Re: So what does (inout int = 0) do?

[Andrei Alexandrescu via Digitalmars-d]

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