isRef, isLazy, isOut

[Andrei Alexandrescu]

Walter Bright wrote:
> Chris Nicholson-Sauls wrote:
>> Walter Bright wrote:
>>> Andrei Alexandrescu wrote:
>>>> I asked in vain for is(var == ref), is(var == out) etc.
>>>
>>> That's because 'is' works with types, not symbols. It didn't fit.
>>
>> is(typeof(var) == ref) ?
> 
> No, because ref is not part of the type. It's a storage class.

Speaking of isLazy - any fresh ideas on dissing lazy? FWIW we can 
replace it with the conversion of expressions to delegates. The 
resulting situation is liable to a few ambiguities but is much better 
than lazy.

Andrei