On Tue, 2012-01-03 at 06:43 +0100, Timon Gehr wrote:
> On 01/03/2012 06:10 AM, Gou Lingfeng wrote:
> > On Mon, 2012-01-02 at 13:18 +0100, Alex Rønne Petersen wrote:
> >> On 02-01-2012 06:25, Gou Lingfeng wrote:
> >>> D's definitions of "is" and "==" have so much redundency. That might
> >>> indicate some flaw. If references and values (for classes and arrays)
> >>> could be clearly distinguished in the syntax, the "is" operator is not
> >>> necessary at all.
> >>
> >> Of course it is. 'is' is strictly identity, while == would call an
> >> overloaded opEquals, if any exists. This difference in semantics is
> >> *very* important when you do *not* want to call opEquals.
> >
> > My impression is that, in D, references are wrapped (protected)
> > pointers, and there are values behind. Although opEquals can be defined
> > anyway, it's usually some function depending on the values, not
> > pointers. If we could clearly show whether we wan't pointer comparison
> > or value comparison, then "a is b" would be "pointer(a)==pointer(b)",
> > and "a==b" "value(a)==value(b)".
> >
> > Or "is" sould evaluate false for "int is int", and true or false for
> > expressions like "int is (ref int)" and "(ref int) is (ref int)". So it
>
> ref is not a type modifier, it is a storage class, therefore, to compare
> to ref int values a,b for identity, use &a is &b.
Or, "&a == &b". If similar notations could apply to other types (arrays
and classes), binary "is" is not needed at all.
On the other hand, if "is" is introduced, it should make some special
sense than "==", like this:
import std.stdio;
int a=1,b=1;
void fun(ref int c,ref int d,ref int e){
writeln("'c is d' evaluates ",c is d,", should be true.");
writeln("'c is e' evaluates ",c is e,", should be false.");
writeln("'c is a' evaluates ",c is a,", should be true.");
writeln("'c is b' evaluates ",c is b,", should be false.");
}
void main() {
writeln("'a is b' evaluates ",a is b,", should be false.");
fun(a,a,b);
}
> > has a consistant meaning everywhere: whether a and b refer to exactly
> > the same memory location. And there's no redundency.
> >
> >
>
> There will always be some redundancy, because a is b implies a == b.
Implication is not redundant, but equivalence is. And the current
definition of "is" for integers is equivalent to "==".
