Thread: How Are C Arrays Represented In Memory?

For comparison, xptr and &xptr here are different:

Code:

#include <stdio.h>

int main()
{
    int x = 123;
    int *xptr = &x;
 
	printf("x       %d\n",x);
	printf("&x      %p\n",&x);
	printf("xptr    %p\n",xptr);
	printf("&xptr   %p\n",&xptr);

	return 0;
}

output:

x       123
&x      0012ff84
xptr    0012ff84
&xptr   0012ff80

I am wondering about the difference between the pointer xptr just above,
and the conceptual pointer x from the array x[].

On line 10 you are printing the value of xptr
on line 11 you are printing it's address.

Hence ... different numbers.

Yes I know, that's what I expect.

With arrays, the array name yields the same address as the "address of" array name.

The x in x[] has the same value as &x.

So what exactly is your question?

An array name, the x in x[] for example can be treated like a pointer.

But the address of that "pointer" equals the value of that pointer.
Why the difference between an array name when used as a pointer, and an actual pointer?

Pointers are not arrays ... however, in some cases, arrays are treated as pointers.

That is x[i] acts like adding an offset to the pointer and dereferencing it... functionally it equates to *(x + i);

Thanks, that explained it:

Arrays and Pointers

I've used array names as a pointers before, but never even considered the address
of array name. Never needed it.