char*f="char*f=%c%s%c;main(){printf(f,34,f,34,10);}%c";
f is a global pointer to the char array which contains the string
"char*f=%c%s%c;main(){printf(f,34,f,34,10);}%c"
Now in main function you are printing this string with arguments 34,f,34,10.
ASCII value of " is 34.
->So in f, the first %c is replaced by ".
->The next %s is replaced by string f.
->the second %c is replaced by " and
->last %c is replaced by backspace. The last %c is actually I feel not
required. So the code can be:
char*f="char*f=%c%s%c;main(){printf(f,34,f,34);}";main(){printf(f,34,f,34);}
I hope it helps. Try to do it manually on paper. You would be able to
understand it.
-Piyush
On Sun, Aug 28, 2011 at 11:46 AM, rahul sharma <rahul23111...@gmail.com>wrote:
> plz expalin char*f=" "
>
>
> On Aug 28, 11:12 am, Piyush Grover <piyush4u.iit...@gmail.com> wrote:
> > it's a quine problem.
> >
> > char*f="char*f=%c%s%c;
> > main(){
> > printf(f,34,f,34,10);}%c";
> >
> > main()
> > {
> > printf(f,34,f,34,10);
> >
> > }
> >
> > I have used whitespaces to make it understand.
> >
> > On Sun, Aug 28, 2011 at 11:39 AM, rahul sharma <rahul23111...@gmail.com
> >wrote:
> >
> >
> >
> >
> >
> >
> >
> > > program whose output is the program itself???????????????
> >
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