How to avoid inout type constructor with Optional type wrapper undoing string type

[aliak]

On Friday, 27 July 2018 at 14:52:20 UTC, Steven Schveighoffer 
wrote:
> On 7/23/18 2:39 PM, aliak wrote:
>> Hi,
>> 
>> I'm playing around with an Optional wrapper type. It stores a 
>> type T and a bool that defines whether a value is defined or 
>> not:
>> 
>> struct Optional(T) {
>>    T value;
>>    bool defined = false;
>>    this(U : T)(auto ref inout(U) value) inout {
>>      this.value = value;
>>      this.defined = true;
>>    }
>> }
>
> Don't use inout here. The point of inout on the constructor is 
> to *transfer* the mutability of the parameter to the struct 
> instance. But you want to simply copy the type into the struct 
> (an immutable(Optional!T) is quite useless, no?)
>
> Just use U, not inout(U), and don't put inout on the 
> constructor.
>
> -Steve

But then it only works for mutable Optional right? Why would an 
immutable(Optional!T) be useless? Data can be "forever" empty or 
a certain value.