gcc not compiling

[Matthew Saltzman]

On Mon, 31 Oct 2005, Derek Martin wrote:

> On Mon, Oct 31, 2005 at 01:30:44PM -0600, STYMA, ROBERT E (ROBERT) wrote:
>> The syntax:
>> int main(int argc, char **argv)
>> works, but most of the C books I have seen recommend
>> the *argv[] version.
>
> 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! */

> 	}
>
> The results may surprise programming students:
>
> 	$ gcc -o ptr ptr.c
> 	ptr.c: In function `main':
> 	ptr.c:2: warning: return type of 'main' is not `int'
> 	[ddm at archonis ~]
> 	$ ./ptr
> 	T
> 	T
> 	T
> 	T
> 	i
> 	i
> 	i
> 	i
>
>
> In reality, foo[x] is just "syntactic sugar" for *(foo + x).  This
> is called pointer math, and works properly regardless of the size and
> type of foo, so long as it was declared properly before being used.
>
> [I'm very bored today...]
>
>

-- 
 		Matthew Saltzman

Clemson University Math Sciences
mjs AT clemson DOT edu
http://www.math.clemson.edu/~mjs