std::any object cast to reference type, change its value, but original object is not changed

Question

I was trying to see if std::any object can cast to reference type, and see whether changing the casted reference means to change original object. As below:

struct My {
    int m_i;
    My() : m_i(1) {}
    My(const My& _) : m_i(2) {}
    My(My&& m) : m_i(3) {};
    My& operator = (const My& _) { m_i = 4; return *this; }
    My& operator = (My&& _) { m_i = 5; return *this; }
};

int main() {
    any a = My();
    My& b2 = any_cast<My&>(a);
    b2.m_i = 6;
    cout << any_cast<My>(a).m_i << endl;
    return 0;
}

It prints 2. For my I expected that, as long as b2 is a reference, I hope changing b2.m_i will effect a.m_i, right? But result seems not as my expectation.

Where did I get wrong, is my expectation valid?

Thanks!

Answer 1

Look at your example without any any:

#include <any>
#include <iostream>
using std::cout;
using std::endl;
using std::any;
using std::any_cast;

struct My {
    int m_i;
    My() : m_i(1) {}
    My(const My& _) : m_i(2) {}
    My(My&& m) : m_i(3) {};
    My& operator = (const My& _) { m_i = 4; return *this; }
    My& operator = (My&& _) { m_i = 5; return *this; }
};

int main() {
    My a = My();
    My& b2 = a;
    b2.m_i = 6;
    cout << static_cast<My>(a).m_i << endl;
}

Output is:

2

Because static_cast<My>(a) is creating a temporary copy of a and your copy assigns 2 to the member. You can use static_cast<My&>(a) to not make a copy.

After removing your somewhat weird copy and assignment, you can also get the same result with any:

struct My {
    int m_i = 1;
};

int main() {
    any a = My();
    My& b2 = any_cast<My&>(a);
    b2.m_i = 6;
    cout << any_cast<My>(a).m_i << endl;  // copy
    cout << any_cast<My&>(a).m_i << endl; // no copy
}

Output:

6
6