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

What is the reason behind disallowing binding an rvalue to an lvalue reference?

No answer to this question can be complete without a reference to the invaluable and distinguished source, The Design and Evolution of C++ by Bjarne Stroustrup.

In section 3.7 Bjarne writes:

I made one serious mistake, though, by allowing a non-const reference to be initialized by a non-lvalue. For example:

void incr(int& rr) { rr++; }

void g()
{
    double ss = 1;
    incr(ss);    // note: double passed, int expected
                 // (fixed: error in Release 2.0)
}

Because of the difference in type the int& cannot refer to the double passed so a temporary was generated to hold an int initialized by ss's value. Thus incr() modified the temporary, and the result wasn't reflected back to the calling function.

I highly recommend The Design and Evolution of C++ for understanding many of the "why questions" one might have, especially regarding the rules that were laid down prior to the C++98 standard. It is an informative and fascinating history of the language.