Why does it show that my function returns an int when it returns a char*? - yacc

The yacc code:
%{
#include<stdio.h>
#include<string.h>
%}
%union {
char* dval;
}
%token <dval> NUM VAR
%type <dval> E P
%left '+' '-'
%left '*' '/'
%%
statement : P {printf("\nt = %s\n \n",$1);}
;
P: E
;
E : E '+' E {strcpy($$,gencode($1,"+",$3));}
| E '-' E {strcpy($$,gencode($1,"-",$3));}
| E '*' E {strcpy($$,gencode($1,"*",$3));}
| E '/' E {strcpy($$,gencode($1,"/",$3));}
| '(' E ')' {strcpy($$,$2);}
| NUM {strcpy($$,$1);}
| VAR {strcpy($$,$1);}
;
%%
**The lex code:**
%{
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include"y.tab.h"
int n=0;
char *ch="t";
%}
%%
[0-9]+ {strcpy(yylval.dval,yytext); return NUM;}
[a-z]+ {strcpy(yylval.dval,yytext); return VAR;}
\n {return 0;}
. {return yytext[0];}
%%
void yyerror(char* str)
{
printf("\n%s",str);
}
char* gencode(char *first,char *op,char *second)
{
char *t;
char x[5];
t=(char*) malloc(sizeof(char)*5);
strcpy(t, ch);
itoa(n, x);
strcat(t, x);
printf("\n%s = %s %s %s\n",t,first,op,second);
n++;
t[strlen(t)]='\0';
return t;
}
main()
{
yyparse();
return 0;
}
For some reason gcc outputs the error: warning: passing argument 2 of ‘strcpy’ makes pointer from integer without a cast [enabled by default].
Where as the second argument of strcpy is the function gencode which returns a char* and not an integer.

You need to declare gencode in the beginning of the yacc file. Otherwise, the compiler sees it as undefined and assumes it returns an int.

You need to declare gencode in the .y file, either in a header you #include or directly in the .y file.
You have a much bigger problem, in that you declare dval (the type used for most rules) as a char *, but then you never initialize it to anything before writing to it with strcpy, so your code will crash (or worse) when strcpy tries to write to random addresses. You can fix this by replacing the strcpy calls in the lex file with yylval.dval = strdup(yytext) and the calls in the yacc file with just assignments to $$ (so $$ = gencode(... or whatever)

Related

How to fix implicit declaration of function 'yyerror' problem

Hi I am making a program that does simple arithmetic operations using Lex and yacc, but I am having a problem with a specific error.
ex1.y
%{
#include <stdio.h>
int sym[26];
%}
%token INTEGER VARIABLE
%left '+' '-'
%left '*' '/' '%'
%%
program:
program statement '\n'
|
;
statement:
expr {printf("%d\n", $1);}
| VARIABLE '=' expr {sym[$1] = $3;}
;
expr:
INTEGER
| VARIABLE { $$ = sym[$1];}
| expr '+' expr { $$ = $1 + $3;}
| expr '-' expr { $$ = $1 - $3;}
| expr '*' expr { $$ = $1 * $3;}
| expr '/' expr { $$ = $1 / $3;}
| '(' expr ')' { $$ = $2;}
;
%%
main() { return yyparse();}
int yyerror(char *s){
fprintf(stderr,"%s\n",s);
return 0;
}
ex1.l
%{
#include <stdlib.h>
#include "y.tab.h"
%}
%%
/* variables */
[a-z] {
yylval = *yytext -'a';
return VARIABLE;
}
/* integers */
[0-9]+ {
yylval = atoi(yytext);
return INTEGER;
}
/* operators */
[-+()=/*\n] { return *yytext;}
/* skip whitespace */
[ \t] ;
/* anything else is an error */
. yyerror("invalid character");
%%
int yywrap (void){
return 1;
}
when I execute bellow instruction
$bison –d -y ex1.y
$lex ex1.l
$gcc lex.yy.c y.tab.c –o ex1
The following error occurs:
ex1.l: In function ‘yylex’:
ex1.l:28:1: warning: implicit declaration of function ‘yyerror’; did you mean ‘perror’? [-Wimplicit-function-declaration]
28 |
| ^
| perror
y.tab.c: In function ‘yyparse’:
y.tab.c:1227:16: warning: implicit declaration of function ‘yylex’ [-Wimplicit-function-declaration]
1227 | yychar = yylex ();
| ^~~~~
y.tab.c:1402:7: warning: implicit declaration of function ‘yyerror’; did you mean ‘yyerrok’? [-Wimplicit-function-declaration]
1402 | yyerror (YY_("syntax error"));
| ^~~~~~~
| yyerrok
I don't know what is wrong with my code. I would appreciate it if you could tell me how to fix the above error.
The version of bison you are using requires you to declare prototypes for yylex() and yyerror. These should go right after the #include <stdio.h> at the top of the file:
int yylex(void);
int yyerror(char* s);
I would use int yyerror(const char* s) as the prototype for yyerror, because it is more accurate, but if you do that you'll have to make the same change in the definition.
You use yyerror in your lex file, so you will have to add its declaration in that file as well.
main() hasn't been a valid prototype any time this century. Return types are required in function declarations, including main(). So I guess you are basing your code on a very old template. There are better starting points in the examples in the bison manual.
(And don't expect it to be easy to work with parser generators if you have no experience with C.)

