declaration IDENTIFIER EQ expression
;
expression : numval SEMICOLON
| PRINT BO numval BC SEMICOLON {printf("Printing");}
;
numval : NUMBER {$$ = $1;}
| NUMBER PLUS NUMBER {$$ = $1 + $3;}
| NUMBER MINUS NUMBER {$$ = $1 - $3;}
| NUMBER MULTIPLY NUMBER {$$ = $1 * $3;}
| NUMBER DEVIDE NUMBER {$$ = $1 / $3;}
| NUMBER POWER NUMBER {int i;int j = $1;for(i = 1; i < $3; i++){j=j*$1;};$$ = j;}
;
%%
int getVariableValue(char varID) {
int i, j, localTemp;
for (i=0;i<((sizeof(varIDs)/sizeof(varIDs[0])));i++) {
if (varID == varIDs[i]) {
localTemp = varValues[i];
}
}
return localTemp;
}
int setVariableValue(char varID, int varValue) {
int i, varPresent = 0;
for (i=0;i<((sizeof(varIDs)/sizeof(varIDs[0])));i++) {
if (varID == varIDs[i]) {
varValues[i] = varValue;
varPresent = 1;
}
}
if (varPresent == 0) {
for (i=0;i<((sizeof(varIDs)/sizeof(varIDs[0])));i++) {
if (&(varIDs[i]) == NULL) {
if (&(varValues[i]) == NULL) {
varIDs[i] = varID;
varValues[i] = varValue;
}
else {
missingVarIDError(varID, varValue);
}
}
else {
notEnoughStorageError(varID, varValue);
}
}
}
}
int missingVarIDError(char *id, int val){
printf("\nERROR ON LINE %d : \nIdentifier '%s' not found, but assigned location DOES have a value; %s",yylineno,id,val);
exit(0);
}
int notEnoughStorageError(char *id, int val){
printf("\nERROR ON LINE %d : \nIdentifier '%s' did not fit in StorageArray, '%3' not stored!",yylineno,id,val);
exit(0);
}
int main (void) {
return yyparse ( );
return 0;
}
Input file;
x=4;
y=2+6;
X=2;
z=5;
print(4);

Simple Lex/Yacc Calculator not printing output

I'm trying to understand how compilers and programming languages are made. And to do so I thought about creating a simple calculator which does just addition and subtraction. Below are the Lex and Yacc files which I wrote.
calc.yacc file:
%{
#include <stdio.h>
#include <stdlib.h>
extern int yylex();
void yyerror(char *);
%}
%union { int number; }
%start line
%token <number> NUM
%type <number> expression
%%
line: expression { printf("%d\n", $1); };
expression: expression '+' NUM { $$ = $1 + $3; };
expression: expression '-' NUM { $$ = $1 - $3; };
expression: NUM { $$ = $1; };
%%
void yyerror(char *s) {
fprintf(stderr, "%s", s);
exit(1);
}
int main() {
yyparse();
return 0;
}
calc.lex file:
%{
#include <stdio.h>
#include <stdlib.h>
#include "y.tab.h"
%}
%%
[0-9]+ {
yylval.number = atoi(yytext);
return NUM;
}
[-+] { return yytext[0]; }
[ \t\f\v\n] { ; }
%%
int yywrap() {
return 1;
}
It compiles nicely but when I run it and type something like 2 + 4 then it gets stuck and doesn't print the answer. Can somebody explain why? My guess is that my grammar is not correct (but I don't know how).
I came to the same idea like rici and changed your samples appropriately:
file calc.l:
%{
#include <stdio.h>
#include <stdlib.h>
#include "calc.y.h"
%}
%%
[0-9]+ {
yylval.number = atoi(yytext);
return NUM;
}
[-+] { return yytext[0]; }
"\n" { return EOL; }
[ \t\f\v\n] { ; }
%%
int yywrap() {
return 1;
}
file calc.y:
%{
#include <stdio.h>
#include <stdlib.h>
extern int yylex();
void yyerror(char *);
%}
%union { int number; }
%start input
%token EOL
%token <number> NUM
%type <number> expression
%%
input: line input | line
line: expression EOL { printf("%d\n", $1); };
expression: expression '+' NUM { $$ = $1 + $3; };
expression: expression '-' NUM { $$ = $1 - $3; };
expression: NUM { $$ = $1; };
%%
void yyerror(char *s) {
fprintf(stderr, "%s", s);
exit(1);
}
int main() {
yyparse();
return 0;
}
Compiled & tested in cygwin on Windows 10 (64 bit):
$ flex -o calc.l.c calc.l
$ bison -o calc.y.c -d calc.y
$ gcc -o calc calc.l.c calc.y.c
$ ./calc
2 + 4
6
2 - 4
-2
234 + 432
666
Notes:
Minor issue: According to the build commands, I had to change the #include for the generated token table. (A matter of taste.)
I introduced the EOL token in the lex source as well as in the line rule of the parser.
While testing I recognized that the 2nd input ended everytimes in a syntax error. I needed a while until I recognized that the grammer was actually limited now to accept precisely one line. Thus, I inserted the recursive input rule in the parser source.

