(see also:
https://groups.google.com/forum/#!msg/julia-users/pDP37YGR-zc/IL4AJKc3AQAJ)
On Sun, Oct 11, 2015 at 6:04 PM, Isaiah Norton <isaiah.nor...@gmail.com>
wrote:
> You are calling `symbol` on an object, which results in a fully-qualified
> name when called inside a module:
>
> julia> module Foo
> abstract a
> f() = symbol(a)
> end
>
> julia> Foo.f()
>
> symbol("Foo.a")
>
>
> (or try adding `@show superSymb` inside your function)
>
> Creating a symbol from a type instance here isn't really necessary because
> you can splice in `$supertype` directly. (see the "Metaprogramming" section
> of the manual)
>
> Having said that: calling a function to create a type is not
> recommended/idiomatic. Instead, you could call `@eval` at the top level in
> your module (possibly in a for loop). There are a handful of examples of
> this in base, for example in "linalg/triangular.jl".
>
>
>
>
> On Sun, Oct 11, 2015 at 12:01 PM, Andrew Keller <
> andrew.keller...@gmail.com> wrote:
>
>> I'm using Julia 0.4.0 on Mac OS X 10.10.5. I'd like to put some code into
>> a module, but I'm having some trouble with namespaces. The following fails
>> (`UndefVarError: test.a not defined`) when enclosed inside `module test`.
>> When outside the module, e.g. pasted into the REPL, the code works fine.
>> Could someone point me to relevant reading material or explain what is
>> going on? It seems I can avoid the problem by putting the string "a" in
>> the dictionary instead of the abstract type, but I want to know why I am
>> unable to do things as written. Thank you for your patience as I am new to
>> the language.
>>
>> module test
>>
>> abstract a
>>
>> dict = Dict("key" => a)
>>
>> function createType(typeName::ASCIIString,supertype::DataType)
>>
>> typeSymb = symbol(typeName)
>> superSymb = symbol(supertype)
>> @eval immutable ($typeSymb){T} <: $superSymb
>>
>> num::Float64
>>
>> end
>>
>> end
>>
>> createType("b",dict["key"])
>>
>> end
>>
>
>
