ES4 draft: Name

Waldemar Horwat waldemar at google.com
Fri Mar 21 14:40:17 PDT 2008


Lars Hansen wrote:
> The operators |==| and |===| compare as equal two |Name| objects that 
> are separately created from the same (qualifier,identifier) pair; 
> similarly, the operators |!=| and |!==| compare as not equal two |Name| 
> objects that are separately created from (qualifier,identifier) pairs.

This is a non-sequitur.  I agree with the intent, but, as stated this would indicate that:

  !(new Name(123) == new Name("123")) from the first clause

Worse, as written, both
  new Name("abc") == new Name("abc")
and
  new Name("abc") != new Name("abc")
would be true!

What you want to state instead is that |Name| objects are equal iff their qualifier and identifier properties are equal.


What happens if you call new Name() with either zero or three or more arguments?


var x:Name = new Name(my_namespace, "xyz");
var y:Name = Name(x);

Now x is as you'd expect, but y is a Name whose namespace is null and identifier is "MyNamespace::xyz" (or whatever toString does on x).


The relational operators are broken on Name objects.  For example,

  a <= b

is not equivalent to

  a == b || a < b

For another example, you can have two Name objects for which none of

  a < b, a == b, a > b

is true.  This makes it difficult to sort these, put them into ordered containers, etc.

The counterexamples to these invariants are the very x and y from the prior example :-).


    Waldemar



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