Rebinding semantics for assignment of optional references

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## Rebinding semantics for assignment of optional references

If you assign to an uninitialized `optional<T&>` the effect is to bind (for the first time) to the object. Clearly, there is no other choice.

    int x = 1 ;
    int& rx = x ;
    optional<int&> ora ;
    optional<int&> orb(x) ;
    ora = orb ; // now 'ora' is bound to 'x' through 'rx'
    *ora = 2 ; // Changes value of 'x' through 'ora'
    assert(x==2);

If you assign to a bare C++ reference, the assignment is forwarded to the referenced object; it's value changes but the reference is never rebound.

    int a = 1 ;
    int& ra = a ;
    int b = 2 ;
    int& rb = b ;
    ra = rb ; // Changes the value of 'a' to 'b'
    assert(a==b);
    b = 3 ;
    assert(ra!=b); // 'ra' is not rebound to 'b'

Now, if you assign to an initialized `optional<T&>`, the effect is to rebind to the new object instead of assigning the referee. This is unlike bare C++ references.

    int a = 1 ;
    int b = 2 ;
    int& ra = a ;
    int& rb = b ;
    optional<int&> ora(ra) ;
    optional<int&> orb(rb) ;
    ora = orb ; // 'ora' is rebound to 'b'
    *ora = 3 ; // Changes value of 'b' (not 'a')
    assert(a==1);
    assert(b==3);

#### Rationale

Rebinding semantics for the assignment of initialized `optional` references has been chosen to provide consistency among initialization states even at the expense of lack of consistency with the semantics of bare C++ references. It is true that `optional<U>` strives to behave as much as possible as `U` does whenever it is initialized; but in the case when `U` is `T&`, doing so would result in inconsistent behavior w.r.t to the lvalue initialization state.

Imagine `optional<T&>` forwarding assignment to the referenced object (thus changing the referenced object value but not rebinding), and consider the following code:

    optional<int&> a = get();
    int x = 1 ;
    int& rx = x ;
    optional<int&> b(rx);
    a = b ;

What does the assignment do?

If `a` is uninitialized, the answer is clear: it binds to `x` (we now have another reference to `x`). But what if `a` is already initialized? it would change the value of the referenced object (whatever that is); which is inconsistent with the other possible case.

If `optional<T&>` would assign just like `T&` does, you would never be able to use Optional's assignment without explicitly handling the previous initialization state unless your code is capable of functioning whether after the assignment, `a` aliases the same object as `b` or not.

That is, you would have to discriminate in order to be consistency.

If in your code rebinding to another object is not an option, then is very likely that binding for the fist time isn't either. In such case, assignment to an uninitialized `optional<T&>` shall be prohibited. It is quite possible that in such scenario the precondition that the lvalue must be already initialized exist. If it doesn't, then binding for the first time is OK while rebinding is not which is IMO very unlikely. In such scenario, you can assign the value itself directly, as in:

    assert(!!opt);
    *opt=value;