On Thursday, 11 December 2014 at 13:55:55 UTC, Marc Schütz wrote:
On Thursday, 11 December 2014 at 12:48:05 UTC, Manu via
Digitalmars-d wrote:
On 8 December 2014 at 07:29, Walter Bright via Digitalmars-d
<[email protected]> wrote:
On 12/7/2014 6:12 AM, Dicebot wrote:
But from existing cases it doesn't seem working good enough.
For example,
not
being able to represent idiom of `scope ref int foo(scope
ref int x) {
return x;
}` seems very limiting.
scope ref int foo(ref int x);
will do it.
Will it? It looks like foo can't be called with scope data?
This is a point that most people don't seem to understand yet,
and which wasn't obvious for me either, at the beginning:
* A `ref` parameter means that it cannot escape the function,
_except_ by return.
* A `scope ref` parameter means that it cannot escape the
function _ever_, not even by return.
* A `scope ref` return means that it cannot leave the current
statement.
Therefore, a `scope ref` return value can be passed on to the
next function as a `ref` argument. If that function again
returns a reference (even if not explicitly designated as
`scope`), the compiler will treat it as if it were `scope ref`.
OH! I had totally misunderstood that. Cheers for the explanation.
