"I think that j
is a pointer to an array"
Yes, it is. And that's also the reason why *j
output the same address as outputting g
would do. In this case an array decays into the pointer and that's why even outputting j
yields the same result.
The fact that the same address is outputted might make you think that j
and *j
are the same pointers, however they are not. Their type is different (a fact that actually matters):
int g[] = {9,8}; // int[]
int (*j)[2] = &g; // int (*)[2]
so using *j
becomes equivalent to using g
directly, just like *(*j)
becomes equivalent to *g
.
And &(*j)
is nothing but j
, which is initialized with an address of an array (an address taken from decayed g
, i.e. an address of the first element of this array).
So why j
and *j
outputs the same address and *(*j)
outputs the value of first element?
Because of the type of j
being int (*)[2]
. A simple example:
int g[] = {9,8};
int (*j)[2] = &g; // j points to the first element as well
cout << *((int*) j);
outputs 9
.