Re: Implementing isEqual: and hash
Re: Implementing isEqual: and hash
- Subject: Re: Implementing isEqual: and hash
- From: "Adam R. Maxwell" <email@hidden>
- Date: Sat, 23 Aug 2008 12:05:44 -0700
On Aug 23, 2008, at 10:26 AM, Jeff Johnson wrote:
On Aug 23, 2008, at 11:38 AM, Adam R. Maxwell wrote:
On Aug 23, 2008, at 5:13 AM, Michael Ash wrote:
On Sat, Aug 23, 2008 at 7:41 AM, Graham Cox
<email@hidden> wrote:
I have a class for which equality can be defined as having the
same internal
string value (which happens to be a UUID-turned-string). I can
easily
implement isEqual: based on that but the docs say I also need to
implement
-hash. Any pointers on what is a good way to do that? Could I
just safely
defer to the -hash returned by the string in question?
The implementation of -hash should *always* match the implementation
of -isEqual:.
And as Jim alluded, -hash must not depend on mutable properties; if
the hash changes while the object is in a hashing collection,
you'll end up with random crashes.
Right. Thus, it's a bad idea to use mutable properties in isEqual:.
If you find yourself tempted to do that, impement an
isEqualToMyClass: method rather than isEqual:.
CF collection callbacks are also handy in cases like this, since you
can use equality functions that may not be correct for the general
case of isEqual:.
If you compare primitives in -isEqual:, you should combine them
(using
xor or the like) in -hash. If you compare objects by calling -
isEqual:
on them, you should combine their hashes (using xor or the like). If
you do some of each, combine them all.
What's the motivation for combining hashes in this case? I've
wondered what is the best thing to do when isEqual: is based on
comparing multiple ivars; I typically just use one of them for the
hash.
It's a documented requirement of the isEqual: and hash methods that
the hash must be the same when isEqual: returns YES. Thus, whatever
logic that returns YES in isEqual: must have some kind of match in
hash.
If I have
@interface Test : NSObject
{
id ivar1;
id ivar2;
}
@end
@implementation Test
- (BOOL)isEqual:(id)other
{
if ([other isKindOfClass:[self class]] == NO) return NO;
return ([ivar1 isEqual:(Test *)other->ivar1] && [ivar2 isEqual:
(Test *)other->ivar2]);
}
- (unsigned)hash { return [ivar1 hash]; }
@end
I believe it's sufficient to use [ivar1 hash], since the object is
only equal if ivar1 is equal in both objects. I was just curious to
know what you gain by using ([ivar1 hash] ^ [ivar2 hash]); is it
possible to know in general if it reduces collisions? Presumably that
depends on the hash table implementation as well.
--
Adam
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