I understand now. Thanks for you two!
On Tue, May 5, 2015 at 2:45 PM, Alex <[email protected]> wrote:
> So in the example above, the type parameter is redundant. However, in the
>> following example, type parameter is not redundant, is it?
>>
>> ```
>> f(x::AbstractArray(Real)) != f{T<:Real}(x::AbstractArray{T}) ?
>> ```
>>
>
> It is not redundant because parametric types are invariant:
> julia> AbstractArray{Float64} <: AbstractArray{Real}
> false
>
> julia> Rational{Int64} <: Rational{Integer}
> false
>
> From this you can see that `f(x::AbstractArray{Real})` will only apply to
> arrays where all elements are of type `Real` (e.g. `zeros(Real, 2)`). The
> other definition applies to arrays where all elements are of some concrete
> subtype of `Real` (e.g. `zeros(Float64,2)`, `zeros(Int,2)`, ...).
>
> Best,
>
> Alex.
>
>
>
>> On Tue, May 5, 2015 at 10:35 AM, Alex <[email protected]> wrote:
>>
>>> Is the type parameter redundant?
>>>>
>>>
>>> Yes, you always have `AbstractArray{T} <: AbstractArray == true`. See
>>> the section on parametric abstract types in the manual
>>> <http://julia.readthedocs.org/en/latest/manual/types/?highlight=abstract#parametric-abstract-types>.
>>> This also works for (normal) composite types, e.g. `Rational{Int} <:
>>> Rational == true`.
>>>
>>> Best,
>>>
>>> Alex.
>>>
>>
>>
