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

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);

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;

Tags:

C++

C