What is __dtor() supposed to do?

[Steven Schveighoffer]

On Tue, 19 Jul 2011 09:42:32 -0400, Cristi Cobzarenco  
<cristi.cobzarenco at gmail.com> wrote:

> This is somewhat in reference to a bug that I was having trouble with
> (Issue 5667 - http://d.puremagic.com/issues/show_bug.cgi?id=5667).
> Right now __dtor() simply represents the function defined between the
> curly braces of ~this(), not the compiler generated function that
> calls ~this() and the destructors of members in proper order. To
> illustrate:
>
> import std.stdio;
>
> struct A { ~this() { writeln( "A.~this()"); } }
> struct B {
>    ~this() { writeln("B.~this()"); }
>
>    A member;
> }
>
> struct C {
>    A member;
> }
>
> void main() {
>      A a;
>      B b;
>      C c;
>
>      writeln( "using __dtor" );
>      a.__dtor(); // prints "A.~this()" as expected
>      b.__dtor(); // prints "B.~this()", but not "A.~this()
>      // c.__dtor(); // does not compile because there is no ~this()  
> defined in C
>
>      writeln( "using TypeInfo_Struct.destroy" )
>      typeid(B).destroy(&a); // prints both A.~this() and B.~this()
>      typeid(C).destroy(&c); // compiles & prints A.~this();
>
>      writeln( "going out of scope" );
> }
>
> Basically my question is - is this the required behaviour? Or should
> __dtor and destroy() do the same thing? If they should, then the bug I
> mentioned is a compiler bug. Otherwise, the fix I proposed to druntime
> would work.

destroy calls the xdtor member made by the compiler.  The code for this  
function as far as I can tell is generated by the compiler, so I would say  
this is *definitely* a compiler bug.  IMO, __dtor should map to the same  
thing.

There is no reason I can think of to separate out the __dtor and the calls  
to the destructors for any sub members.  Can you?

-Steve