cout with pointers

[hamid1455]

Say I have a pointer like this:

double meow = 0;
double * pDouble = &meow;

Then cout << &pDouble; will print the memory address that pDouble is pointing to, right?

So then what is the memory address that I get when I output just plain old pDouble?

Also, outputting *pDouble gives me the value of the pointer or the value of what the pointer is pointing to?

Thanks in advance guys.

[linuxp]

pDouble = &meow = address of meow
*pDouble = meow = value of meow
&pDouble = address of pDouble itself

[hamid1455]

so I could make a pointer that points to another pointer?

[nullptr]

Yes. And you could make a pointer that points to a pointer that points to a pointer if you wanted to. int************ p; is valid C++ code, though I can't think why you would need that many levels of pointers.