So what does (inout int = 0) do?

[Timon Gehr via Digitalmars-d]

On 15.04.2016 23:03, Andrei Alexandrescu wrote:
> On 04/15/2016 04:47 PM, Steven Schveighoffer wrote:
>> There's no difference between a function that declares its variables
>> inout within its parameters or one that declares them locally.
>
> So now we get to things like:
>
> void fun() {
>    inout int ohHello = 42;
>    ...
> }
>
> How to explain such a construct? Not to mention globals of that type are
> not allowed.
>
>
> Andrei
>

It's an int that has not decided yet whether it wants to be mutable, 
const or immutable and goes out of scope before it is able to make up 
its mind.