IsExpression question

[Stewart Gordon]

Sean Kelly wrote:
> Kirk McDonald wrote:
>> Even after writing most of a library founded on the thing, I still 
>> don't get something. What exactly is the expression
>>
>>     is(T == function)
>>
>> testing? I would expect it to test whether T is a function pointer 
>> type (like how "is(T == delegate)" tests whether T is a delegate 
>> type), but it doesn't.
> 
> It's testing whether T is a function type, not a function pointer type. 
>  For template code, it's mostly useful in this sort of situation:
> 
> template isFunctionType( alias ref ) {
>     const bool isFunctionType = is( ref == function );
> }
> 
> Though it can be used to detect a function pointer type like so:
> 
> template isFunctionPointerType( T ) {
>     const bool isFunctionPointerType = is( typeof(*T) == function );
> }
<snip>

The spec states:

"If TypeSpecialization is one of  typedef  struct  union  class 
interface enum  function  delegate  then the condition is satisifed if 
Type is one of those."

However, that doesn't excuse the behaviour of function being 
inconsistent with the behaviour of delegate.  This would make more sense:

     template isFunctionPointerType(T) {
         const bool isFunctionPointerType = is(T == function);
     }

     template isFunction(alias ref) {
         const bool isFunction = is(typeof(&ref) == function);
     }

But I do wonder what you can really do with a function in a template 
without knowing at least how many parameters it has.

Indeed, the function as a kind of data type ought not to exist in D.  At 
the moment, it appears to exist only in the form of IsExpression that 
you've used.  We have function pointers, indicated by the function 
keyword, and function delegates, indicated by the delegate keyword.  So 
whatever is(... == delegate) does, is(... == function) logically ought 
to act correspondingly.

Stewart.