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.
>>>
>>>
