As a general rule, whenever you are tempted to insert an abstract type
(including type unions) into the type parameter position of a formal method
argument--which is to say, `x::SomeType{ThisTypeIsAbstract}`, you want to
transform that into a type parameter of the method instead, so
`f{T<:ThisTypeIsAbstract}(..., x::SomeType{T}, ...)`.
On Thursday, April 2, 2015 at 3:37:13 PM UTC-5, Michael Francis wrote:
>
> Nice, that will go a long way to solving the issue I have.
>
> Thanks!
>
> On Thursday, April 2, 2015 at 4:30:57 PM UTC-4, Patrick O'Leary wrote:
>>
>> However, you can define
>>
>> f{T<:Union(Int64, Float64)}(v::Vector{T}) = v
>>
>> to cover that case with a single method definition.
>>
>> On Thursday, April 2, 2015 at 3:28:34 PM UTC-5, Michael Francis wrote:
>>>
>>> Thanks, that is what I suspected.
>>>
>>> On Thursday, April 2, 2015 at 4:13:31 PM UTC-4, Mauro wrote:
>>>>
>>>> Types in Julia are invariant:
>>>>
>>>> If T1<:T2 is true
>>>> then A1{T1}<:A1{T2} is false (unless T1==T2).
>>>>
>>>> Now setting T1=Float64 and T2=Union( Int64, Float64) means
>>>> Vector{Float64}<:Vector{Union( Int64, Float64)} is false. Thus the no
>>>> method error.
>>>>
>>>> On Thu, 2015-04-02 at 21:32, Michael Francis <[email protected]>
>>>> wrote:
>>>> > Is the following expected behavior for union types, I'm assuming yes
>>>> ?
>>>> >
>>>> > v = Union( Int64,Float64 )[]
>>>> > push!( v, 1.2)
>>>> > push!( v, 1) # works
>>>> >
>>>> > f( v::Union(Int64,Float64)...) = v
>>>> > f( [ 1.2 ]...) # Works
>>>> >
>>>> > f( v::Vector{Union( Int64, Float64)}) = v
>>>> > f( [ 1.2 ]) # 'f' has not method matching f(::Array{Float64,1})
>>>> >
>>>> > At first glance I'd expect the last case to work as well.
>>>>
>>>>
