How does object pointer works in c++?

Question

I was trying to understand how a objects are stored in memory (heap) and how reference variable works. I wrote a simple program to understand the concept. But Why I am getting dereferenced address i.e. *(&cat) but not the &cat when function callBar() returns the address?

#include<iostream>
using std::cout;
using std::string;
class Cat{
        public: string name;
};
Cat* callBar(Cat *cat){
        cat = new Cat();
        cout<<&cat<<"\n";
        cout<<*(&cat)<<"\n";
            
        return cat;
}
int main(void){
        Cat *c = new Cat();
        cout<<&c<<"\n";
        cout<<callBar(c)<<"\n";
        cout<<*(&c)<<"\n";
        return 0;
}

The output of the code is:

0x7ffd38985d70
0x7ffd38985d48
0x56368fbd22b0
0x56368fbd22b0
0x56368fbd1e70

I created the instance of the class in the main and printed its address. I passed this object to another function and then printed its address again, which gave me new reference. Atlast, I deferenced it.

It is still not clear to me why I am getting dereferenced address i.e. *(&cat) but not the &cat when function returns?

There are no object references in the code provided. Maybe the question should be: How does object pointers work in c++? — Eljay
– Eljay, Commented Jul 12, 2020 at 14:46

Tony Delroy · Accepted Answer · 2020-07-12 14:58:28Z

2

It is still not clear to me why I am getting dereferenced address i.e. *(&cat) but not the &cat when function returns?

Inside callBar the relevant bits are...

Cat* callBar(...unused argument...){
    cat = new Cat();
    cout << &cat << "\n";
    cout << *(&cat) << "\n";
    return cat;
}

You're not getting &cat back because your return statement is return cat; not return &cat;.

But - the rest of your question - why are you getting *(&cat)? The & operator there takes the address of the Cat* cat local variable, then dereferences it with *: those operations cancel out, so *(&cat) == cat.

edited Jul 12, 2020 at 14:58

answered Jul 12, 2020 at 14:36

Tony Delroy

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3 Comments

Saurav Jaiswal Over a year ago

How does it returns unchanged? I mean I am creating a new object on the same reference variable cat. cat = new Cat(); And If it cancel out then the addresses should match?

Eljay Over a year ago

@SauravJaiswal • You are creating a new object on the same callBar local parameter variable cat. That has the effect of setting cat to a different value, not of changing the caller's c argument.

Tony Delroy Over a year ago

Sorry - I didn't even notice that line in the function - too sleepy! :-/ Point still stands though - after you assign that value to cat, *(&cat) is the same thing.

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