Issue with typeof

[Moritz Maxeiner via Digitalmars-d-learn]

On Monday, 20 March 2017 at 16:55:38 UTC, StarGrazer wrote:
> typeof(&method) fails unless method is static. Says & requires 
> this.
>
> But for typeof, it shouldn't matter and should pass.
>
> So how to get a function pointer to a non static member 
> function(of an interface)?

One warning beforehand:
If the method is not static, a function pointer by itself is 
useless, you should use a delegate; non-static methods have a 
hidden this parameter (OOP), merely taking the address of such a 
method (i.e. a function pointer) would leave you with something 
you could not invoke, since you can't explicitly pass the hidden 
this to a function pointer.

>
> I've tried creating the type like
>
> T t;
> typeof(&t.method) fptr;

The correct syntax for the above would be either

typeof(&T.method) fptr;

or

typeof(t.method)* fptr;

, both of which declare fptr as a function pointer. You won't be 
able to use fptr the way I think you expect, however:

fptr = &t.method;

will not work, because the RHS is a delegate (which uses the 
address of t as the hidden this parameter implicitly) and the LHS 
requires a function pointer.
For most cases, this is exactly what auto storage class is for:

auto fptr = &t.method;

If you need the type for another function's parameter's type, 
then that won't work, of course. Two solutions I see there are
1) Make that function templatized
2) Create a template that can be used like this (you'll want to 
consult std.traits for this):

void foo (delegateOf!(T.method) fptr) {}