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));
}

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

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.

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