>
> 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]
> <javascript:>> 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.
>>
>
>
