[swift-evolution] Compile-time generic specialization

[Abe Schneider]

Sorry, I forgot to copy in its definition:

typealias T<S:Storage> = Tensor<S>

As a quick sanity check I changed all `T.` syntax to `Tensor<S>` and got the same behavior.

Thanks!

> On Feb 20, 2017, at 3:58 PM, David Sweeris <davesweeris at mac.com> wrote:
> 
> 
> On Feb 20, 2017, at 12:23, Abe Schneider via swift-evolution <swift-evolution at swift.org <mailto:swift-evolution at swift.org>> wrote:
> 
>> However, if I define an operation to on the Tensor:
>> 
>> class SomeOp<S:Storage> {
>>     typealias StorageType = S
>>     var output:Tensor<S>
>>     
>>     init() {
>>         output = Tensor<S>(size: 10)
>>     }
>>     
>>     func apply() -> Tensor<S> {
>>         let result = T.cos(output)
>>         return result
>>     }
>> }
>> 
>> let op1 = SomeOp<FloatStorage>()
>> let result3 = op1.apply() // calls default `cos` instead of FloatStorage version
>> 
>> 
>> 
>> So one question I have is why doesn’t the correct version of `cos` get called? Before it was because there wasn’t a vtable available to figure out which function to call. However, in this case since the function was defined in the class, I would assume there would be (I also tried variants of this with an accompanying protocol and non-static versions of the function).
>> 
>> 
>> I can get `SomeOp` to work correctly if I create specializations of the class:
>> 
>> extension SomeOp where S:FloatStorage {
>>     func apply() -> Tensor<S> {
>>         let result = T.cos(output)
>>         return result
>>     }
>> }
>> 
>> extension SomeOp where S:IntStorage {
>>     func apply() -> Tensor<S> {
>>         let result = T.cos(output)
>>         return result
>>     }
>> }
>> 
>> 
>> However, this doesn’t seem like a good design to me, as it requires copying the same code for each StorageType introduced.
> 
> Where is T defined? What happens if you replace "T" with "Tensor<S>"?
> 
> - Dave Sweeris 

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