Re: gcc not compiling

[Les Mikesell <lesmikesell gmail com>]

On Mon, 2005-10-31 at 16:36, Matthew Saltzman wrote:
> >
> > I meant to comment that these two different notations are functionally
> > identical; an array name is nothing more than a pointer.  For example,
> > if we have the following code:
> >
> >
> > 	#include <stdio.h>
> > 	void main (void){
> >
> > 		char foo[] = "This is a string\n";
> > 		char *bar = foo;
> >
> > 		/* these expressions all print "T\n" */
> > 		printf("%c\n", foo[0]);
> > 		printf("%c\n", *bar);
> > 		printf("%c\n", bar[0]);
> > 		printf("%c\n", *foo);
> 
> /* ...and my favorite... */
> 
>  		printf("%c\n", 0[foo]);
> >
> > 		/* these expressions all print "i\n" */
> > 		printf("%c\n", foo[2]);
> > 		printf("%c\n", *(bar + 2));
> > 		printf("%c\n", bar[2]);
> > 		printf("%c\n", *(foo + 2));
> 
>  		printf("%c\n", 2[foo]);
> 
> /* That's right, folks, [] is commutative! */

/* Or, if you have a strong stomach, simplify to:
  printf("%c\n", "This is a string\n"[2]) ;
  printf("%c\n", 2["This is a string\n"] ) ;
/* It's all simple addition...*/

> > 	}

-- 
  Les Mikesell
   lesmikesell gmail com