Why am I not reading the first member of the class when I point to the object with a pointer?-CodePudding

I'm experimenting with smart pointers and I wrote the code below:

struct Buffer
{
    char Data[128];
};

class SmartPtr {
    char * dataPtr;

public:
    SmartPtr(Buffer& b)
    {
        dataPtr = b.Data;
    }

    ~SmartPtr()
    {
        cout << "desctructor called" << endl;
    }

    void operator=(Buffer & b)
    {
        dataPtr = b.Data;
    }

    char*& operator*()
    {
        return dataPtr;
    }
};

I though that when using a pointer of the same type as the type of the first member of a class I can access that member.
What I'm trying to do is accessing dataPtr by getting the address of the object p.
I'm expecting p and dataPtr to have the same address.

The following code prints garbage.

static Buffer buff;

int main(void)
{
    strcpy(buff.Data, "Hello world");

    SmartPtr p = buff;
    char* s = (char*)&p;
    cout << "Data: " << s << endl;
    
    return 0;
}

Output:

Data: Ç☺┴─≈
desctructor called

If I change that to a double pointer I get the expected result

static Buffer buff;

int main(void)
{
    strcpy(buff.Data, "Hello world");

    SmartPtr p = buff;
    char** s = (char**)&p;
    cout << "Data: " << *s << endl;
    
    return 0;
}

Output:

Data: Hello world
desctructor called

Is it because with a single pointer I dereference the address of the actual pointer as a variable instead of the address it contains?

CodePudding user response：

In the first code, you are taking the address of p (and thus the address of p.dataPtr), casting it to char*, and then printing it as-is. So, operator<< is misinterpreting the raw memory of p itself as-if it were a null-terminated string, which it is not, so you get garbage.

In the second code, you are taking the address of p (and thus the address of p.dataPtr), casting it to char**, dereferencing it to access its value as a char*, and then printing that. So, operator<< is printing the data that p.dataPtr is pointing at as a null-terminated string, which it is, so you get the correct result.