Confused about pointers and their addresses

Question

I have created some pointer in code but the results are not what I expected.

Here is the simple code:

int main(int argc, char const* argv[])
{
    int myInt = 23;
    int* ptr = &myInt;

    char* buffer = new char[8];
    memset(buffer, 0, 8);

    char** ptr2 = &buffer;

    std::cout << "ptr address is " << ptr << std::endl;
    std::cout << "buffer pointer is pointing to address " << buffer << std::endl;
    std::cout << "ptr2 pointer is pointing to address " << ptr2 << std::endl;
    std::cout << "Dereferencing ptr2 = " << *ptr2 << std::endl;
    return 0;
} 

And here are the results of running the code:

ptr address is 0x7ffde215a14c

buffer pointer is pointing to address

ptr2 pointer is pointing to address 0x7ffde215a150

Dereferencing ptr2 =

I am wondering why the buffer pointer address does not show and why the dereferencing of ptr2 also shows nothing and yet the pointer (ptr2) pointing to the buffer pointer is showing that address. None of this makes any sense to me.

Answer 1

The stream << operator is explicitly overloaded for all kinds of char* to print it as a null-terminated string. To print the pointer you need to cast it:

std::cout << "buffer pointer is pointing to address " << reinterpret_cast<void*>(buffer) << std::endl;