It'll work (here), but it will be wrong in general.
you actually want:
Base.eltype{T<:Foo}(::Type{T}) = eltype(super(T))
(i believe that this definition is already built in to 0.4, so in 0.4, it
will "just work" given your first method definition below. but in 0.3, you
may need to write this definition yourself)
In particular, Keno's code will fail given other possible type definitions,
such as
type Foo2 <: Foo{Int}; end
or
type Foo3{A, T} <: Foo{T}; end
On Tue, Mar 3, 2015 at 10:03 PM Sheehan Olver <[email protected]> wrote:
>
> Beautiful!! Now to rewrite the 100 for loops I made over every possible
> choice of T :)
>
>
>
>
> On 4 Mar 2015, at 2:01 pm, Keno Fischer <[email protected]>
> wrote:
>
> Base.eltype{T<:Foo}(::Type{T})=T.parameters[1]
>
> should work
>
> On Tue, Mar 3, 2015 at 9:58 PM, Sheehan Olver <[email protected]>
> wrote:
>
>> Suppose I have
>>
>> abstract Foo{T}
>> immutable Foo1{T} <:Foo{T} end
>> immutable Foo2{T} <:Foo{T} end
>>
>>
>>
>> How can I write a function that returns T given the type? E.g.: I want
>> to do
>>
>> Base.eltype(::Type{Foo{T}})=T
>>
>>
>> But this won't work for eltype(Foo1{T})
>>
>>
>>
>>
>
>
