Using out qualifier instead of *, how to pass null?

[Derek Parnell]

On Fri, 02 Feb 2007 00:38:30 +0100, xs0 wrote:

> Derek Parnell wrote:
>> On Thu, 01 Feb 2007 03:00:01 -0500, Rick Mann wrote:
>> 
>>> In wrapping this C API, I've been going back and forth
>>> between converting "Foo*" output parameters to "out Foo". 
>>> I prefer the latter, as I don't have to take the address
>>> at the call site, but I don't seem to be able to pass null
>>> (which the API allows, meaning, "I'm not interested in
>>> this return value").
>>>
>>> Am I just stuck with Foo*? Alternatively, I can overload
>>> the C function with D wrappers, I suppose. Hmm. That's
>>> probably best.
>> 
>> Is not possible to use a 'null' object ... ?
>> 
>> class Foo {}
>> void Bar(out Foo f) 
>> { 
>>   if (f is null)
>>       ...  
>>   else
>>       ...
>> }
>> Foo NullFoo;
>> Foo RealFoo = new Foo;
>> 
>> Bar(NullFoo);
>> Bar(RealFoo);
> 
> hmm.. doesn't out cause (f is null) to be always true in that code?

LOL... yes it does. I was concentrating on the problem - "I don't seem to
be able to pass null", but missed that obvious gotcha. This code below
actually works though...

// ------------
import std.stdio;

class Foo
{
    char[] m_id;

    this(char[] x) { m_id = x.dup; }
}
void Bar(inout Foo f)
{
  if (f is null)
      writefln("NULL");
  else
      writefln("Got %s", f.m_id);
}
void main()
{
    Foo NullFoo;
    Foo FooA = new Foo("Fred");
    Foo FooB = new Foo("Betty");

    Bar(NullFoo);
    Bar(FooB);
    Bar(FooA);
}
// ------------

-- 
Derek
(skype: derek.j.parnell)
Melbourne, Australia
"Down with mediocrity!"
2/02/2007 11:04:56 AM