Why do immutable variables need reference counting?

[Salih Dincer]

On Monday, 11 April 2022 at 03:24:11 UTC, Ali Çehreli wrote:
> 
> The output:
>
> 0 is (about to be) alive!
> 0 is (already) dead.
> 1 is (about to be) alive!
> 1 is (already) dead.
> 2 is (about to be) alive!
> 2 is (already) dead.

It worked for me in a different way.

> 1 is (about to be) alive!
2 is (already) dead.
2 is (already) dead.
2 is (already) dead.

> 2 is (already) dead.
2 is (already) dead.
Hello D!
1 is (already) dead.

Because I changed the code like this.

```d

struct S
{
   . . .

   string toString() { return ""; }
}

//S test(inout S s)/*
S test(S s)//*/
{
   s.i = 2;
   return s;
}

void main()
{
   immutable s = S(1);
   test(s).writeln;
   "Hello".writefln!"%s D!";
}
```

If the inout is set, it does not allow compilation.

Thanks, SDB79