Thank you, Avik. I appreciate your help.
On Wednesday, April 22, 2015 at 1:34:51 PM UTC-4, Avik Sengupta wrote:
>
>
> The answer to your first question is easy. subtypes() only display the
> direct subtypes. ASCIIString is a subtype of DirectIndexString, which in
> turn is a subtype of String
>
> julia> subtypes(String)
> 8-element Array{Any,1}:
> Base.GenericString
> DirectIndexString
> RepString
> RevString{T<:AbstractString}
> RopeString
> SubString{T<:AbstractString}
> UTF16String
> UTF8String
>
> julia> subtypes(DirectIndexString)
> 2-element Array{Any,1}:
> ASCIIString
> UTF32String
>
> The answer to your second question is more contentious, though a valid
> answer might be that this is simply a design choice. A generic discussion
> is here:
> http://en.wikipedia.org/wiki/Covariance_and_contravariance_%28computer_science%29
>
> and there are various discussions in the mailing list about why this choice
> has been made in Julia.
>
> Regards
> -
> Avik
>
> On Wednesday, 22 April 2015 18:20:46 UTC+1, Test This wrote:
>>
>> Avik and Patrick,
>>
>> Thanks to both of you for clarifying this and for the alternative.
>>
>> I have a couple of related questions:
>>
>> - Why is ASCIIString not listed when I do subtypes(String)?
>> - Why is Dict{ASCIIString, Int} not a subtype of Dict{String, Int}, if
>> ASCIIString is a subtype of String?
>>
>> Anyhow, for now, your answers and links are very helpful.
>>
>>
>>
>>
>> On Wednesday, April 22, 2015 at 10:38:27 AM UTC-4, Patrick O'Leary wrote:
>>>
>>> (The crusade continues)
>>>
>>> Never fear though, this doesn't mean you have to write more code! Julia
>>> supports the use of type variables to express generics. So in your case,
>>> instead of:
>>>
>>> function func(a::Params, b::String, c::Dict{String, Array{Int, 1}},
>>> d::Dict{String, Array{Int, 1}})
>>> ...
>>> end
>>>
>>> which has the aforementioned invariance problem, you can write
>>>
>>> function func{T<:String}(a::Params, b::T, c::Dict{T, Array{Int, 1}},
>>> d::Dict{T, Array{Int, 1}})
>>> ...
>>> end
>>>
>>> which defines a family of methods for any subtype of String. These
>>> methods are called, in Julia terms, "parametric methods", and are discussed
>>> in the manual here:
>>> http://docs.julialang.org/en/release-0.3/manual/methods/#parametric-methods
>>>
>>> Hope this helps,
>>> Patrick
>>>
>>>
>>> On Wednesday, April 22, 2015 at 8:54:12 AM UTC-5, Avik Sengupta wrote:
>>>>
>>>> So a one line answer to this is julia container types are invariant.
>>>>
>>>> Lets take this step by step
>>>>
>>>> julia> f(x::String) = "I am $x"
>>>> f (generic function with 2 methods)
>>>>
>>>> julia> f("abc")
>>>> "I am abc"
>>>>
>>>> julia> g(x::Dict) = "I am a dict of type: $(typeof(x))"
>>>> g (generic function with 1 method)
>>>>
>>>> julia> g(Dict("abc"=>1))
>>>> "I am a dict of type: Dict{ASCIIString,Int64}"
>>>>
>>>> julia> h(x::Dict{String, Int}) = "I am a dict of {String=>Int}"
>>>> h (generic function with 1 method)
>>>>
>>>> julia> h(Dict("abc"=>1))
>>>> ERROR: MethodError: `h` has no method matching
>>>> h(::Dict{ASCIIString,Int64})
>>>>
>>>>
>>>> Basically, while an "ASCIIString" is a subtype of the abstract type
>>>> "String" , a Dict{ASCIIString, Int} is not a subtype of Dict{String, Int}
>>>>
>>>> See here for more:
>>>> http://docs.julialang.org/en/release-0.3/manual/types/?highlight=contravariant#man-parametric-types
>>>>
>>>> Regards
>>>> -
>>>> Avik
>>>>
>>>> On Wednesday, 22 April 2015 14:14:06 UTC+1, Test This wrote:
>>>>>
>>>>>
>>>>> I defined a function
>>>>>
>>>>> function func(a::Params, b::String, c::Dict{String, Array{Int, 1}},
>>>>> d::Dict{String, Array{Int, 1}})
>>>>> ...
>>>>> end
>>>>>
>>>>> When I run the program, calling this function with func(paramvalue,
>>>>> "H", d1, d2), I get an error saying func has no method matching
>>>>> (::Params, ::ASCIIString, ::Dict{ASCIIString,Array{Int64,1}},
>>>>> ::Dict{ASCIIString,Array{Int64,1}})
>>>>>
>>>>> The code works if I change String to ASCIIString in the function
>>>>> definition. But I thought (and the REPL seems to agree) that
>>>>> any value of type ASCIIString is also of type String. Then, why am I
>>>>> getting this error.
>>>>>
>>>>> Thanks in advance for your help.
>>>>>
>>>>
