Question

In order to stem the argument going on in the comments of an answer I gave recently, I’d like some constructive answers to the following questions:

1. Is a reference’s lifetime distinct from the object it refers to? Is a reference simply an alias for its target?
2. Can a reference outlive its target in a well-formed program without resulting in undefined behaviour?
3. Can a reference be made to refer to a new object if the storage allocated for the original object is reused?
4. Does the following code demonstrate the above points without invoking undefined behaviour?

Example code by Ben Voigt and simplified (run it on ideone.com):

#include <iostream>
#include <new>

struct something
{
    int i;
};

int main(void)
{
    char buffer[sizeof (something) + 40];
    something* p = new (buffer) something;
    p->i = 11;
    int& outlives = p->i;
    std::cout << outlives << "\n";
    p->~something(); // p->i dies with its parent object
    new (p) char[40]; // memory is reused, lifetime of *p (and p->i) is so done
    new (&outlives) int(13);
    std::cout << outlives << "\n"; // but reference is still alive and well
                                   // and useful, because strict aliasing was respected
}

Answer 1

Is a reference’s lifetime distinct from the object it refers to? Is a reference simply an alias for its target?

A reference has its own lifetime:

int x = 0;
{
   int& r = x;
}      // r dies now
x = 5; // x is still alive

A ref-to-const additionally may extend the lifetime of its referee:

int foo() { return 0; }
const int& r = foo();   // note, this is *not* a reference to a local variable
cout << r;              // valid; the lifetime of the result of foo() is extended

though this is not without caveats:

A reference to const only extends the lifetime of a temporary object if the reference is a) local and b) bound to a prvalue whose evaluation creates said temporary object. (So it doesn’t work for members, or local references which are bound to xvalues.) Also, non-const rvalue references extend the lifetime in the exact same fashion. [@FredOverflow]

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Can a reference outlive its target in a well-formed program without resulting in undefined behaviour?

Sure, as long as you don’t use it.

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Can a reference be made to refer to a new object if the storage allocated for the original object is reused?

Yes, under some conditions:

[C++11: 3.8/7]: If, after the lifetime of an object has ended and before the storage which the object occupied is reused or released, a new object is created at the storage location which the original object occupied, a pointer that pointed to the original object, a reference that referred to the original object, or the name of the original
object will automatically refer to the new object and, once the lifetime of the new object has started, can be used to manipulate the new object, if:

• the storage for the new object exactly overlays the storage location which the original object occupied, and
• the new object is of the same type as the original object (ignoring the top-level cv-qualifiers), and
• the type of the original object is not const-qualified, and, if a class type, does not contain any non-static data member whose type is const-qualified or a reference type, and
• the original object was a most derived object (1.8) of type T and the new object is a most derived object of type T (that is, they are not base class subobjects).

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Does the following code demonstrate the above points without invoking undefined behaviour?

Tl;dr.