C pointers and addresses
- by yCalleecharan
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...
