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