Re: Why do immutable variables need reference counting?

[Salih Dincer via Digitalmars-d-learn]

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