Warning: 2 shift/reduce conflicts [-Wconflicts-sr] err

%{
#include<stdio.h>
#include<stdlib.h>
int regs[30];
%}
%token NUMBER LETTER
%left PLUS MINUS
%left MULT DIV
%%
prog: prog st | ; //when I remove this line the error goes
st : E {printf("ans %d", $1);}| LETTER '=' E {regs[$1] = $3; printf("variable contains %d",regs[$1]);};
E : E PLUS E{$$ = $1 + $3;} //addition
| E MINUS E{$$ = $1 - $3 ;} //subtraction
| MINUS E{$$ = -$2;}
| E MULT E{$$ = $1 * $3 ;}
| E DIV E { if($3)$$= $1 / $3; else yyerror("Divide by 0");}
/*|LBRACE E RBRACE{$$= $2;}
| RBRACE E LBRACE{yyerror("Wrong expression");} */
| NUMBER {$$ = $1;}
| LETTER {$$ = regs[$1];}
;
%%
int main(void)
{
printf("Enter Expression: ");
yyparse();
return 0;
}
int yyerror(char *msg)
{
printf("%s", msg);// printing error
exit(0);
}
I am not able to resolve the conflicts. Also I am getting a segmentation fault when I run it with some edits. I am using yacc and lex for the same.
The two shift-reduce conflicts are the result of the fact that you don't require any explicit separator between statements. Because of that, a = b - 3 could be interpreted as one statement or as two (a = b; - 3). The second interpretation may not seem very natural to you but it is easily derived by the grammar.
In addition, your use of unary minus leads to an incorrect parse of -2/3 as -(2/3) instead of (-2)/3. (You may or may not find this serious, since it has few semantic consequences with these particular operators.) This particular issue and a correct resolution is discussed in the bison manual, and in many many other internet resources.
Both of these explanations are made a bit more visible if you use the -v command line option to bison to produce a description of the parser. See Understanding your parser (again, in the bison manual).

How to Read Multiple Lines of input file for arithmetic yacc program?

