• Namespace (14/14) Jun 06 2012 If i have this code:
• Kenji Hara (13/27) Jun 06 2012 Because name lookup with alias this is implemented as simple

"Namespace" <rswhite4 googlemail.com> writes:

If i have this code:

class Bar {
public:
Foo GetFoo() {
     return this._foo;
}

alias GetFoo this;
}

to allow access to Foo methods from Bar i get the error, that 
"GetFoo" isn't a property if i use -property.
Why?
The solution is to set  property before "Foo GetFoo()" but why 
must GetFoo a property? What if i don't want to declare GetFoo as 
a property? I cannot use it with alias this?

"Kenji Hara" <k.hara.pg gmail.com> writes:

On Wednesday, 6 June 2012 at 18:12:39 UTC, Namespace wrote:
 If i have this code:

 class Bar {
 public:
   Foo GetFoo() {
     return this._foo;
   }

   alias GetFoo this;
 }

 to allow access to Foo methods from Bar i get the error, that 
 "GetFoo" isn't a property if i use -property.
 Why?
 The solution is to set  property before "Foo GetFoo()" but why 
 must GetFoo a property? What if i don't want to declare GetFoo 
 as a property? I cannot use it with alias this?

Because name lookup with alias this is implemented as simple 
rewriting of expressions.

   auto bar = new Bar();
   bar.hoge;

If class Bar doesn't have member hoge, it is rewritten as:

   bar.GetFoo.hoge;

After that, if Bar.GetFoo.hoge is a property, bar.GetFoo is 
converted to bar.GetFoo().hoge as like other property functions.
If Bar.GetFoo is not a property and you use -property switch, 
compiler shows "not a property" error because GetFoo is not a 
property.

There is no magic.