Is it possible to return the current object if it is an r-value reference?

Question

I have recently learned about r-value references. In order to more thoroughly experiment I decided to write a simple DenseMatrix class. My question is is it possible to write any function ( Transpose for this example ) such that for auto A = B.Transpose() separate matrix is returned, but for auto A = (B + C).Transpose() the result of the Transpose is calculated in place?

Answer 1

Yes, you can overload the Transpose member function on the ref-qualification of the object it's being called on:

class DenseMatrix {
 
  DenseMatrix Transpose() const & {  // #1 called on l-values
     auto copy = *this;
     // transpose copy
     return copy;
  }

  DenseMatrix&& Transpose() && {  // #2 called on r-values
     // transpose *this
     return std::move(*this);
  }
};

So you get the result:

B.Transpose();        // calls #1
(B + C).Transpose();  // calls #2

Here's a demo.

Note that you could implement the l-value overload in terms of the r-value overload, like this:

DenseMatrix Transpose() const & {  
  auto copy = *this;
  return std::move(copy).Transpose();
}

Here's a demo.