Creating a.out from yacc.

Creating a.out from yacc.

Post by N » Sat, 14 Jun 2003 21:29:05



Hi.
I am having a bit of trouble linking my lex and yacc file and creating
an executable. i'll describe what i've done and if someone has an idea
were i went wrong i'd love to hear.
1.lex ll.lex
   created lex.yy.c
2. i then ran yacc on  yc.yacc (which inculded #include "lex.yy.c")
   which created y.tab.c
3. i wrote a main function
int main()
{
return (yyparse());
Quote:}

4. i then ran
gcc main.c y.tab.c -ll

5. i got a file a.out
6. i did chmod +x a.out
7. now when  i write
a.out <input
i get an answer:
a.out: command not found.

????????
Ideas?

Thanks
Noa.

 
 
 

Creating a.out from yacc.

Post by Jim Dabel » Sat, 14 Jun 2003 21:40:36



> a.out: command not found.

Many systems don't include the current directory as part of the path.  Run
it like this:

./a.out < input

(notice the leading ./)

Also, lots of systems include the current directory as part of the path, but
only for non-root users.  Logging in as root to do ordinary tasks is not a
good idea, for all sorts of different reasons (just in case you are doing
this).

--
Jim Dabell

 
 
 

1. YACC + YACC = YACC

Hi,

I'd like to use YACC and LEX twice in an executable. I'd like to read
two different file formats at different times in the program. I link
two objects, each created by yacc. The later object overwrites the
previous definitions.

Is there any way to make the yacc funtions unique?

-frank


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