Extended Type Design.

[Andrei Alexandrescu (See Website For Email)]

Derek Parnell wrote:
> On Tue, 20 Mar 2007 16:01:35 -0700, Walter Bright wrote:
> 
> 
>> A symbol is a name to which is 'bound' a value.
>  ...
> 
>> Here, we bind a new value to the symbol x:
>>      x = 4;
> 
> I used to use the verb 'to assign' for this concept. I guess that's still
> okay or must I modernize <G>

You may want to modernize. "Assign" doesn't quite catch the notion of 
indirect reference, and from a couple of posts I understand that this is 
a source of confusion.

A very useful way to see "int x = 4;" is that the symbol x is bound to 
the Platonic number 4. The 4 itself cannot change. You can rebind x by, 
say, writing ++x. That unbinds x from Plato 4 and binds it to Plato 5. 
Once this is clear, the notions of values and references clarifies a lot.

>> static int x = 3;
>>
>> '3' is the value.
>> 'int' is the type.
>> 'x' is the symbol.
>> 'static' is the storage class.
>>
>>
>> A storage class originally meant where the symbol is stored, such as in 
>> the data segment, on the stack, in a register, or in ROM. It's been 
>> generalized a bit since then. The main way to tell a storage class apart 
>> is that:
>> 1) a storage class applies to the symbol
> 
> "to the symbol"?  Don't you mean "to the data that the symbol represents"?
> In the case above, the symbol is 'x', and I don't think 'x' is being stored
> anywhere except in the compiler's internal tables, and I'm sure 'static'
> isn't referring to the compiler's internals.

The meaning was simple. When you say "poor fella Jim", "poor" applies to 
Jim, not to "fella".

>> 'invariant' is a guarantee that any data of that type will never change. 
> 
>  class Foo 
>  { 
>    int a; 
>    char[] s; 
>    this(char[] d)
>    {
>        s = d.dup;
>        a = s.length;
>    }
>  }
>  invariant Foo f = new Foo("nice");
> 
>  f.a = 1; // fails??? changing f's data
>  f.s = "bad"; // fails??? changing f's data
>  f.s.length = 1; // fails??? changing f's data
>  f.s[0] = 'r'; // okay ??? not changing f's data

They all fail. The last fails because invariance is transitive. All that 
could be done is to rebind f to another invariant object.


Andrei