Hi, I always thought that *&p = p = &*p in C. I tried this code:

 #include <stdio.h>
 #include <stdlib.h>

 char a[] = "programming";
 char *ap = &a[4];  

int main(void)
{

 printf("%x %x %x\n", ap, &*(ap), *&(ap));   /* line 13 */
 printf("%x %x %x\n\n", ap+1, &*(ap+1), *&(ap+1));   /* line 14 */
}

The first printf line (line 13) gives me the addresses:

40b0a8 40b0a8 40b0a8

which are the same as expected. But when I added the second printf line, Borland complains:

"first.c": E2027 Must take address of a memory location in function main at line 14

I was expecting to get:

40b0a9 40b0a9 40b0a9.

It seems that the expression *&(ap+1) on line 14 is the culprit here. I thought all three pointer expressions on line 14 are equivalent. Why am I thinking wrong?

A second related question: The line

char *ap = a;

points to the first element of array a. I used

char *ap = &a[4];  

to point to the 5th element of array a.

Is the expression

char *ap = a;

same as the expression

char *ap = &a[0];

Is the last expression only more verbose than the previous one?

Thanks a lot...