views:

66

answers:

2

I have 2 scenarios.

This fails:

class F<X>
{
  public X X { get; set; }
}

error CS0102: The type 'F<X>' already contains a definition for 'X'

This works:

class F<X>
{
  class G
  {
    public X X { get; set; }
  }
}

The only logical explanation is that in the second snippet the type parameter X is out of scope, which is not true...

Why should a type parameter affect my definitions in a type?

IMO, for consistency, either both should work or neither should work.

Any other ideas?

PS: I call it 'lexical', but it is probably not the correct term.

Update:

As per Henk's answer, here is a non-generic version displaying the same behavior, but perhaps easier to grok.

Fails:

class F
{
  class X { }
  public X X { get; set; }
}

Works:

class X { }
class F
{
  public X X { get; set; }
}

From what I can see, the C# compiler creates a lexical scope at type definition boundries.

It also implies that types and member names live in the same 'location' (or namespace in terms of LISP).

+2  A: 

X is defined as a type in the scope of F. It is similar to this:

class F
{
  public void X();

  public int X(); // Bad, trying to redefine X.

  class G
  {
    public string X(); // OK, different scope
  }
}

F.X isn't out of scope in G, but this doesn't stop G from defining a new X.

Marcelo Cantos
Sorry, your X is a member, mine is a type parameter. I can understand that.
leppie
Sorry to be obtuse, @leppie. Are you agreeing or disagreeing with me?
Marcelo Cantos
I am disagreeing. :)
leppie
+3  A: 

The class G introduces a distinctive naming scope. If you omit the default rules, the 2 versions become:

public F<X>.X F<X>.X { get; set; }    // error
public F<X>.X F<X>.G.X { get; set; }  // OK
Henk Holterman
Thanks, that seems to make a little sense :) And why: class F { class X { } public X X { get; set; } }also fails.
leppie