I was just playing with pointers and found this weird thing. I created the pointer p and displayed its address as I did in the following program, the address of p and &p is different, why are they different?
int main()
{
int *p, s[5];
cout << p <<endl;
cout << &p <<endl;
p = s;
cout << p <<endl;
cout << &p <<endl;
}
P is a pointer that means it holds an address to an integer. So when you print p it displays address of the integer it points to. Whereas when you print &p you are actually printing the address of p itself rather than the address it points to.
s is an array, say, of type int[10], and you try to print it, it will silently convert to a pointer to its first element, which is of type int*. The type of &p, on the other hand, is int(*)[10]. As you can see from the type, &p points to the entire array, not just the first element. But since an array and its first element start at the same address in memory, you get the same output. We have a related FAQ which you might find interesting.As Jeeva said, pointer p also occupies a part of memory(the yellow part in this pic, its address is 0x300) and in this area, it holds the value which is the address of the array s. So p == 0x100 and &p == 0x300
If p were equal to &p, the pointer would point to itself, which is only possible with void pointers:
void* p = &p;
assert(p == &p);
I have yet to see a practical use for this, though :)
Oh wait, it also happens in self-referencing recursive data structures, which can be quite useful:
struct X
{
X* p;
void print()
{
std::cout << " p = " << p << '\n';
std::cout << "&p = " << &p << '\n';
}
};
X x;
x.p = &x;
x.print();
Think linked container, where the last node has itself as its successor. Output on my system:
p = 0x7fffb4455d40
&p = 0x7fffb4455d40
Why? Because an object usually starts at the same address in memory as its first data member.