lex and yacc : a simple calculator with syntax error

a simple calculator support only + - * / and integer. I use GNU/Linux.
hoc1.l:
%{
#include "y.tab.h"
extern int yylval;
%}
%%
[ \t] { ; }
[0-9]+ { sscanf(yytext, "%d", &yylval); printf("\nget %d\n", yylval); return NUMBER; }
\n {return 0;}
%%
int yywrap(void) {
return 1;
}
hoc1.y
%{
#include<stdio.h>
#define YYSTYPE int
%}
%token NUMBER
%left '+' '-'
%left '*' '/'
%%
list:
| list '\n'
| list expr '\n' {printf("\t%d\n",$2);}
;
expr: NUMBER { $$ = $1; }
| expr '+' expr {$$ = $1+$3;}
| expr '-' expr {$$ = $1-$3;}
| expr '*' expr {$$ = $1*$3;}
| expr '/' expr {$$ = $1/$3;}
;
%%
int main(void)
{
yyparse();
return 0;
}
int yyerror(char *s) {
fprintf(stderr, "*%s*\n", s);
return 0;
}
runtime-error:
% ./hoc
8+9
get 8
+
get 9
*syntax error*
why and how to sovle it, thx!
You forgot to include your operators in your lex file, and you should return nonzero on a successful token read: returning 0 intuitively means there was no match by yylex. Remove the line in your lex file handling the newline character and replace it with the following:
[-+*/\n] { return *yytext; }
. { yyerror("unrecognized character"); return 0; }
Now it should work. Returning *yytext allows your yacc grammar to parse an expression successfully, e.g. if you get a '+', return it to allow the grammar to parse properly.

What does return 0 do in the code and why is the yywrap function written without a body?

I AM USING BISON AND FLEX.
What does return 0 do in case of the kcalc.l file that I have posted?
And I am not getting the use of yywrap without a body (i mean not literally but an empty body).The code is of a calculator without any variable managing and basic operations that can be done like addition subtraction multiplication division and handling of unary minus operator. I have been studying through the lex and yacc specifications but did not get any answer for the query I asked .
Kcal.y
%{
#include <stdio.h>
%}
%token Number
%left '-' '+'
%left '*' '/'
%nonassoc UMINUS
%%
statement: expression
{ printf(" result = %d\n", $1);} ;
expression: expression '+' expression
{ $$ = $1 + $3;
printf("Recognised'+'expression\n");
}
| expression '-' expression
{ $$ = $1 - $3;
printf("Recognised '-' expression\n");
}
| expression '*' expression
{ $$ = $1 * $3;
printf("Recognised '*' expression\n");
}
| expression '/' expression
{ if ($3 == 0)
printf ("divide by zero\n");
else
$$ = $1 / $3;
printf("Recognised '/' expression\n");
}
| '-' expression %prec UMINUS
{
$$ = - $2;
printf("Recognised paranthesized expression\n");
}
| '(' expression ')'
{
$$ = $2;
printf("Recognised paranthesized expression");
}
| Number { $$ = $1;
printf("Recognised a no.\n");
}
;
%%
int main(void)
{
return yyparse();
}
int yyerror (char *msg)
{
return fprintf(stderr,"Yacc :%s", msg);
}
yywrap()
{
}
 
kcalc.l
%{
#include "y.tab.h"
extern int yylval;
%}
%%
[0-9]+ { yylval = atoi(yytext);
printf("accepted the number : %d\n", yylval);
return Number; }
[ \t] { printf("skipped whitespace \n");}
\n { printf("reached end of line\n");
**return 0;**
}
. { printf("found other data \" %s\n", yytext);
return yytext[0];
}
%%
The return 0 notifies the end-of-input to the parser, so apparently the expression should be contained on a single line. The empty body of yywrap is just wrong. If you use -Wall with the gcc compiler it will give two warnings for yywrap:
kcal.y:54: warning: return type defaults to ‘int’
kcal.y:55: warning: control reaches end of non-void function
The first one because no result type for the function is specified (K&R style C), so it is assumed it should return an int. The second warning because it lacks a return statement for such an int.
Since a newline terminates the input, the chances of yywrap ever being called are slim. But it will be called if the input does not contain a newline. If by sheer accident the (more or less random) return value of yywrap were to be interpreted as 0 the tokenizer would end up in an infinite loop of repeatedly calling yywrap.

Syntax error in Bison after one token is processed

I am trying to come up to speed on Flex and Bison. I can parse one token with a very simple "language" but it fails on the second, even though the token is legitimate.
test.l:
%{
#include <stdio.h>
#include "test.hpp"
%}
%%
[0-9]+ {printf("Number entered\n"); return INTEGER_NUMBER;}
[a-zA-Z]+ {printf("plain text entered: '%s'\n",yytext); return PLAIN_TEXT;}
[ \t] ;
. ;
%%
test.y
%{
#include <stdio.h>
extern "C" {
int yyparse(void);
int yylex(void);
int yywrap() { return 1; }
extern int yylineno;
extern char* yytext;
extern int yylval;
}
/* #define YYSTYPE char * */
void yyerror(const char *message)
{
fprintf(stderr, "%d: error: '%s' at '%s', yylval=%u\n", yylineno, message, yytext, yylval);
}
main()
{
yyparse();
}
%}
%token PLAIN_TEXT INTEGER_NUMBER
%%
test : text | number;
text : PLAIN_TEXT
{
/*printf("plain text\n");*/
};
number : INTEGER_NUMBER
{
/*printf("number\n");*/
};
%%
Results:
$ ./test
cat
plain text entered: 'cat'
dog
plain text entered: 'dog'
1: error: 'syntax error' at 'dog', yylval=0
$ ./test
34
Number entered
34
Number entered
1: error: 'syntax error' at '34', yylval=0
Why am I getting this syntax error?
Your test.y seems to lack the grammar for the case that several tests
continue.
So, how about adding the grammar like the following?
%%
tests : test | tests test; /* added */
test : text | number;
...