为什么不能使用右值初始化左值引用？

Question

我可以做const A& a = A()和A&& a = A() ，但是为什么我不能做A& a = A() ？

Answer 1

The rationale is that it rarely makes sense to mutate a temporary. Any attempt to do so is likely to be a programming error rather than a deliberate decision.

The prototypical example is as follows. Assume binding to non-const reference is allowed.

void foo(int& x, int& y); // sets x and y
int xx, yy;
foo(xx, yy); // xx and yy are set
// now make a small change...
void foo(long& x, long& y); // upgrade
int xx, yy; // forgot to change here 
foo(xx, yy); // silently breaks

While one does sometimes want to mutate a temporary, it's often for an entirely different reason from that of mutating an lvalue. Rvalue references were invented to accommodate these cases.

Answer 2

By "initialising an lvalue reference" you refer to defining the lavalue which the reference is referring to from then on.
So the lvalue-reference will propagate any access to the rvalue "initialisation".
Ie you would attempt to access an rvalue-expression like an lvalue.
It is not possible in all cases to access an rvalue like an lvalue.

That is why.

Answer 3

An rvalue is a temporary - it's going to die soon. Creating a lvalue reference to such an object would be a recipe for disaster since the reference would become a dangling reference to a dead object very quickly.