C++ : Slicing to base rvalue?

Question

I came across the following code and I couldn't find on google why the following statement is valid C++ :

Base&& b = Derived();

Please explain or give reference

Here is a sample code :

#include <iostream>
using namespace std;

class Base{
public:
    virtual ~Base(){}
    virtual void say_hi() { cout << "hi base"; }
};
class Derived : public Base{
public:
    virtual ~Derived(){}
    virtual void say_hi() { cout << "hi derived"; }
};

int main(int argc, const char * argv[]) {
    Base&& b = Derived();
    b.say_hi();

    return 0;
}

prints :

hi derived

Would you understand why `Derived d; Base& b = d;` would be valid? If so, can you [edit] your question to explain where you feel the difference is? — Angew is no longer proud of SO, Jan 20 '16 at 15:27

score 8 · Accepted Answer · edited May 23 '17 at 11:53

This is binding a temporary to an r-value reference. Temporaries may be bound to both constant l-value references and r-value references.

As for why it's properly calling the derived function, that's because you're calling a virtual function. Dynamic dispatch is taking place as normal.

This is sort of the same as using a function call with an r-value reference parameter:

void callHi(Base&& b){
  b.say_hi();
}

...

callHi(Derived{}); // ultimately calls derived say_hi method

You mention slicing. For slicing to occur, the sequence of events is a bit more complex than simply calling a virtual function.

References are references, plain and simple. +1 – erip Jan 20 '16 at 15:32 — erip, Jan 20 '16 at 15:32

C++ : Slicing to base rvalue?

1 Answers1