[swift-users] Can we `override var hashValue`?

[Zhao Xin]

>
> There is no reason to compare the shape, it is a constant in each of

these types.  (So I am not sure what your point is.)


Sorry. The `let shape` should be `var shape`. I just wanted to make the
subclass to be something more than the super class.

If two values are equal, their hash values should be equal.  As long
> as your override implementation guarantees this, you can override
> hashValue.


But the question is how? If this must be guaranteed by the subclass, how to
writing the override? Or it just can't be done?

You should also understand that the ==(Apple, Apple) and ==(Banana,
> Banana) are not overrides for ==(Fruit, Fruit), and they would not be
> called through dynamic dispatch when you have, for example, two apples
> typed as fruits.


In fact, in my example code, `apple` and `banana` is instance of `Apple`
and `Banana`. They are not using `let apple:Fruit = Apple()`. The `==` used
the *`Fruit` version* as it was the only appropriate one. My big question
is, since they used the `*Fruit` version*, and the *`Fruit` version of
`hashValue`* could guarantee the `hashValue` equality, isn't that enough?

Zhaoxin


On Sat, Sep 3, 2016 at 7:02 AM, Dmitri Gribenko <gribozavr at gmail.com> wrote:

> On Sat, Sep 3, 2016 at 1:31 AM, Zhao Xin via swift-users
> <swift-users at swift.org> wrote:
> > func ==(lhs: Apple, rhs: Apple) -> Bool {
> >     return lhs.name == rhs.name && lhs.shape == rhs.shape
> > }
> >
> > func ==(lhs: Banana, rhs: Banana) -> Bool {
> >     return lhs.name == rhs.name && lhs.shape == rhs.shape
> > }
>
> There is no reason to compare the shape, it is a constant in each of
> these types.  (So I am not sure what your point is.)
>
> > My question is, apple equals banana, but their hashValues (in their own
> > types)  don't. What's wrong here? Is that means we shouldn't override
> > hashValue in subclass in Swift?
>
> This means you should not override hashValue in this particular way.
> If two values are equal, their hash values should be equal.  As long
> as your override implementation guarantees this, you can override
> hashValue.
>
> You should also understand that the ==(Apple, Apple) and ==(Banana,
> Banana) are not overrides for ==(Fruit, Fruit), and they would not be
> called through dynamic dispatch when you have, for example, two apples
> typed as fruits.
>
> Dmitri
>
> --
> main(i,j){for(i=2;;i++){for(j=2;j<i;j++){if(!(i%j)){j=0;break;}}if
> (j){printf("%d\n",i);}}} /*Dmitri Gribenko <gribozavr at gmail.com>*/
>
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