Why is a static struct's dtor called at the exit of a function?

[Jonathan M Davis]

On Thursday 21 July 2011 11:52:58 Steven Schveighoffer wrote:
> On Wed, 20 Jul 2011 15:57:44 -0400, Jonathan M Davis <jmdavisProg at gmx.com>
> 
> wrote:
> > On Wednesday 20 July 2011 21:54:19 Andrej Mitrovic wrote:
> >> import std.stdio;
> >> 
> >> void main()
> >> {
> >> 
> >>     writeln("test");
> >>     test();
> >>     writeln();
> >>     writeln("test");
> >>     test();
> >> 
> >> }
> >> 
> >> struct Foo
> >> {
> >> 
> >>     int state;
> >>     this (int i)
> >>     {
> >>     
> >>         state = i;
> >>     
> >>     }
> >>     
> >>     ~this()
> >>     {
> >>     
> >>         writeln("dtor");
> >>     
> >>     }
> >> 
> >> }
> >> 
> >> void test()
> >> {
> >> 
> >>     static Foo foo = Foo(1);
> >>     writeln(foo.state);
> >> 
> >> }
> >> 
> >> Prints:
> >> test
> >> 1
> >> dtor
> >> 
> >> test
> >> 1
> >> dtor
> >> 
> >> Shouldn't static variables be left untouched? I tried to use a struct
> >> to handle an external device context, however the dtor would be kept
> >> called between function calls, so this caused crashes for me.
> >> 
> >> Using a class instead fixes my issue of course.
> >> 
> >> I thought "static" for variables implied it has lifetime until the end
> >> of the application, regardless of whether the memory for the variable
> >> was allocated on the heap or is in the .data segment or somewhere
> >> else.
> > 
> > static variables are supposed to have essentially the same lifetime as
> > the
> > program. So, it definitely looks like a bug.
> 
> slight correction -- static variables are thread local (must be marked
> shared to be shared between threads), so the destructor should be called
> on Thread destruction.

Ah, yes. There is that slight change in D from C++.

> However, I don't think that any destructors are called on thread
> destruction...

That doesn't sound good.

- Jonathan M Davis