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 addre

Answer 1

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);

Answer 2

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 3

Having this C source:

int a = 10;
int * ptr = &a;

Use this

printf("The address of ptr is %p\n", (void *) &ptr);

to print the address of ptr.

Please note that the conversion specifier p is the only conversion specifier to print a pointer's value and it is defined to be used with void* typed pointers only.

From man printf:

p

The void * pointer argument is printed in hexadecimal (as if by %#x or %#lx).

Answer 4

If you are trying to compile these codes from a Linux terminal, you might get an error saying

expects argument type int

Its because, when you try to get the memory address by printf, you cannot specify it as %d as its shown in the video. Instead of that try to put %p.

Example:

// this might works fine since the out put is an integer as its expected.
printf("%d\n", *p); 

// but to get the address:
printf("%p\n", p); 

Answer 5

You can use the %p formatter. It's always best practice cast your pointer void* before printing.

The C standard says:

The argument shall be a pointer to void. The value of the pointer is converted to a sequence of printing characters, in an implementation-defined manner.

Here's how you do it:

printf("%p", (void*)p);

Answer 6

&a gives address of a - &p gives address of p.

int * * p_to_p = &p;