Goodness, I'll get this right one of these times. Sorry for spouting off
so much disinformation! I'll try to make up for it by adding more examples
to the documentation.
const _foo_seed = UInt === UInt64 ? 0x64c74221932dea5b : 0x80783eb4
Base.hash(f::Foo, h::UInt) = hash(f.x, h += _foo_seed)
The canonical hash function to define is the one that takes a second `UInt`
argument. Here's what the help has to say about hashing multiple values:
"New types should implement the 2-argument form, typically by calling the
2-argument `hash` method recursively in order to mix hashes of the contents
with each other (and with `h`)". And that's exactly how most base
functions do it.
On Thursday, July 16, 2015 at 12:35:25 PM UTC-4, Seth wrote:
>
>
>
> On Thursday, July 16, 2015 at 9:25:01 AM UTC-7, Matt Bauman wrote:
>>
>> On Thursday, July 16, 2015 at 12:19:25 PM UTC-4, milktrader wrote:
>>>
>>> Also, back to the OP question, is the correct solution to simply define
>>>
>>> Base.hash(f::Foo) = f.x
>>>
>>
>> No, I'd define Base.hash(f::Foo) = hash(f.x, 0x64c74221932dea5b), where
>> I chose the constant by rand(UInt). This way it won't collide with other
>> types. I really need to spend a bit more time with my interfaces chapter
>> and add this info there.
>>
>>
> How would you do this (in a performant way) for a type that has two
> internal values? That is, I have
>
> immutable Foo{T}
> x::T
> y::T
> end
>
>
> Obviously, the hash should be based on both values, right? I could do
>
> Base.hash(f::Foo) = hash(hash(f.x, f.y), 0x64c74221932dea5b)
>
> But that calls hash twice. (Is this even necessary with immutables?)
>
>
