How to get address of a pointer in c/c++?

Question

How to get address of a pointer in c/c++?

Eg: I have below code.

int a =10; int *p = &a; 

So how do I get address of pointer p? Now I want to print address of p, what should I do?

print("%s",???) what I pass to ???.

Answer 1

To get the address of p do:

int **pp = &p; 

and you can go on:

int ***ppp = &pp; int ****pppp = &ppp; ... 

or, only in C++11, you can do:

auto pp = std::addressof(p); 

To print the address in C, most compilers support %p, so you can simply do:

printf("addr: %p", pp); 

otherwise you need to cast it (assuming a 32 bit platform)

printf("addr: 0x%u", (unsigned)pp); 

In C++ you can do:

cout << "addr: " << pp; 

Answer 2

int a = 10; 

To get the address of a, you do: &a (address of a) which returns an int* (pointer to int)

int *p = &a; 

Then you store the address of a in p which is of type int*.

Finally, if you do &p you get the address of p which is of type int**, i.e. pointer to pointer to int:

int** p_ptr = &p; 

just seen your edit:

to print out the pointer's address, you either need to convert it:

printf("address of pointer is: 0x%0X\n", (unsigned)&p); printf("address of pointer to pointer is: 0x%0X\n", (unsigned)&p_ptr); 

or if your printf supports it, use the %p:

printf("address of pointer is: %p\n", p); printf("address of pointer to pointer is: %p\n", p_ptr);