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The following program
void display(const int& a) { cout << a ; } will work if called with a literal like this
display(5); but without the const it won't work.
So how can a const reference keep pointing to an R-Value (anonymous variable)?
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If you could pass an r-value by (non-const) reference, you could also be able to assign to it from inside the function. Therefore the rule that if r-values are to be passed by reference, this has to be a const reference.
A non-const reference cannot point to a literal. You cannot bind a literal to a reference to non-const (because modifying the value of a literal is not an operation that makes sense) and only l-values can be bound to references to non-const.
When we create our own copy constructor, we pass an object by reference and we generally pass it as a const reference. One reason for passing const reference is, we should use const in C++ wherever possible so that objects are not accidentally modified.
The important difference is that when passing by const reference, no new object is created. In the function body, the parameter is effectively an alias for the object passed in. Because the reference is a const reference the function body cannot directly change the value of that object.
For your final question:
how can a const reference keep pointing to an R-Value (anonymous variable)
Here is the answer. The C++ language says that a local const reference prolongs the lifetime of temporary values until the end of the containing scope, but saving you the cost of a copy-construction (i.e. if you were to use an local variable instead).
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