Does an Array variable point to itself?

Question

I tried some code to check the behavior of array and pointers. Its as follows.

#include 
main(){
int s[]={1,2};
int *b=s;
printf(\"%d, %d, %d\\n\"         


        

Answer 1

OK, let say the following is how the memory looks when you execute

int s[]={1,2};
int *b=s;

s[]
+-----+-----+
|  1  |  2  |
+-----+-----+
100   104 
^
|
|  int *b = &s
+-----+
| 100 |
+-----+
200

s is an array. What that means is, it is a contiguous memory location which is associated with a variable s and each element is accessed by offsetting the array variable name.

So when you use s it actually boils down to the address of the array (which is 100 in this case). And when you do *s, it boils down to *(s+0) which is equivalent of s[0] and so *s represents the contents stored in the zeroth location (in this case s[0] is 1). When do do an &s, this will print the address of the s (which is100` in this case).

Note that, here s and &s represents an address; *s and s[x] represents an integer.

The same applies to the pointer. So, b prints the content it has, which is the address of s (which is 100 in this case). &b prints the address of b, which is 200 in this case. And, *b prints the content of the first element of the array s which is 1.

I have modified you program to make it print the address.

#include 

int main(void)
{
    int s[]={1,2};
    int *b=s;
    printf("%p, %p, %d\n", (void *)s, (void *)&s, *s);
    printf("%p, %p, %d\n", (void *)b, (void *)&b, *b);
    return 0;
}

Output:

0xbfc4f3e4, 0xbfc4f3e4, 1
0xbfc4f3e4, 0xbfc4f3ec, 1

EDIT: %p expects void *. Added the same!