#C++ This behavior is actually specified in the standard (and that's why it was deprecated).

#include<iostream>
#include<memory>
int main()
{
    std::auto_ptr<int> a(new int(0));
    std::cout<<a.get()<<'\n';
    std::auto_ptr<int> b = a;
    std::cout<<a.get()<<'\n';
}

Output

some address
0

The process that causes this is the same as Abhijit's answer but without requiring a std::move and the same as marinus' answer but using a standard class instead of defining it myself.

Edit: I'm adding some explanation. In the output, "some address" will actually be a hex value for the address of the allocated integer. std::auto_ptr releases its stores pointer when assigned to another auto_ptr and sets its internal pointer to 0. Calling get() retrieves access to the stores pointer.

C++

This behavior is actually specified in the standard (and that's why it was deprecated).

#include<iostream>
#include<memory>
int main()
{
    std::auto_ptr<int> a(new int(0));
    std::cout<<a.get()<<'\n';
    std::auto_ptr<int> b = a;
    std::cout<<a.get()<<'\n';
}

Output

some address
0

The process that causes this is the same as Abhijit's answer but without requiring a std::move and the same as marinus' answer but using a standard class instead of defining it myself.

Edit: I'm adding some explanation. In the output, "some address" will actually be a hex value for the address of the allocated integer. std::auto_ptr releases its stores pointer when assigned to another auto_ptr and sets its internal pointer to 0. Calling get() retrieves access to the stores pointer.

#C++ This behavior is actually specified in the standard (and that's why it was deprecated).

#include<iostream>
#include<memory>
int main()
{
    std::auto_ptr<int> a = new int(0);
    std::cout<<a.get()<<'\n';
    std::auto_ptr<int> b = a;
    std::auto_ptr<<a.get()<<'\n';
}

Output

some address
0

The process that causes this is the same as Abhijit's answer but without requiring a std::move and the same as marinus' answer but using a standard class instead of defining it myself.

Edit: I'm adding some explanation. In the output, "some address" will actually be a hex value for the address of the allocated integer. std::auto_ptr releases its stores pointer when assigned to another auto_ptr and sets its internal pointer to 0. Calling get() retrieves access to the stores pointer.

#C++ This behavior is actually specified in the standard (and that's why it was deprecated).

#include<iostream>
#include<memory>
int main()
{
    std::auto_ptr<int> a(new int(0));
    std::cout<<a.get()<<'\n';
    std::auto_ptr<int> b = a;
    std::cout<<a.get()<<'\n';
}

Output

some address
0

The process that causes this is the same as Abhijit's answer but without requiring a std::move and the same as marinus' answer but using a standard class instead of defining it myself.

Edit: I'm adding some explanation. In the output, "some address" will actually be a hex value for the address of the allocated integer. std::auto_ptr releases its stores pointer when assigned to another auto_ptr and sets its internal pointer to 0. Calling get() retrieves access to the stores pointer.

#C++ This behavior is actually specified in the standard (and that's why it was deprecated).

#include<iostream>
#include<memory>
int main()
{
    std::auto_ptr<int> a = new int(0);
    std::cout<<a.get()<<'\n';
    std::auto_ptr<int> b = a;
    std::auto_ptr<<a.get()<<'\n';
}

Output

some address
0

The process that causes this is the same as Abhijit's answer but without requiring a std::move and the same as marinus' answer but using a standard class instead of defining it myself.

#C++ This behavior is actually specified in the standard (and that's why it was deprecated).

#include<iostream>
#include<memory>
int main()
{
    std::auto_ptr<int> a = new int(0);
    std::cout<<a.get()<<'\n';
    std::auto_ptr<int> b = a;
    std::auto_ptr<<a.get()<<'\n';
}

Output

some address
0

The process that causes this is the same as Abhijit's answer but without requiring a std::move and the same as marinus' answer but using a standard class instead of defining it myself.

Edit: I'm adding some explanation. In the output, "some address" will actually be a hex value for the address of the allocated integer. std::auto_ptr releases its stores pointer when assigned to another auto_ptr and sets its internal pointer to 0. Calling get() retrieves access to the stores pointer.