Grabbing the address of the pointer variable itself, instead of the address that we are pointing to,-CodePudding

Let's consider I have the following x pointer variable in my function:

  int* x = new int { 666 };

I know I can print its value by using the * operator:

std::cout << *x << std::endl;

And that I can even print the address of where the 666 is being stored on the heap, like this:

std::cout << (uint64_t)x << std::endl;

But what I'd like to know is on whether it's also possible to grab the address of the x variable itself, that is, the address of the region of memory in the stack containing the pointer to the heap int containing 666?

Thanks

CodePudding user response：

Just use another &

std::cout << (uint64_t)&x << std::endl;

e.g.:

int v = 666; // variable
int * x = &v; // address of variable
int ** p = &x; // address of pointer x

and so on

int *** pp = &p;
int **** ppp = &pp;

and how to access to it:

std::cout << ****ppp << " == " << ***pp << " == " 
    << **p << " == " << *x << " == " << v << std::endl;