On 03-01-2012 07:53, Gou Lingfeng wrote:
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.
That's uglier than 'is'.
On the other hand, if "is" is introduced, it should make some special
sense than "==", like this:
Introduced? It's a feature in the language today.
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 "==".
- Alex