I am new to compilers and learning to make calculator that inputs multiple line equations (one equation each line) from a .txt file. And I am facing the problem of segmentation fault.
YACC Code :
%{
#include <stdio.h>
#include <string.h>
#define YYSTYPE int /* the attribute type for Yacc's stack */
extern int yylval; /* defined by lex, holds attrib of cur token */
extern char yytext[]; /* defined by lex and holds most recent token */
extern FILE * yyin; /* defined by lex; lex reads from this file */
%}
%token NUM
%%
Begin : Line
| Begin Line
;
Line : Calc {printf("%s",$$); }
;
Calc : Expr {printf("Result = %d\n",$1);}
Expr : Fact '+' Expr { $$ = $1 + $3; }
| Fact '-' Expr { $$ = $1 - $3; }
| Fact '*' Expr { $$ = $1 * $3; }
| Fact '/' Expr { $$ = $1 / $3; }
| Fact { $$ = $1; }
| '-' Expr { $$ = -$2; }
;
Fact : '(' Expr ')' { $$ = $2; }
| Id { $$ = $1; }
;
Id : NUM { $$ = yylval; }
;
%%
void yyerror(char *mesg); /* this one is required by YACC */
main(int argc, char* *argv){
char ch;
if(argc != 2) {printf("useage: calc filename \n"); exit(1);}
if( !(yyin = fopen(argv[1],"r")) ){
printf("cannot open file\n");exit(1);
}
yyparse();
}
void yyerror(char *mesg){
printf("Bad Expression : %s\n", mesg);
exit(1); /* stop after the first error */
}
LEX Code :
%{
#include <stdio.h>
#include "y.tab.h"
int yylval; /*declared extern by yacc code. used to pass info to yacc*/
%}
letter [A-Za-z]
digit [0-9]
num ({digit})*
op "+"|"*"|"("|")"|"/"|"-"
ws [ \t\n]
other .
%%
{ws} { /* note, no return */ }
{num} { yylval = atoi(yytext); return NUM;}
{op} { return yytext[0];}
{other} { printf("bad%cbad%d\n",*yytext,*yytext); return '?'; }
%%
/* c functions called in the matching section could go here */
I am trying to print the expression along with result.
Thanks In Advance.
In your parser, you have:
Line : Calc {printf("%s",$$); }
Now $$ is the semantic value which the rule is computing, and you haven't assigned anything to it. So it would not be unreasonable to assume that it is undefined, which would be bad, but in fact it does have a value because of the default rule $$ = $1;. All the same, it would be much more readable to write
printf("%s", $1);
But that's not correct, is it? After all, you have
#define YYSTYPE int
so all semantic types are integers. But you're telling printf that $1 is a string (%s). printf will believe you, so it will go ahead and try to dereference the int as though it were a char*, with predictable results (i.e., a segfault).
You are probably using a compiler which is clever enough to notice the fact that you are trying to print an int with a %s format code. But either you haven't asked the compiler to help you or you are ignoring its advice.
Always compile with warnings enabled. If you are using gcc or clang, that means putting -Wall in the command line. (If you are using some other compiler, find out how to produce warnings. It will be documented.) And then read the warnings and fix them before trying to run the program.
There are several other errors and/or questionable practices in your code. Your grammar is inaccurate (why do you use fact as the left-hand operand of every operator?), and despite your comment, your lexical scanner ignores newline characters, so there is no way the parser can know whether expressions are one per line, two per line, or spread over multiple lines; that will make it hard to use the calculator as a command-line tool.
There is no need to define the lex macro digit; (f)lex recognizes the Posix character class [[:digit:]] (and others, documented here) automatically. Nor is it particularly useful to define the macro num. Overuse of lex macros makes your program harder to read; it is usually better to just write the patterns out in place:
[[:digit:]]+ { yylval = atoi(yytext); return NUM; }
which would be more readable and less work both for you and for anyone reading your code. (If your professor or tutor disagrees, I'd be happy to discuss the matter with them directly.)

Runtime "syntax error" from lex and yacc

I cannot figure out why I am getting these results.
++
+add
+syntax error 2
++
+add
+syntax error 4
The ++ is my input and lex echoes each character and yacc prints add whenever it gets a +. It's giving me this error on every other + it gets. Doesn't matter how I give the input, I get the same results if I hit enter on every +.
lex
%{
#include "y.tab.h"
int chars = 0;
%}
%%
"+" {ECHO; chars++; return ADD;}
. {ECHO; chars++;}
\n {ECHO;}
%%
yacc
%{
#include <stdio.h>
extern int chars;
void yyerror (const char *str) {
printf ("%s %d\n", str, chars);
}
%}
%token ADD
%%
symbol : ADD {printf ("add\n");}
;
%%
int main () {
while (1) {
yyparse ();
}
}
Your grammar only accepts a 'sentence' that consists of a single token, +. When you type a second +, you induce a syntax error; your grammar doesn't allow ADD followed by ADD. Your next token after the + must be EOF for the grammar to accept your input. (Because of the . and \n rules, you can type all sorts of other stuff at the code, but there can only be one + in the input.)

