Function pointer variable not recognized as function by is-operator

[Jonathan M Davis]

On Sunday, October 07, 2012 10:42:49 Timon Gehr wrote:
> On 10/07/2012 10:35 AM, Jonathan M Davis wrote:
> > On Sunday, October 07, 2012 10:25:41 Tommi wrote:
> >> The following compiles, which I'm pretty sure must be a bug,
> >> right? Just checking to be sure I won't be polluting the bug
> >> tracker.
> >> 
> >> void main()
> >> {
> >> 
> >>       auto f = (int i) {};
> >>       static assert (!is(f == function)); // should fail
> >>       static assert (!is(f == delegate));
> >> 
> >> }
> > 
> > It's not a bug. is(f == function) checks for functions, not function
> > pointers.
> > 
> > http://stackoverflow.com/questions/11067972
> > 
> > - Jonathan M Davis
> 
> Actually is(f==function) checks whether or not f is a function type.
> The ==function part is unimportant, because f is not even a type.

Ah. That's true. I should have caught that. Regardless, the types of function 
pointers still won't be true for == function - e.g. is(typeof(f) == function) 
won't be true - because it's checking for functions, not function pointers.

- Jonathan M Davis