Can we print the address of object of class in C++?

Question

How can we Print the address of object of class in the member function of that class in C++?

class A
{
     int x;
   private:
     A(){x=2;}
}
int main()
{
   A B;
   return 0;
}

How to print the address of B in member function or in main().

Note that your code throws a compile error since your constructor is private. The question is still valid with a public constructor. — talekeDskobeDa, Jan 15 '19 at 13:04

score 12 · Answer 1 · answered Dec 26 '13 at 14:47

12

Just add

#include <iostream>

at the top of the file and the following line at the bottom of main:

std::cout << &B << std::endl;

answered Dec 26 '13 at 14:47

Michael Schlottke-Lakemper

8,763
6
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score 5 · Answer 2 · answered Dec 26 '13 at 14:51

5

Inside main:

std::cout << &B << std::endl;

Inside member function:

A () {
    x=2;
    //this is a pointer to the this-object
    std::cout << this << std::endl;
}

Don't forget to include <iostream> for output.

answered Dec 26 '13 at 14:51

ben

5,671
4
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55

score 3 · Answer 3 · answered Dec 26 '13 at 14:47

3

Easiest way: printf("%p", &B);

answered Dec 26 '13 at 14:47

egur

7,830
2
27
47

score 2 · Answer 4 · answered Dec 26 '13 at 14:46

2

Did you try

std::cout << &B;

answered Dec 26 '13 at 14:46

m0skit0

25,268
11
79
127

score 2 · Answer 5 · answered Dec 26 '13 at 14:47

2

To get the address of an object, use the address-of operator &.

answered Dec 26 '13 at 14:47

Philipp

67,764
9
118
153

score 2 · Answer 6 · answered Dec 17 '18 at 17:27

With C++20 you can use std::to_address:

#include <memory>
#include <iostream>

class A{};

int main()
{
    A B;
    std::cout << std::to_address(&B) << '\n'
}

Which has the advantage of being able to accept raw pointers, smart pointers, iterators, and pointer-like objects. Making it useful for cases that require generic code.

score 1 · Answer 7 · answered Dec 26 '13 at 14:46

1

You can try

cout << &B

answered Dec 26 '13 at 14:46

gturri

13,807
9
40
57

score 1 · Answer 8 · answered May 22 '21 at 12:30

1

#include <iostream>
using namespace std;
class A{
     public:
     A(){
      cout<<this<<endl;
     }
};

int main()
{
   A* a = new A();
   cout<<a<<endl;

   return 0;

}

answered May 22 '21 at 12:30

Rakibul Islam

29
5

score 0 · Answer 9 · answered Dec 26 '13 at 14:47

0

If you are trying to print the memory address you need to convert B to a pointer when printing. So

cout << &B << endl;

The ampersand, when used before a variable, calls the pointer of the variable instead of the variable itself.

answered Dec 26 '13 at 14:47

Jesse Laning

490
4
12

score 0 · Answer 10 · answered Dec 26 '13 at 14:50

If you mean in THAT specific class you have defined. You can't. Your constructor is private and you forgot a semi-colon at the end of your class definition, therefore, your code won't compile.

Ignoring these issues to display the pointer of an object in main you can use

std::cout << &B;

Or in a member function you can use

std::cout << this;

(Don't forget to include iostream)

score 0 · Answer 11 · answered Jan 05 '17 at 08:28

0

this line has to be add-cout<<"address is:"<<&B;

#include <iostream>
using namespace std;

    class A
    {
     int x;

    };

  int main()
   {
   A B;
   cout<<"address is:"<<&B;
    return 0;
   }

answered Jan 05 '17 at 08:28

Kavita Jain

90
1
9

score 0 · Answer 12 · answered Dec 17 '18 at 16:36

I see that there are a lot of answers for your question already, but I don't know why none of them point that you have a private constructor for class A, and the way the object B is being instantiated should throw a compilation error. At least, it did in my case. There is a whole lot of concepts around using private constructors, https://stackoverflow.com/a/2385764/3278350 gives a good idea.

Can we print the address of object of class in C++?

12 Answers12