Why am I getting conflicts: 1 shift/reduce

I'm new to bison and I'm getting a "conflicts: 1 shift/reduce" error. Can anyone shed some light on this?
Here's the y file.
test.y:
%{
#include <stdio.h>
#include <string.h>
#define YYERROR_VERBOSE
#define YYDEBUG 1
void yyerror(const char *str);
int yywrap();
%}
%union
{
int integer;
char *string;
}
%token <string> VAR_LOCAL
%token <integer> LIT_NUMBER
%token <string> LIT_STRING
%token WS_LINEBRK
//%token SYMB_EQL
%token SYMB_PLUS
%token SYMB_MINUS
%token SYMB_MUL
%token SYMB_DIV
%%
/*
// Sample input
num = 10
str = "this is a string"
*/
inputs: /* empty token */
| literal
| variable
| inputs stmt WS_LINEBRK
;
stmt: variable "=" exps
;
exps: variable op literal
| variable op variable
| literal op literal
| literal op variable
;
op: SYMB_PLUS | SYMB_MINUS | SYMB_MUL | SYMB_DIV ;
variable: VAR_LOCAL
{
printf("variable: %s\n", $1);
}
;
literal:
number | string
;
string: LIT_STRING
{
printf("word: %s\n", $1);
}
;
number: LIT_NUMBER
{
printf("number: %d\n", $1);
}
;
%%
void yyerror(const char *str)
{
fprintf(stderr,"error: %s\n",str);
}
int yywrap()
{
return 1;
}
main()
{
yyparse();
}
Here's the lex file
test.l:
%{
#include <stdio.h>
#include <stdlib.h>
#include "y.tab.h"
int line_no = 0;
%}
%%
[a-z][a-zA-Z0-9]* {
// local variable
yylval.string=strdup(yytext);
return VAR_LOCAL;
}
[0-9]+ {
//number literal
yylval.integer=atoi(yytext);
return LIT_NUMBER;
}
= return SYMB_EQL;
\+ return SYMB_PLUS;
\- return SYMB_MINUS;
\* return SYMB_MUL;
\/ return SYMB_DIV;
\"[-+\!\.a-zA-Z0-9' ]+\" {
// word literal
yylval.string=strdup(yytext);
return LIT_STRING;
}
\n {
// line break
printf("\n");
return WS_LINEBRK;
}
[ \t]+ /* ignore whitespace */;
%%
bison -r test.y will write a file test.output with a detailed description of the generated state machine that allows you to see what's going on - such as the state where the shift/reduce conflict occurs.
In your case, the problem is in the start state (corresponding to your start nonterminal, inputs). Say the first token is VAR_LOCAL. There's two things your parser could do:
It could match the variable case.
It could also match the inputs stmt WS_LINEBRK case: inputs matches the empty string (first line), and stmt matches variable "=" exps.
With the one token of lookahead that bison parsers use, there's no way to tell. You need to change your grammar to get rid of this case.
To fix the grammar, as Fabian has suggested, move the variable and literal to the end of exps from inputs
inputs:
| variable
| literal
exps:
...
| variable
| literal
That allows x= y,x="aliteral" syntax.
To allow for empty input lines, change the /* empty token */ rule to WS_LINEBREAK:
inputs: WS_LINEBRK
| stmt WS_LINEBRK
| inputs stmt WS_LINEBRK
;
On another note, since the scanner still looks for the SYMB_ EQUAL ; but the parser no longer defines it (its commented out), something needs to be done in order to compile. One option is to uncomment the %token definition and use SYMB_ EQUAL instead of the literal "=" in the parser .y file.