CODEWITHSUNDEEP
flex-lexer
Maansi
Maansi

Reputation: 87

Why the following LEX program is not printing "No. of tokens"

My code is printing the identifiers,separators and all other things except it is not printing the number of tokens.Can't point out the problem.

%{
int n=0;
%}
%%
"while"|"if"|"else"|"printf" {
n++;
printf("\t keywords : %s", yytext);}
"int"|"float" {
n++;printf("\t identifier : %s", yytext);
}

"<="|"=="|"="|"++"|"-"|"*"|"+" {
n++;printf("\t operator : %s", yytext);
}

[(){}|, ;]     {n++;printf("\t seperator : %s", yytext);}

[0-9]*"."[0-9]+ {
n++;printf("\t float : %s", yytext);
}

[0-9]+ {  
n++;printf("\t integer : %s", yytext);
}
.;
%%
int main(void)
{
yylex();
printf("\n total no. of tokens = %d\n",n);
}
int yywrap()
{
return 0;
}

Upvotes: 0

Views: 656

Answers (1)

rici
rici

Reputation: 241731

If yywrap() returns 0, the lexer assumes that yywrap() has somehow arranged for yyin to have more data, and the lexer will continue to read input. So your lexer will never terminate.

If you want to signal that there is no more data, you need to return 1 from yywrap().

It's probably better to avoid the need for yywrap by placing 

%option noyywrap

in the flex prologue.

I usually use %option noinput nounput noyywrap, which eliminates some compiler warnings assuming you ask for compiler warnings, which you should always do. Also %option nodefault can help you find lex specification bugs, since it will complain if some input does not have a matching rule. (The default (f)lex action on unrecognised input is to simply write the unmatched character to standard output. That's not usually very helpful, and unlike an error message, it is very easy to miss.) Finally, %option 8bit is only necessary if you request a lexer optimised for speed rather than table-size. But it doesn't hurt to add it, and it might save you from an embarrassing bug if you (or someone) someday decides to try the faster scanner skeleton. (Not recommended, except in very special circumstances.)

Upvotes: 2

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