Why can you indirectly bind an rvalue to an lvalue reference but not directly?

Question

From what I\'ve read and seen you cannot bind an expression that is an rvalue to an lvalue reference. What I have seen however is that you can bind an rvalue to an rvalue refere

Answer 1

It's a fundamental rule of C++ and it prevents bugs:

int foo();

int& x = 3;      // whoops
int& y = foo();  // whoops (sometimes)

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"Rvalue references" (a set of types; not to be confused with actual rvalues) were created at least in part so that you can still do this if you really want to:

int&& x = 3;     // oh, go on then *sigh*
int&& y = foo(); // you'd better be sure!

In the previous examples, I bind (or attempt to bind) objects "referenced" by an rvalue expression to a reference.

Now, I shall bind the object named by an lvalue expression to a reference:

int i = foo();

int& x = i;      // no problem michael

And to make sure that you really meant to obtain an rvalue reference from an lvalue expression, introducing the incredibly poorly-named std::move:

int&& x = std::move(i);  // doesn't move anything

These later rules came much, much later than the original, fundamental rule that has doubtless prevented many bugs over the past twenty years.

Note that Visual Studio has historically accepted T& x = bar() where T is a user-defined type; go figure.