Why does this compile (method in class without return type)

[Jonathan M Davis via Digitalmars-d-learn]

On Wednesday, 3 May 2017 at 07:34:03 UTC, Daniel Kozák wrote:
> V Wed, 03 May 2017 06:54:15 +0000
> nkm1 via Digitalmars-d-learn <digitalmars-d-learn at puremagic.com>
> napsáno:
>
>> Consider:
>> 
>> import std.stdio;
>> 
>> class A
>> {
>>      final print() { writeln(this); } // no return type
>> }
>> 
>> class B : A
>> {
>>      final void print() { writeln(this); }
>> }
>> 
>> void main()
>> {
>>      auto b = new B;
>>      b.print();
>> 
>>      A a1 = b;
>>      a1.print();
>> 
>>      A a2 = new A;
>>      a2.print();
>> }
>> 
>> That compiles:
>> 
>> $ dmd -de -w -g ./main.d
>> $ main
>> main.B
>> main.B
>> main.A
>> 
>> with dmd 2.074 on linux:
>> 
>> $ dmd --version
>> DMD64 D Compiler v2.074.0
>> Copyright (c) 1999-2017 by Digital Mars written by Walter 
>> Bright
>> 
>> Is that a bug? (in the compiler). I'm learning D, and I'm half 
>> way through Andrei's book; I also read the documentation (on 
>> D's website) and I think that shouldn't compile?
>
> print in A is template:
>
> import std.stdio;
>
> class A
> {
>     template print() {
>         void print()
>         {
>             writeln("A version");
>         } // no return type
>     }
> }

How is it a template in the original example?

final print() { writeln(this); } // no return type

does not have the extra set of parens required to turn it into a 
template. It _does_ use inference, just like

static a = 42;

uses inference and

final auto print() { writeln(this); }

uses inference, but it shouldn't be a template any more than

static a = 42;

is a template.

- Jonathan M Davis