What's meaning of &&pointer or &*(pointer_to_pointer) in C++

Question

int *p; int **pp; int a = 9; p = &a; pp = &p; cout << "&a: " << &a cout << "&p: " << &p cout << "&pp: " << &pp cout << "pp : " << pp cout << "*pp: " << *pp cout << "&*pp: " << &*pp

&&p and &&pp aren't defined in c++ so they are wrong using, but what &*pp is meaning? Is &*pp equalent to &&a?

When the program is starting, the result is as follows:

&a: 00AEFAE4 &p: 00AEFAFC &pp: 00AEFAF0 pp : 00AEFAFC *pp: 00AEFAE4 &*pp: 00AEFAFC (=&p ???)

On the other hand, why is &*pp equalent to &p?

C++ provides templates so that direct use of pointers can be avoided — Nahuel Fouilleul
– Nahuel Fouilleul, Commented Jan 24, 2018 at 5:25
Too little context to make an answer. Please expand. What code generates your output? &*pp looks like you dereference pp to p and then take the address again so effectively it is &p, but can't be sure without more information. — user4581301
– user4581301, Commented Jan 24, 2018 at 5:28
The code is very simple, my question is; why is &*pp equalent to &p? — ATES
– ATES, Commented Jan 24, 2018 at 5:36

Warda Baig · Accepted Answer · 2018-01-24 07:13:25Z

you asked what does &*pp means here pp is a double pointer which contains address of the pointer (*p) by writing &*pp firstly you are dereferencing value of **pp

*(pp) which will become value inside pp which is the address of pointer p by writing &(*pp) now you are trying to get address of the dereferenced value which is (address of p) that will ultimetely become address of p

in simple words where you use both of these operators together they cancel out each other and give you the value present in the pointer in this case &* will be cancelled out and you will get value of pp which is address of p i tried to make it as clear as i could.... hope that helps :)

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