> In C++, it's an rvalue reference , which can be effectively thought of as an lvalue
hmm...this doesn't sound quite right? the comma operator's result in C++ is not an rvalue reference - it takes on exactly the value category of its right operand (which in this case is an lvalue)
hmm...this doesn't sound quite right? the comma operator's result in C++ is not an rvalue reference - it takes on exactly the value category of its right operand (which in this case is an lvalue)