immuable method address and export method address.

[Igor Stepanov]

Hello, I've two simple questions:

1. I have a structure:

struct Foo
{
   public int bar() const {return 0;}
   public int bar() immutable {return 1;}
}

How can I get immutable bar address?

When I've wrote next code I've got a const bar address. Is it a 
bug?

void main()
{
     immutable Foo boo;
     int delegate() immutable dg = &boo.bar;
}

Next question:
If I write туче code, it'll be builded successfully

void main()
{
     foreach(cur; __traits(getOverloads, Foo, "bar"))
     {
         void* p = &cur;
         writeln(cur); //prints a function pointer.
     }
}

If I change protection of bar to export I've got a error:

struct Foo
{
   export int bar() const {return 0;}
   export int bar() immutable {return 1;}
}

void main()
{
     foreach(cur; __traits(getOverloads, Foo, "bar"))
     {
         void* p = &cur; //Error: need 'this' to access member bar
         writeln(cur);
     }
}

How should it be?